- M208_1
Vectors and conics
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978-1-4730-0611-9 (epub)IntroductionThe idea of vectors and conics may be new to you. In this course we look at some of the ways that we represent points, lines and planes in mathematics.In Section 1 we revise coordinate geometry in two-dimensional Euclidean space,^{2}, and then extend these ideas to three-dimensional Euclidean space, ^{3}. We discuss the equation of a plane in ^{3}, but find that we do not have the tools to determine the equation of a plane, and leave this until Section 3.In Section 2 we introduce the idea of a vector, and look at the algebra of vectors. Vectors give us a way of looking at points and lines, in the plane and in ^{3}, which is sometimes more useful than Cartesian coordinates, although the two are closely related.In Section 3 we introduce the idea of the dot product of two vectors, and then use it to determine the general form of the equation of a plane in ^{3}.In Section 4 we explain the origin of conics, as the curves of intersection of double cones and planes in ^{3}. The focus–directrix definitions of the non-degenerate conics, the ellipse, the parabola and the hyperbola, are given. We observe that conics are precisely the subsets of the plane determined by an equation of degree two.This OpenLearn course is an adapted extract from an Open University course M208 Pure MathematicsAfter studying this course, you should be able to:recognise the equation of a line in the planedetermine the point of intersection of two lines in the plane, if it existsrecognise the one-one correspondence between the set of points in three-dimensional space and the set of ordered triples of real numbersrecognise the equation of a plane in three dimensionsexplain what are meant by a vector, a scalar multiple of a vector, and the sum and difference of two vectors.1 Coordinate geometry: points, planes and lines1.1 Points, lines and distances in two-dimensional Euclidean spaceIn coordinate geometry we generally use rectangular (or *Cartesian*) coordinate axes, as illustrated below, to describe the Euclidean plane. We can represent any point in the plane *uniquely* by an ordered pair of coordinates (*a*, *b*); that is, any point in the plane has precisely one possible pair of coordinates with respect to the chosen axes.
This gives a one-one correspondence between the points of the plane and ordered pairs of real numbers. Often we do not bother to distinguish explicitly between the points and their representation as ordered pairs; we simply write (*a*, *b*) to denote the point *A*.The plane together with an origin and a pair of *x*-, *y*-axes is often called *two-dimensional Euclidean space*, denoted by the symbol ^{2}.1.2 LinesThe equation of any line in ^{2}, except a line parallel to the *y*-axis, can be written in the formwhere *m* is the *gradient* or *slope* of the line, and *c* is its *y-intercept*; that is, (0, *c*) is the point at which the line crosses the *y*-axis. (See the first sketch below.)In the particular case that the line cuts the *y*-axis at the origin, its equation has the simple formas *c* = 0 in this case. (See the second sketch below.)Another special case occurs when *m* = 0. Then the line is parallel to the *x*-axis, and its equation is of the formwhere *c* is the *y*-intercept. (See the third sketch below.)Finally, the equation of a line parallel to the *y*-axis cannot be written in the form *y* = *mx* + *c*, but it can be written aswhere (*a*, 0) is the point at which the line crosses the *x*-axis. (See the final sketch below.)
In both cases (1.1) and (1.2) above, the equation of a line in the plane can be written in the formfor some real numbers *a*, *b* and *c*, where *a* and *b* are not both zero.Thus any line in ^{2} has an equation of the form (1.3); conversely, any equation of the form (1.3) represents a line in ^{2}.Equation of a lineThe general equation of a line in ^{2} is*ax* + *by* = *c*,where *a*, *b* and *c* are real, and *a* and *b* are not both zero.Exercise 1Determine the equation of the line with gradient −3 that passes through the point (2, −1).Using the formula for the equation of a line when given its gradient and one point on it, we find that the equation of this line isWe can rearrange this in the formExercise 2For each of the following pairs of points, determine the equation of the line through them.**(a)** (1, 1) and (3, 5).**(b)** (0, 0) and (0, 8).**(c)** (0, 0) and (4, 2).**(d)** (4, −1) and (2, −1).**(a)** Since (1, 1) and (3, 5) lie on the line, its gradient is Then, since the point (1, 1) lies on the line, its equation must be *y* − 1 = 2(*x* − 1), or *y* = 2*x* − 1.**(b)** Both these points have *x*-coordinate 0, so they lie on the line with equation *x* = 0, the *y*-axis.**(c)** Since the origin lies on the line, its equation must be of the form *y* = *mx*, where *m* is its gradient.Since (4, 2) lies on the line, its coordinates must satisfy the equation of the line. Thus 2 = 4 *m*, so Hence the equation of this line is **(d)** Both these points have *y*-coordinate −1, so they lie on the line with equation *y* = −1.1.3 Parallel and perpendicular linesWe often wish to know whether two lines are *parallel* (that is, they never meet) or *perpendicular* (that is, they meet at right angles).Two distinct lines, *y* = *m*_{1}*x* + *c*_{1} and *y* = *m*_{2}*x* + *c*_{2}, are parallel if and only if they have the same gradient; that is, if and only if *m*_{1} = *m*_{2}. For example, the lines *y* = −2*x* + 7 and *y* = −2*x* − 3 are parallel since both have gradient −2, whereas the lines *y* = −2*x* + 7 and *y* = 2*x* − 3 are not parallel since their gradients are not equal (they are −2 and 2, respectively).Two lines ℓ_{1} and ℓ_{2} with equations *y* = *m*_{1}*x* + *c*_{1} and *y* = *m*_{2}*x* + *c*_{2}, respectively, where *m*_{1} and *m*_{2} are both non-zero, are perpendicular if and only if *m*_{1}*m*_{2} = −1. If the lines are perpendicular, then one (ℓ_{1}, say) must slope up from left to right and the other (ℓ_{2}, say) must slope down from left to right, as shown below.
Let the lines intersect at *P*, and let *Q* be a point on ℓ_{1} to the right of *P*. Suppose that *Q* is *a* units to the right of *P* and *b* units up from *P*, as illustrated above. Let *R* be the point on ℓ_{2} obtained by rotating *PQ* anticlockwise through an angle of /2; then *R* is *b* units to the left of *P* and *a* units up from *P*, as shown.Then the gradient of ℓ_{1} is and the gradient of ℓ_{2} is It follows that The proof of the converse is similar.Two distinct lines with equations *y* = *m*_{1}*x* + *c*_{1} and *y* = *m*_{2}*x* + *c*_{2}, where *m*_{1} and *m*_{2} are both non-zero, are**parallel** if and only if *m*_{1} = *m*_{2} and *c*_{1} ≠ *c*_{2};**perpendicular** if and only if *m*_{1}*m*_{2} = −1.Example 1Determine which of the following lines are parallel, and which are perpendicular to each other.The gradients of the given lines are, respectively:−2,
, 2, , −2 and Thus the lines ℓ_{1} and ℓ_{5} are parallel, and the lines ℓ_{2} and ℓ_{4} are perpendicular.Example 3Determine which of the following lines are parallel, and which are perpendicular to each other.Since the gradient of a line whose equation is in the form *y* = *mx* + *c* is *m*, the gradients of the given lines are, respectively:−2, 2, , , and −2Thus ℓ_{1} and ℓ_{6} are parallel, and ℓ_{4} and ℓ_{5} are parallel; ℓ_{1} and ℓ_{4} are perpendicular, ℓ_{1} and ℓ_{5} are perpendicular, ℓ_{2} and ℓ_{3} are perpendicular, ℓ_{4} and ℓ_{6} are perpendicular, and ℓ_{5} and ℓ_{6} are perpendicular.1.4 Intersection of two linesTwo arbitrary lines in ^{2} may have a single point of intersection, may be parallel, or may coincide. The first two possibilities are illustrated below. Can we tell from the equations of the lines which of the three possibilities occurs?
We deal with this question by considering some examples. Given the equations of two lines, it is often useful to sketch the lines, since this will not only show whether the lines do intersect, but it will give us some idea of the location of any point of intersection. Then, since the coordinates of the point of intersection must satisfy the equations of both lines, we can find these coordinates by solving the two equations as simultaneous equations in *x* and *y*.Example 2Determine whether the following pairs of lines intersect each other at a single point, are parallel, or coincide. If the lines intersect at a single point, find its coordinates.**(a)** *y* = *x* + 1 and *y* = 3*x* − 1.**(b)** *y* = 2*x* − 2 and *y* = 2*x* + 1.**(c)** *y* =
*x* − 2
and 2*y* = *x* − 4.**(a)** These lines have gradients 1 and 3, and *y*-intercepts 1 and −1, respectively; this enables us to sketch the lines, as shown below. Clearly the lines intersect, at some point in the right half-plane.
At the point of intersection, *y* = *x* + 1 and *y* = 3*x* − 1;
thus *x* + 1 = 3*x* − 1, from which we deduce that *x* = 1.
Substituting this value of *x* into either equation gives *y* = 2. So the point of intersection of the two lines is (1, 2). (You can always check your solution of simultaneous equations by substituting your answer back into the original equations.)
**(b)** These lines both have gradient 2, and their *y*-intercepts are −2 and 1, respectively. We can thus sketch the lines, as shown below.
Since the lines have equal gradient and their *y*-intercepts are different, they are parallel and so do not intersect. (If we try to solve the equations *y* = 2*x* − 2, *y* = 2*x* + 1 as a pair of simultaneous equations, we are led to the contradiction that −2 = 1. Thus there can be no point (*x*, *y*) that lies on both lines.)**(c)** At first glance these two equations may appear to be different. However, if the second equation is divided throughout by 2 (which leaves unchanged the actual line that it represents), we obtain the first equation. So the two equations are equivalent: they represent the same line. (In a sense, the two lines intersect at each of their points.)The line has gradient and *y*-intercept −2, so we can sketch it as shown below. (Two lines coincide if the equation of one can be rearranged to be a multiple of the equation of the other.)Example 4Determine whether the following pairs of lines intersect each other at a single point, are parallel, or coincide. If the lines intersect at a single point, find its coordinates.**(a)** and *y* = 2*x* − 1.**(b)** *y* = 2*x* − 1 and .**(c)** 4*y* = 2*x* + 6 and 6*y* = 3*x* + 9.**(d)** *y* = 7*x* − 1 and *y* = −*x*.
**(a)** These lines have gradients and 2, and *y*-intercepts 1 and −1, respectively. This enables us to sketch the lines below. They intersect at a single point, in the right half-plane. At the point of intersection, Substituting for *x* in the equation of the second line, we obtain Thus the point of intersection of the lines is **(b)** These lines both have gradient 2. However, since they have different *y*-intercepts, −1 and , they do not coincide and so they are parallel.
**(c)** These two lines coincide, since the equation of the second line is the equation of the first multiplied by .
**(d)** These lines have gradients 7 and −1, and *y*-intercepts −1 and 0, respectively. This enables us to sketch the lines. They intersect at a single point, in the right half-plane.
At the point of intersection Substituting for *x* in the equation of the second line, Thus the point of intersection of the lines is 1.5 Distance between two points in the planeNext, we find the formula for the distance between two points *P* (*x*_{1}, *y*_{1}) and *Q*(*x*_{2}, *y*_{2}) in the plane. In the diagram below we have drawn *P* and *Q* in the first quadrant, but the formula we derive holds wherever the points are in the plane.
We can construct a right-angled triangle *PNQ* as shown; the line *PN* is parallel to the *x*-axis, the line *QN* is parallel to the *y*-axis, the angle *PNQ* is a right angle, and *PQ* is the hypotenuse of the triangle.The length of *PN* is |*x*_{2} − *x*_{1}| and the length of *QN* is |*y*_{2} − *y*_{1}|.*Note:* For any real number *x*, It follows from Pythagoras' Theorem that *PQ*^{2} = *PN*^{2} + *QN*^{2}, so Distance Formula in two-dimensional Euclidean spaceThe distance between two points (*x*_{1}, *y*_{1}) and (*x*_{2}, *y*_{2}) in the plane is For example, it follows from the formula above that the distance between the points (1, 2) and (3, −4) is Example 5Find the distance between each of the following pairs of points in the plane.**(a)** (0, 0) and (5, 0).**(b)** (0, 0) and (3, 4).**(c)** (1, 2) and (5, 1).**(d)** (3, −8) and (−1, 4).We use the formula for the distance between two points in the plane. This gives the following distances.**(a)**
**(b)**
**(c)**
**(d)**
1.6 Points, planes, lines and distances in three-dimensional Euclidean spaceWe now study three-dimensional space, ^{3}. This is a space with which you are familiar, of course, as ‘the real world’ is a three-dimensional space.We define a coordinate system in three-dimensional space via three mutually perpendicular axes. *Mutually perpendicular* means that any two of the axes are at right angles to each other.First, we choose a point *O* as the origin, and then we choose an *x*-axis and a *y*-axis at right angles to each other, as described earlier.Next, we draw a third line through the origin, perpendicular both to the *x*-axis and to the *y*-axis; this line is called the *z*-axis. We choose the positive direction of the *z*-axis such that the *x*-, *y*- and *z*-axes form a so-called *right-handed system of axes*. This means that if you hold the thumb and first and second fingers of your right hand at right angles to each other, and label them *x*, *y* and *z* (in that order), you can turn your hand in such a way that your fingers point in the positive directions of the corresponding axes.Finally, we choose a unit of distance along each axis.
We have seen already that there is a one-one correspondence between the points of the plane and the ordered pairs (*a*, *b*) of real numbers. Similarly, there is a one-one correspondence between the points of three-dimensional space and the ordered triples (*a*, *b*, *c*) of real numbers. The point with coordinates (*a*, *b*, *c*) is reached from the origin by moving a distance *a* in the direction of the *x*-axis, a distance *b* in the direction of the *y*-axis, and a distance *c* in the direction of the *z*-axis.For instance, the points with coordinates (1, 2, 3) and (1, −2, −3) are as shown in the diagram below.
It may help you to visualise axes and coordinates in three dimensions by thinking of the corner of a room, with the *x*- and *y*-axes along the foot of the two walls, and the *z*-axis along the vertical join of the walls. To become familiar with this notation, choose axes as above for the room you are in, and determine the approximate coordinates of the corners of the room (using appropriate units of distance).
Example 6On a single diagram, sketch the *x*-, *y*- and *z*-axes and the points with coordinates (−1, 2, −1) and (0, 1, 2).
In view of the one-one correspondence between points in three-dimensional space and ordered triples of real numbers, often we do not distinguish between the points and their representation as ordered triples; thus we simply write (*a*, *b*, *c*) to denote the point represented by this triple.We call three-dimensional space, together with an origin and a set of *x*-, *y*- and *z*-axes, *three-dimensional Euclidean space*.1.7 Planes in three-dimensional Euclidean spaceWe now look at the general form of the equation of a plane in ^{3}.Three planes whose equations are easy to find are those that contain a pair of axes. For example, the (*x*, *y*)-plane is the plane that contains the *x*-axis and the *y*-axis. Points which lie in this plane are precisely those points (*x*, *y*, *z*) in ^{3} for which *z* = 0, so the equation of the (*x*, *y*)-plane is *z* = 0.
Example 7Find the equations of the (*y*, *z*)-plane and the (*x*, *z*)-plane.Points (*x*, *y*, *z*) that lie in the (*y*, *z*)-plane all have *x* = 0; so *x* = 0 is the equation of this plane.Similarly, points (*x*, *y*, *z*) that lie in the (*x*, *z*)-plane all have *y* = 0; so *y* = 0 is the equation of this plane.Example 8Sketch the planes whose equations are as follows.**(a)** *z* = 2**(b)** *y* = −1**(a)**
**(b)**
You saw in Section 1.2 that the general form of the equation of a line in ^{2} is *ax* + *by* = *c*, where *a*, *b* and *c* are real, and *a* and *b* are not both zero. In ^{3} the analogue of this equation is the equation of a plane.We have already met the planes with equations *x* = 0, *y* = 0 and *z* = 0. Each of these equations is a special case of a more general equation
*ax* + *by* + *cz* = *d *,
where *a*, *b*, *c* and *d* are real, and *a*, *b* and *c* are not all zero. In fact, the equation of any plane in ^{3} is of this form.Equation of a planeThe general equation of a plane in ^{3} is *ax* + *by* + *cz* = *d*, where *a*, *b*, *c* and *d* are real, and *a*, *b* and *c* are not all zero.(We shall prove this in Section 3.3).Example 3Determine the general form of the equation of a plane that passes through the origin.Let the plane have equation *ax* + *by* + *cz* = *d*Since (0, 0, 0) lies in the plane, its coordinates must satisfy the equation of the plane; thus *a* × 0 + *b* × 0 + *c* × 0 = *d*, so *d* = 0. Also, the plane whose equation is *ax* + *by* + *cz* = 0 clearly passes through the origin (0, 0, 0).Hence the general form of the equation of a plane that passes through the origin is *ax* + *by* + *cz* = 0 where *a*, *b* and *c* are real and not all zero.
The (*x*, *y*)-, (*y*, *z*)- and (*x*, *z*)-planes all pass through the origin, and their equations are all of the form shown above.Example 9For each of the following points, determine the general form of the equation of a plane that passes through the point.**(a)** (1, 2, 3)**(b)** (−1, −4, 2)**(a)** Let the equation of the plane be of the form *ax* + *by* + *cz* = *d*Since (1, 2, 3) lies in the plane, its coordinates must satisfy the equation of the plane; thus *a* × 1 + *b* × 2 + *c* × 3 = *d*,so *d* = *a* + 2*b* + 3*c*.Thus the general form of the equation of a plane through (1, 2, 3) is *ax* + *by* + *cz* = *a* + 2*b* +3*c***(b)** Let the equation of the plane be of the form *ax* + *by* + *cz* = *d*Since (−1, −4, 2) lies in the plane, its coordinates must satisfy the equation of the plane; thus *a* × (−1) + *b* × (−4) + *c* × 2 = *d*,so *d* = −*a* - 4*b* + 4*c*.Thus the general form of the equation of a plane through (−1, −4, 2) is *ax* + *by* + *cz* = −*a* − 4*b* + 2*c*.1.8 Intersection of two planesWe saw earlier that two arbitrary lines in ^{2} may intersect, be parallel, or coincide. In an analogous way, two arbitrary planes in ^{3} may intersect, be parallel, or coincide.
In general, if two distinct planes intersect, then the set of common points is a line that lies in both planes.For example, the (*x*, *y*)-plane and the (*x*, *z*)-plane intersect in the *x*-axis, which lies in both planes.Similarly, the planes *x* − *y* = 0 and *x* + *y* + *z* = 1 intersect in a line; the points (*x*, *y*, *z*) on this line all satisfy both of the equations *x* − *y* = 0 and *x* + *y* + *z* = 1.
Two planes in ^{3} may be parallel, and so cannot intersect. For example, the plane with equation *z* = 0 is the (*x*, *y*)-plane, and the plane with equation *z* = 1 is a plane parallel to the (*x*, *y*)-plane, passing through the point (0, 0, 1) on the *z*-axis. These two planes do not intersect; every point in the plane *z* = 1 lies at distance 1 above the plane *z* = 0.Finally, two planes may coincide. For example, the planes with equationscoincide, since the second equation is simply the first multiplied by the number .In general, two planes are coincident if the equation of one can be rearranged to be a multiple of the equation of the other.Example 10Determine whether the planes with equations *z* = 2 and *y* = −1 intersect, are parallel, or coincide. Illustrate your answer with a sketch.Hint: You have already sketched these two planes in Exercise 8.From the sketches in Exercise 8, the two planes are clearly not parallel or coincident; hence they must intersect in a line that lies in both planes.
1.9 Distance between points in three-dimensional Euclidean spaceYou saw in Section 1.5 that the distance between two points (*x*_{1}, *y*_{1}) and (*x*_{2}, *y*_{2}) in the plane is given by We can establish a similar formula for the distance between two points in ^{3}, as follows.Let *P* (*x*_{1}, *y*_{1}, *z*_{1}) and *Q*(*x*_{2}, *y*_{2}, *z*_{2}) be two points in ^{3}. Let *M* be the foot of the perpendicular from *Q* to the plane through *P* that is parallel to the (*x*, *y*)-plane; then *M* has coordinates (*x*_{2}, *y*_{2}, *z*_{1}). Next, let *N* be the point in this plane with coordinates (*x*_{1}, *y*_{2}, *z*_{1}).Then the triangles *PQM* and *PMN* are both right-angled triangles, with right angles at *M* and *N*, respectively, as shown below.
The length of *PN* is |*y*_{2} − *y*_{1}|, the length of *NM* is |*x*_{2} − *x*_{1}|, and the length of *MQ* is |*z*_{2} − *z*_{1}|. It follows from Pythagoras' Theorem that *PM*^{2} = *NM*^{2} + *PN*^{2}, so *PM*^{2} = (*x*_{2} − *x*_{1})^{2} + (*y*_{2} − *y*_{1})^{2}.*Note:* This part of the discussion is similar to the derivation of the Distance Formula in ^{2} (see Section 1.5).Using Pythagoras' Theorem again, we obtain *PQ*^{2} = *PM*^{2} + *MQ*^{2}sothat is,Distance Formula in three-dimensional Euclidean spaceThe distance between two points (*x*_{1}, *y*_{1}, *z*_{1}) and (*x*_{2}, *y*_{2}, *z*_{2}) in ^{3} isFor example, it follows from this formula that the distance between the points (1, 2, 3) and (4, −2, 15) is Example 11For each of the following pairs of points in ^{3}, find the distance between them.**(a)** (1, 1, 1) and (4, 1, −3).**(b)** (1, 2, 3) and (3, 0, 3).We use the formula for the distance between two points in ^{3} This gives the following distances. 1.10 Further exercisesExample 12Determine the equation of the line through each of the following pairs of points. Show that both equations can be written in the form *ax* + *by* = *c*, for some real numbers *a*, *b* and *c*, where *a* and *b* are not both zero.**(a)** (−2, −4) and (1, 6).**(b)** (0, 0) and (7, 3).**(a)** Since (−2, −4) and (1, 6) lie on the line, its gradient isIt follows that the equation of the line iswhich can be simplified to that is, 10*x* − 3*y* = −8.This equation is of the desired form, with *a* = 10, *b* = −3 and *c* = −8. (Any multiple of these numbers is also a valid answer.)**(b)** Since the line passes through the origin and the point (7, 3), it has an equation of the form *y* = *mx*, for some *m*. The coordinates of (7, 3) must satisfy the equation *y* = *mx*. Thus 3 = 7*m*, so
.Hence the equation of the line isThis can be written as 3*x* − 7*y* − 0,which is of the desired form, with *a* = 3, *b* = −7 and *c* = 0.Example 13Determine the values of *k* for which the lines 3*x* + 4*y* + 7 = 0 and 2*x* + *ky* = 3 are **(a)** parallel, **(b)** perpendicular.The gradients of the lines 3 *x* + 4*y* + 7 = 0 and 2 *x* + *ky* = 3 are and , respectively.Thus the lines are**(a)** parallel if
, that is,
;**(b)** perpendicular if , that is, .
Example 14Sketch the lines with the following equations, on a single diagram:
Example 15Determine the coordinates of the points of intersection of the lines in Exercise 14.Let *A* be the point of intersection of the lines *y* = −3*x* and
. We equate the two expressions for *y* to obtainMultiplying through by 3 gives −9*x* = *x* + 6,soSince *A* lies on the line *y* = −3*x*, it follows that
So the point *A* has coordinatesNext, let *B* be the point of intersection of the lines and *y* − 3 = 3(*x* − 3). We rewrite the second equation, and equate the two expressions for *y* to obtainMultiplying through by 3 and collecting terms givesso *x* = 3.Since *B* lies on the line it follows that *y* = 3. So the point *B* has coordinates (3, 3).Finally, let *C* be the point of intersection of the lines *y* = −3*x* and *y* − 3 = 3(*x* − 3). We rewrite the second equation, and equate the two expressions for *y* to obtain −3*x* = 3(*x* − 3) + 3.Collecting terms gives 6*x* = 6,so *x* = 1.Since *C* lies on the line *y* = −3*x*, it follows that *y* = −3. So the point *C* has coordinates (1, −3).Example 16Find the distances between the vertices of the triangle formed by the points of intersection found in Exercise 15.We use the Distance Formula given Section 1.5.Since , *B* = (3, 3) and *C* = (1, −3),*Remark:* In the triangle *ABC*, *AB*^{2} + *AC*^{2} = *BC*^{2},so *BAC* is a right angle.Example 17Determine whether each of the pairs of planes given by the following equations intersect, are parallel, or coincide.**(a)** *x* = 1 and *y* = 2.**(b)** *z* = 1 and *z* = 3.Illustrate your answer to each part with a sketch.**(a)** The planes with equations *x* = 1 and *y* = 2 are parallel to the (*y*, *z*)-plane and the (*x*, *z*)-plane, respectively. They intersect in a line, as shown. **(b)** The planes with equations *z* = 1 and *z* = 3 are both parallel to the (*x*, *y*)-plane. They are parallel to each other, as shown.Example 18Determine the distance between the points (1, −2, 3) and (−2, 3, −1) in ^{3}.We use the Distance Formula given in Section 1.9. The required distance is thus2 Vectors2.1 DefinitionsIn this section we introduce an alternative way of describing points in the plane ^{2} or in three-dimensional space ^{3}; namely, by using vectors.You meet vectors such as velocity and acceleration in everyday life; these have in common the fact that to specify each fully, we must give both its size and also its direction. For example, if you are driving along a road, then you are interested both in how fast your car is moving (that is, its speed) and also in what direction. Physical quantities that are uniquely determined only if we know both their size and their direction are called *vectors*. We shall use the term *magnitude* for the size of the physical quantity represented.For example, a velocity of 50 km h^{−1} north-east differs from a velocity of 90 km h^{−1} north-east, and both of these differ from a velocity of 50 km h^{−1} south-west.By contrast, some physical quantities (such as temperature and volume) have only a magnitude – they have no direction associated with them. We call such quantities *scalars*.DefinitionsA **vector** is a quantity that is determined by its magnitude and direction. A **scalar** is a quantity that is determined by its magnitude.We can *represent* a vector geometrically by a line segment in ^{2} or in ^{3}. The length of the line segment is a measure of the magnitude of the vector, and the direction of the line is the same as the direction of the vector; we indicate the direction of the line with an arrow, as shown below. A vector represented by a line segment from *A* to *B* is often written as
; the magnitude of the vector is proportional to the distance between *A* and *B*, and the arrow indicates the direction of the vector.
Often we use single letters, such as **a**, **b**, **p**, **q** or **v**, to denote vectors. Vectors are usually distinguished in print by the use of a bold typeface, or in hand-written work by underlining the letters (for example, __v__). This is a useful convention, as it reminds us that they are vector quantities.We use two pairs of vertical lines to denote the magnitude of a vector; thus we denote the length of a vector **v** by the symbol ‖**v**‖.There is just one vector which does not fit conveniently into the above definition; namely, the zero vector. This arises quite naturally, however. For, just as we can imagine a car travelling at 50 km/h NE (we abbreviate north-east to NE, and so on), it makes sense to say that a car is stationary; in this case, the magnitude of the velocity of the car is zero, but we can assign any direction that we please to its velocity.DefinitionThe **zero vector** is the vector whose magnitude is zero, and whose direction is arbitrary. It is denoted by the symbol **0**.In order to gain a feeling for how we should build up an arithmetic for vectors that agrees with our intuition, we shall now look at some geometric examples of vectors in the plane.For convenience, we shall use compass bearings to describe the directions of the vectors.
In the diagram above, **v** is the vector 3 cm NE. This vector is different from the vectors **a**, **b**, **e** and **f** as its direction differs from their directions. Also, **v** is a different vector from **c** although it has the same direction as **c**, since its magnitude is different from that of **c**. However, **v** and **d** are the same vector, since they have the same magnitude and direction. Note that the geometric representations of equal vectors do not need to start at the same point and finish at the same point. They need only be the same length and in the same direction.DefinitionTwo vectors **a** and **b** are **equal** ifthey have the same magnitude (‖**a**‖ = ‖**b**‖)

andthey are in the same direction.

We write **a** = **b**.Now, although the vectors **v** and **b** are unequal, they are closely related: they have the same magnitude (3 cm), and their directions are exactly the opposite of each other (NE and SW). We say that **b** is ‘minus **v**’, and write this as **b** = −**v**. If we write **v** in terms of a line segment *AB* in the form
, then −**v** can be expressed as
.
DefinitionThe **negative** of a vector **v** is the vector with the same magnitude as **v**, but the opposite direction. It is denoted by −**v**.2.2 Multiplication by a scalarIn the collection of vectors sketched in Section 2.1, although **v** is not equal to **c**, the vectors **v** and **c** are closely related: **c** is a vector in the same direction as **v**, but it is twice as long as **v**. Thus it is natural to write 2**v** for **c**, since we can think of a journey represented by **c** as being a journey **v** followed by a second journey **v**.In an analogous way, we can write for **e**, the vector whose magnitude is times that of **f** and whose direction is that of −**f**.Scalar multiple of a vectorLet *k* be a scalar and **v** a vector. Then *k***v** is the vector whose magnitude is |*k*| times the magnitude of **v**, that is, ‖*k***v**‖ = |*k*| ‖**v**‖, and whose direction isthe direction of **v** if *k* > 0,

the direction of −**v** if *k* < 0.

If *k* = 0, then *k***v** = **0**.Example 19For each of the vectors shown below, decide whether it is a multiple of any of the other vectors; if it is, write down an equation of the form **v**_{1} = *k***v**_{2} that specifies the relationship between them.
The vector **d** is in the same direction as **a**, but none of the other vectors is; also, the length of **d** is two-thirds that of **a**. HenceNext, **e** is along the same line as **b** but in the opposite direction; none of the others is along the same line. Also, the length of **e** is three times that of **b**. HenceFinally, **c** and **f** are not multiples of any of the other vectors.Example 20For the vector **d** in Exercise 19, sketch 3**d** and −2**d**.The vector 3**d** is in the same direction as **d**, but its magnitude is three times that of **d**; the vector −2**d** is in the direction opposite to that of **d**, and its magnitude is twice that of **d**.
2.3 Addition of vectorsWe saw above that the vector 2**v** can be regarded as the vector **v** ‘followed by’ the vector **v**; we can also quite naturally describe this vector as being the ‘sum’, **v** + **v**, of the vector with itself.Analogously, if **p** is the vector 2 cm E and **q** is the vector 3 cm NE, we can think of the ‘sum’ **p** + **q** of the vectors as follows. Starting from a given point, *O* say, draw the vector **p**; starting from its finishing point, draw the vector **q**. Then if the final point reached is *S*, **p** + **q** is the vector
.
Triangle Law for addition of vectorsThe sum **p** + **q** of two vectors **p** and **q** is obtained as follows.Starting at any point, draw the vector **p**.Starting from the finishing point of the vector **p**, draw the vector **q**.Then the sum **p** + **q** is the vector from the starting point of **p** to the finishing point of **q**.There is an equivalent way of visualising the sum of vectors geometrically. Instead of following **p** with **q** to find the third side of a triangle, as above, we obtain exactly the same vector if we draw **p** and **q** with the same starting point, complete the parallelogram of which these are adjacent sides, and take the diagonal of the parallelogram from the common starting point (see below).
Parallelogram Law for addition of vectorsThe sum **p** + **q** of two vectors **p** and **q** is obtained as follows.Starting at the same point, draw the vectors **p** and **q**.Complete the parallelogram of which these are adjacent sides.Then the sum **p** + **q** is the vector from the starting point of **p** and **q** to the opposite corner of the parallelogram.The order in which we add vectors does not matter: the sum **q** + **p** is precisely the same vector as the sum **p** + **q**. It turns out also that the sums (**p** + **q**) + **r** and **p** + (**q** + **r**) are equal. Thus addition of vectors is commutative and associative. (We omit the proof.)Finally, we can define subtraction of vectors in terms of addition and scalar multiplication. For example, if **p** and **q** are the vectors 3.8 cm E and 2.4 cm NE, respectively, then −**q** is the vector 2.4 cm SW, and we define the difference **p** − **q** to be the vector obtained by drawing first **p** and then −**q**.
DefinitionThe **difference**, **p** − **q**, of two vectors **p** and **q** is **p** − **q** = **p** + (−**q**).In general, **q** − **p** does not equal **p** − **q**; in fact, as you would expect, **q** − **p** = −(**p** − **q**).Since the vector −**q** has the same magnitude as **q** but the opposite direction, we can draw **p** − **q** by using either of the two constructions which we use for adding vectors.Example 21For the vectors **p** and **q** shown below, sketch **p** + **q**, **p** − **q** and .
First we sketch the vectors **p**, **q**, −**q**, 2**p** and .
We use the Triangle Law for the addition of vectors to sketch **p** + **q**, **p** − **q** and
.
2.4 Components and the arithmetic of vectorsWe introduce now a different method of representing vectors, which will make the manipulation of vectors much easier. Thus we shall avoid having to solve problems involving vectors by drawing the vectors and making measurements, which is very time-consuming and never very accurate.We can think of a vector as a translation, that is, as representing a movement by a certain amount in a given direction. Then we can use the Cartesian axes in the plane or in ^{3} to describe the translation.For example, consider the vector **v**, in ^{2}, shown below. In view of the Triangle Law for adding vectors, we can think of **v** as the sum of two vectors, one of magnitude 3 units in the direction of the positive *x*-axis and one of magnitude 4 units in the direction of the positive *y*-axis. (In this sense, we can associate the ordered pair of real numbers (3, 4) with the vector **v**.)Similarly, in ^{3}, the vector **w** shown above can be considered as the sum of three vectors, one of magnitude 1 in the direction of the positive *x*-axis, one of magnitude 3 in the direction of the positive *y*-axis, and one of magnitude 2 in the direction of the positive *z*-axis. So we can associate the ordered triple of real numbers (1, 3, 2) with the vector **w**.Thus we can express a vector in ^{2} in terms of multiples of *unit vectors* (vectors of magnitude 1 unit) in the directions of the *x*- and *y*-axes, and a vector in ^{3} in terms of multiples of unit vectors in the directions of the *x*-, *y*- and *z*-axes. This expression of a vector as a sum of vectors in standard directions is called *decomposition of the vector into components*. This representation of a vector in terms of components will be useful in Section 3.
DefinitionsIn ^{2}, the vectors **i** and **j** are unit vectors in the positive directions of the *x*- and *y*-axes, respectively. Any vector **p** in ^{2} can be expressed as a sum of the form**p** = *a*_{1}**i** + *a*_{2}**j**, for some real numbers *a*_{1}, *a*_{2};

often we write **p** = (*a*_{1}, *a*_{2}), for brevity. The numbers *a*_{1} and *a*_{2} are the **components** of **p** in the *x*- and *y*-directions, respectively.In ^{3}, the vectors **i**, **j** and **k** are unit vectors in the positive directions of the *x*-, *y*- and *z*-axes, respectively. Any vector **p** in ^{3} can be expressed as a sum of the form**p** = *a*_{1}**i** + *a*_{2}**j** + *a*_{3}**k**, for some real numbers *a*_{1}, *a*_{2}, *a*_{3};

often we write **p** = (*a*_{1}, *a*_{2}, *a*_{3}), for brevity. The numbers *a*_{1}, *a*_{2} and *a*_{3} are the **components** of **p** in the *x*-, *y*- and *z*-directions, respectively.Using these definitions, we can express the above vector **v** as 3**i** + 4**j**, or simply as (3, 4).Also using these definitions, we can express the above vector **w** as **i** + 3**j** + 2**k**, or simply as (1, 3, 2).It is important to remember that the above expression of a vector as an ordered pair or an ordered triple of real numbers simply describes the components of the corresponding translation; it does not mean that we are identifying the vector with the point whose Cartesian coordinates are that pair or triple.Example 22Sketch the following vectors in ^{2} on a single diagram:2**i** − 3**j**, −3**i** + 4**j**, −2**i** − 2**j**.
In the above discussion we expressed vectors in ^{2} and in ^{3} in terms of components relative to unit vectors that are mutually perpendicular. In general, however, we do not need to restrict ourselves in this way. In fact, we can express a vector in ^{2} in terms of *any* two vectors that are not parallel (that is, whose directions are not the same or exactly opposite), and a vector in ^{3} in terms of *any* three non-coplanar vectors. (A set of vectors in ^{3} is *coplanar* if there is some plane in ^{3} that contains all the vectors of the set).Now that we can describe vectors in terms of their components, we return to the operations on vectors which we described geometrically in Sections 2.1 through 2.3, in order to describe the same operations in terms of components.The following observations follow immediately from the definition of components.DefinitionsTwo vectors, both in ^{2} or both in ^{3}, are **equal** if and only if their corresponding components are equal.The **zero vector** in ^{2} is **0** = 0**i** + 0**j** = (0, 0), and the zero vector in ^{3} is **0** = 0**i** + 0**j** + 0**k** = (0, 0, 0).We have described addition of vectors already. To add two vectors **p** and **q** using the Triangle Law, we first draw the vector **p**; then, starting from the finishing point of **p**, we draw the vector **q**. The sum **p** + **q** is the vector from the starting point of **p** to the finishing point of **q**.
*Recall that we can think of a vector as a translation.*In order to see how addition can be described in terms of components, we look at the sum of two vectors in the plane. Let **p** and **q** be the vectors
**p**= *a*_{1}**i** + *a*_{2}**j** = (*a*_{1},*a*_{2}), **q** = *b*_{1}**i** + *b*_{2}**j** = (*b*_{1},*b*_{2}).
Then a translation **p** followed by a translation **q** can be described in terms of the following four successive translations:move *a*_{1} units in the direction of the positive *x*-axis (this translation is vector *a*_{1}**i**);then *a*_{2} units in the direction of the positive *y*-axis (this is *a*_{2}**j**);then *b*_{1} units in the direction of the positive *x*-axis (this is *b*_{1}**i**);then *b*_{2} units in the direction of the positive *y*-axis (this is *b*_{2}**j**).
Since the order in which we perform translations does not matter, the net effect is a translation of *a*_{1} + *b*_{1} units in the positive direction of the *x*-axis, followed by a translation of *a*_{2} + *b*_{2} units in the positive direction of the *y*-axis. In other words, the sum **p** + **q** is the vector
(*a*_{1}, *a*_{2}) + (*b*_{1}, *b*_{2}) = (*a*_{1} + *b*_{1}, *a*_{2} + *b*_{2}).
There is a corresponding formula in ^{3}.Addition of vectorsTo add vectors in ^{2} or in ^{3} given in component form, add their corresponding components:*Note:* There are analogous formulas for vectors expressed in terms of the unit vectors **i** and **j**, or **i**, **j** and **k**.For example, the sum of the vectors (1, −3) and (4, 2) in ^{2} is the vector (1 + 4, −3 + 2) = (5, −1), and the sum of the vectors (2, −1, −3) and (−2, 3, 2) in ^{3} is the vector (2 − 2, −1 + 3, −3 + 2) = (0, 2, −1).In equations (2.1) and (2.2) above, the symbol + is used with two different meanings. On the left-hand side of each equation, + denotes addition of vectors, and on the right-hand side, + denotes addition of real numbers. It is sensible to use the same symbol, as addition of vectors obeys the same rules as addition of real numbers.Example 23Determine the following sums of vectors in ^{2}:We add vectors by adding their corresponding components. Hence (3,2) + (1,−5) = (4,−3),(5,−2) + (−5,2) = (0,0),(4,1) + (0,0) = (4,1),((1,3) + (4,2)) + (−3,1) = (5,5) + (−3,1) = (2,6),(1,3) + ((4,2) + (−3,1)) = (1,3) + (1,3) = (2,6).You saw in Exercise 23 (above) that the sum of (5, −2) and (−5, 2) is the zero vector. In general, in ^{2} the sum of the vectors **p** = (*a*_{1}, *a*_{2}) and (−*a*_{1}, −*a*_{2}) is the zero vector, so that in terms of its components, the vector −**p** is (−*a*_{1}, −*a*_{2}). Similarly, in terms of components, the negative of the vector **p** = (*a*_{1}, *a*_{2}, *a*3) in ^{3} is −**p** = (−*a*_{1}, −*a*_{2}, −*a*3). For example, if **p** = (1, −3), then −**p** = (−1, 3); and if **p** = (2, −1, 3), then −**p** = (−2, 1, −3).
We can then express the difference **p** − **q** of two vectors **p** and **q** in terms of components since, as we saw in Section 2.3, **p** − **q** is defined to be **p** + (−**q**).For example, if **p** = (4, 1) and **q** = (2, −1), then −**q** = (−2, 1) and **p** − **q** = (4 − 2,1 + 1) = (2, 2).Similarly, if **p** = **i** + 2**j** − 3**k** and **q** = 2**i** − **j** + **k**, then −**q** = −2**i** + **j** − **k** and**p** − **q** = −**i** + 3**j** − 4**k**.Negative of a vectorTo find the negative of a vector in ^{2} or in ^{3} given in component form, take the negatives of its components:
−(*a*_{1}, *a*_{2}) = (−*a*_{1}, −*a*_{2}) and −(*a*_{1}, *a*_{2}, *a*_{3}) = (−*a*_{1}, −*a*_{2}, −*a*_{3}).Subtraction of vectorsTo subtract a vector in ^{2} or in ^{3} given in component form, subtract its corresponding components:
For the above formulas, there are analogous formulas for vectors expressed in terms of the unit vectors **i** and **j**, or **i**, **j** and **k**.Example 24For each of the following pairs of vectors **p** and **q**, write down −**p**, −**q** and **p** − **q**.**(a)** **p** = (3, −1) and **q** = (−1, −2).**(b)** **p** = −**i** − 2**j** and **q** = 2**i** − **j**.**(c)** **p** = −**i** + 2**k** and **q** = **i** − 2**j** − **k**.**(a)** Here **p** = (3, −1) and **q** = (−1, −2), so
−**p** = (−3,1),
−**q** = (1,2),
**p**−**q** = (3 + 1,−1 + 2) = (4,1).
**(b)** Here **p** = −**i** − 2**j** and **q** = 2**i** − **j**, so
−**p** =**i**+ 2**j**,
−**q** = −2**i** +**j**,
**p**−**q** = (−1 − 2)**i** + (−2 + 1)**j** = −3**i** −**j**.
**(c)** Here **p** = −**i** + 2**k** and **q** = **i** − 2**j** − **k**, so
−**p** = **i** − 2**k**,
−**q** = −**i** + 2**j** + **k**,
**p**−**q** = −2**i** + 2**j** + 3**k**.
The other operation that we described geometrically earlier was multiplication by a scalar. For example, the vector 2**p** has magnitude twice that of **p** and has the same direction; so, in terms of components, if **p** = (*a*_{1}, *a*_{2}), then 2**p** = (2*a*_{1}, 2*a*_{2}).
In general, if a vector in ^{2} or in ^{3} is multiplied by a scalar *k*, then each of its components is multiplied by *k*. Recall that the product *k***p** is in the same direction as **p** if *k* is positive, and in the opposite direction if *k* is negative.Multiplication by a scalarTo multiply a vector given in component form in ^{2} or in ^{3} by a real number *k*, multiply each component in turn by *k*:
*k*(*a*_{1}, *a*_{2}) = (*ka*_{1}, *ka*_{2}) and *k*(*a*_{1}, *a*_{2}, *a*_{3}) = (*ka*_{1}, *ka*_{2}, *ka*_{3}).*Note:* There are analogous formulas for vectors expressed in terms of the unit vectors **i** and **j**, or **i**, **j** and **k**.For example, if **p** = (2, −1), then 2**p** = (4, −2) and −3**p** = (−6, 3).Example 25For each of the following pairs of vectors **p** and **q**, determine 2**p**, 3**q**, 2**p** + 3**q** and 2**p** − 3**q**.**(a)** **p** = (3, −1) and **q** = (−1, −2).**(b)** **p** = −**i** + 2**k** and **q** = **i** − 2**j** − **k**.**(a)** Since **p** = (3, −1) and **q** = (−1, −2),
2**p** = (6,−2),
3**q** = (−3,−6),
2**p** + 3**q** = (3,−8),
2**p** − 3**q** = (9,4).
**(b)** Since **p** = −**i** + 2**k** and **q** = **i** − 2**j** − **k**,
2**p** = −2**i** + 4**k**,
3**q** = 3**i** − 6**j** − 3**k**,
2**p** + 3**q** = **i** − 6**j** + **k**,
2**p** − 3**q** = −5**i** + 6**j** + 7**k**.
The set of ordered pairs (triples) of real numbers, together with the operations of addition of ordered pairs (triples) and multiplication of an ordered pair (triple) by a scalar, is an example of the algebraic structure known as a *vector space*.DefinitionsThe **vector space** ^{2} is the set of ordered pairs of real numbers with the operations of addition and multiplication by a scalar defined as follows:Similarly, the **vector space** ^{3} is the set of ordered triples of real numbers with analogous operations of addition and multiplication by a scalar.2.5 Position vectorsFinally, we relate the method of specifying points in the plane as an ordered pair of real numbers (that is, via the Cartesian coordinate system) to vectors.In order to do this, we use those vectors whose starting point is the origin; such vectors are called *position vectors*. For example, the position vector (2, −1) is the vector shown below.
DefinitionThe **position vector****p** = *a*_{1}**i** + *a*_{2}**j** (often written as **p** = (*a*_{1}, *a*_{2}), for brevity)is the vector in ^{2} whose starting point is the origin and whose finishing point is the point with Cartesian coordinates (*a*_{1}, *a*_{2}).*Note:* The position vector **p** = *a*_{1}**i** + a_{2}**j** + *a*_{3}**k** (often written as **p** = (*a*_{1}, *a*_{2}, *a*_{3})) in ^{3} can be defined in a similar way.Thus there is a one-one correspondence between the points of the plane and the set of position vectors in ^{2}. We shall make use of this in Section 2.6 to simplify the solution of various geometric problems in the plane.Example 26Let **p** and **q** be the position vectors (5, 3) and (1, 4), respectively.**(a)** Determine the position vectors **p** − **q**, **p** + **q** and .**(b)** Sketch **p**, **q** and each of the position vectors that you found in part (a).**(a)** Since **p** = (5, 3) and **q** = (1, 4),**(b)**
2.6 LinesEarlier, we found the equation of a line in the (*x*, *y*)-plane in the form
*ax* + *by* = *c*,
for some real numbers *a*, *b* and *c*, where *a* and *b* are not both zero. We now find an equivalent equation for a line in terms of vectors.Let *P* and *Q* be two given points with position vectors **p** and **q**, and denote by ℓ the line that passes through *P* and *Q*. How can we find the position vector **r** of a point *R* on ℓ that lies between *P* and *Q*?First, we use the Triangle Law for the addition of vectors to find expressions for the vectors and ; this givesandSince and are parallel, the vector **q** − **r** is parallel to the vector **q** − **p**, so it must be a multiple of **q** − **p**; that is,*Note:* Two vectors are *parallel* if they are in the same direction or in opposite directions.We can rearrange the above equation in the formsoThis is a general formula for the position vector of a point on the line segment *PQ*, in the following sense: each point on ℓ between *P* and *Q* corresponds to a particular value of λ between 0 and 1, and vice versa.In equation (2.3),This is the *midpoint* of the line segment *PQ*.Equation (2.3) also makes sense when λ > 1 and when λ < 0, and not just when λ lies between 0 and 1. In fact,Thus each point on the line ℓ has a position vector of the form of equation (2.3), for some value of λ. In other words, we can regard equation (2.3) as the *vector form of the equation of the line* ℓ, with λ as a parameter.Vector form of the equation of a lineThe equation of the line through the points with position vectors **p** and **q** isExample 27**(a)** Let *P* and *Q* be the points with position vectors **p** = (3, 1) and **q** = (2, 3), respectively. Write down the vector form of the equation of the line ℓ through *P* and *Q*.**(b)** Determine the points on ℓ whose position vectors are given by equation (2.3) when λ takes the values ,
and .**(c)** On a single diagram, sketch *P*, *Q*, the line ℓ through *P* and *Q*, and the three points that you found in part (b).**(a)** We use equation (2.3) to obtain the vector form of the equation of ℓ as**(b)** Using the above formula with , and in turn, we obtain the following position vectors:Thus the three points have Cartesian coordinates , and , respectively.**(c)**The vector form of the equation of the line ℓ passing through the points (3, 1) and (2, 3) isthat is,We can use this equation to determine whether or not a given point lies on the line ℓ. For example, the point (5, −3) lies on ℓ if there is some real number λ such thatEquating the corresponding components, we see that this condition reduces to a pair of simultaneous equations that must be satisfied by such a number λ:These have the solution λ = 3, and hence the point (5, −3) does lie on the line ℓ.Example 28Let *P*, *Q* and ℓ be as in Exercise 27.**(a)** Determine the value of λ corresponding to the point (4, −1) in the vector form of the equation of ℓ: **r** = λ(3, 1) + (1 − λ)(2, 3).**(b)** Use the vector form of the equation of ℓ to prove that the point does not lie on ℓ.**(a)** The vector form of the equation of ℓ isHence at the point (4, −1) on ℓ, we haveEquating corresponding components, we obtain the following pair of simultaneous equations that must be satisfied by the number λ:These have the solution λ = 2.**(b)** Since the vector equation of ℓ is **r** = λ(3, 1) + (1 − λ)(2, 3), the point lies on ℓ if and only if there is some real number λ for whichEquating corresponding components, we obtain the following pair of simultaneous equations that must be satisfied by such a number λ:The first of these equations has solution , and the second has solution
.It follows that there is no real number λ that satisfies equation (S.1), so the point does not lie on ℓ.We can obtain another useful result from our derivation of equation (2.3). We saw above that it follows thatComparing these expressions for and , we conclude that *R* is the point that divides the line *PQ* in the ratio (1 − λ) : λ.
For example, let *PR* : *RQ* = 3: 2. Since the ratio 3 : 2 is just the same as the ratio , we write this equation in the form , so that *PR* : *RQ* = (1 − λ) : λ with . (Here we divide each expression in the ratio by the sum 3 + 2 = 5, in order to obtain the ratio in the standard form (1 − λ) : λ.)Section FormulaThe position vector **r** of the point that divides the line joining the points with position vectors **p** and **q** in the ratio (1 − λ) : λ is*Note:* When **q** = (0, 0), **r** has the simple form **r** = λ**p**; we shall find this fact useful later.Example 4Let *P* and *Q* be the points (−1, 3) and (5, 6). Determine the point *R* that divides the line segment *PQ* in the ratio 1 : 2 (thus *R* is one-third of the way from *P* to *Q*).The position vectors of *P* and *Q* are **p** = (−1, 3) and **q** = (5, 6); let *R* have position vector **r**.Since , we can apply the Section Formula with ; henceThus *R* is the point with coordinates (1, 4).Example 29Let *P* and *Q* be the points (−3, 1) and (7, −4). Determine**(a)** the point *R* that divides *PQ* in the ratio 3 : 2;**(b)** the midpoint *M* of *PQ*.**(a)** The point *R* divides *PQ* in the ratio . Applying the Section Formula with , we find that the position vector of *R* isThus *R* is the point (3, −2).**(b)** The midpoint *M* divides *PQ* in the ratio . Applying the Section Formula with , we find that the position vector of *M* isThus *M* is the point .We finish with an example that uses many of the ideas that you have met in Section 2.Example 5The triangle *OAB* has vertices at *O* (the origin) and at points *A* and *B* with position vectors **a** and **b**, respectively. *Q* is the midpoint of *OB*, and the point *P* is one-third of the way along *BA* from the vertex *B*; *R* is the point of intersection of the lines *OP* and *AQ*. The points *P*, *Q* and *R* have position vectors **p**, **q** and **r**, respectively.Determine **p**, **q** and **r** in terms of **a** and **b**.
The point *P* divides *BA* in the ratio . It follows from the Section Formula with thatNow *Q* is the midpoint of *OB*, so*Note:* Here we use the fact that the position vector of *O* is **0** = (0, 0).To find **r**, we use the fact that *R* is the point of intersection of *OP* and *AQ*. First, since *R* lies on *AQ*, its position vector **r** must be of the formfor some real number λ. (**r** could also be written as λ**a** + (1 − λ)**q**, but with a different value of λ.) Substituting from equation (2.5), we can rewrite the above formula asThe point *R* lies on the line *OP* also, so we can express its position vector **r** as a scalar multiple of **p**. Thus we can use equation (2.4) to write **r** in the form for some real number *k*.The expressions in equations (2.6) and (2.7) for **r** must be equal, so which we can rewrite as We know that the vectors **a** and **b** are not parallel. Hence non-zero multiples of **a** and **b** cannot be parallel. It follows that the only way in which equation (2.8) can hold is for the coefficients of **a** and **b** in equation (2.8) to be zero: that is, we must have and . Hence the following simultaneous equations for λ and *k* must hold:*Note:* This is a common technique in the application of vectors to the solution of geometric problems.From the second equation we deduce that . Then, substituting this value for *k* into the first equation, we find that ; hence , so .Substituting the value for λ into equation (2.6), we conclude thatExample 30The triangle *OAB* has vertices at *O* (the origin) and at points *A* and *B* with position vectors **a** and **b**, respectively. *P* is the midpoint of *AB*, and *Q* is the midpoint of *OA*; *R* is the point of intersection of the lines *OP* and *BQ*. The points *P*, *Q* and *R* have position vectors **p**, **q** and **r**, respectively.**(a)** Determine **p**, **q** and **r** in terms of **a** and **b**.**(b)** Determine the ratio *OR* : *RP*.
**(a)** Since *P* is the midpoint of *AB*, its position vector isSince *Q* is the midpoint of *OA*, its position vector is The point *R* lies on the two lines *OP* and *BQ*, so we can express its position vector **r** in two different ways. Since *R* lies on *OP*,and since *R* lies on *BQ*,These two expressions for **r** must be equal; thusSince **a** and **b** are not parallel, their coefficients in this last equation must both be zero. This gives two equations for *k* and λ:Thus *k* = 2λ, so 2λ − 1 + λ = 0, and hence . Hence **(b)** The position vectors of *O*, *R* and *P* are **0**, and , respectively. ThusIt follows from the Section Formula (with ) that Thus *R* is two-thirds of the way along *OP* from *O*.2.7 Further exercisesExample 31Let **p** = 2**i** − 3**j** + **k** and **q** = −**i** −2**j** −4**k** be two vectors in ^{3}. Determine **p** + **q**, **p** − **q** and 2**p** − 3**q**.Since **p** = 2**i** − 3**j** + **k** and **q** = −**i** −2**j** −4**k**, we haveandExample 32Let **u** and **v** be the position vectors (1, 1) and (2, 1), respectively.**(a)** Determine the position vectors **u** + 2**v**, −**u**, −**u** + 3**v** and **u** − 3**v**.**(b)** On a single diagram, sketch **u**, **v** and the position vectors that you found in part (a).**(a)** Here,and**(b)**Example 33Let **u** = (2, 6) and **v** = (4, 2).**(a)** Determine numbers *α* and *β* such that*Hint:* Obtain simultaneous equations involving *α* and *β* by equating the first and second components of the vectors on each side of equation (2.9).**(b)** Sketch the position vectors **u**, **v** and (3, 4) on a single diagram.**(a)**First we writeThusThen, following the hint, we equate the first and second components of these vectors; this gives two simultaneous equations:These two equations can be solved to giveThus we can write**(b)**Example 34Let *A* and *B* be the points (5, 4) and (−2, −3).Determine the point *R* that divides *AB* in the ratio 2 : 5, and the midpoint *M* of *AB*.The point *R* divides *AB* in the ratio . It follows from the Section Formula, with , that the position vector **r** of *R* isHence the coordinates of *R* are (3, 2).The midpoint *M* of *AB* has position vector **m**, where Hence the coordinates of *M* are
.Example 35**(a)** Determine the vector form of the equation of the line ℓ through the points (2, 3) and (5, −1).**(b)** Hence determine whether the points (7, 2) and (−1, 7) lie on ℓ.**(a)** Let **p** = (2, 3) and **q** = (5, −1). Then the vector form of the equation of the line ℓ isthat is,where λ ∈ .**(b)** The point (7, 2) lies on the line ℓ if there is a value of λ such thatEquating components in turn, we obtain The second equation gives , but this value of λ does not satisfy the first equation. It follows that (7, 2) does not lie on the line ℓ.The point (−1, 7) lies on the line ℓ if there is a value of λ such thatEquating components in turn, we obtainThe second equation gives λ = 2, and this value also satisfies the first equation. It follows that the point (−1, 7) lies on the line ℓ.Example 36The rectangle *OABC* has vertex *O* at the origin, and the position vectors of the vertices *A*, *B* and *C* are **a**, **b** and **c**, respectively. *P* is the midpoint of *BC*, and the point *Q* is one-third of the way along *AB* from *A*. *S* is the point of intersection of the lines *OC* and *QP*.**(a)** Determine the position vectors of *P* and *Q*, in terms of **a** and **c**.**(b)** Determine the vector form of the equation of the line that passes through *P* and *Q*, in terms of **a** and **c**.**(c)** Use the equation that you obtained in part (b), and the fact that *S* lies on the line *OC*, to find the position vector of *S*, in terms of **a** and **c**.
**(a)***P* is the midpoint of *BC*, so its position vector **p** can be written asNow **b** = **a** + **c** by the Parallelogram Law, since *OB* is the diagonal of the parallelogram with adjacent sides **a** and **c**. ThusNext, the position vector **q** of the point *Q* is given by .
Here ;
also ,
since (as they have the same magnitude and direction).
Hence**(b)** The vector form of the line through *P* and *Q* isMultiplying out and collecting terms, we obtain where λ ∈ .**(c)** The point *S* lies on *QP*, so its position vector **s** can be written in the formBut **s** can also be written solely as a multiple of **c**, since *S* is also on *OC* and so **s** is in the same direction as **c**. This means that the coefficient of **a** in equation (S.2) for **s** must be 0.Thus , so λ = 2.It follows that3 Dot product3.1 Definition, properties and some applicationsIn the previous section we saw how to add two vectors and how to multiply a vector by a scalar, but we did not consider how to multiply two vectors. There are two different ways in which we can multiply two vectors, known as the *dot product* (or *scalar product*) and the *vector product*. They are given these names because the result of the first is a scalar and the result of the second is a vector. (We shall not consider vector products in this course.)In the audio section we explain the definition of the dot product and investigate some of its basic properties. In addition, we see how the dot product can be used to find the angle between two vectors, to give a condition for two vectors to be orthogonal (that is, at right angles) and to find the projection of one vector onto another vector.Listen to the audio below as you work through the following frames.Click 'Play' to listen to Audio Clip 1 while viewing frames 1–7 in the main text below.Click 'Play' to listen to Audio Clip 2 while viewing frames 8–12 in the main text below.Example 37We use the formula for the dot product given in Frame 6 Section 3..**(a)****(b)****(c)**Example 38**(a)** When **u** = (2, −3), the length of **u** isso**(b)** When **u** = 5**i** + 12**j**, the length of **u** issoExample 39In each case we use the formula for the angle between two vectors given in Frame 10 Section 3.1 , letting **u** denote the first vector of the pair, **v** the second vector and *θ* the angle between the two vectors.**(a)** Here **u · v** = (1, 4) **·** (5, 2) = 5 + 8 = 13,andThen so**(b)** Here **u · v** = (−2, 2) **·** (1, −1) = −2 −2 = −4,andThenso**(c)** HereandThen soClick 'Play' to listen to Audio Clip 3 while viewing frames 13–17 in the main text below.Example 40
To calculate the projections of **u** onto **v** and **v** onto **u**, we need the following:Then the projection of **v** onto **u** isand the projection of **u** onto **v** is Example 41In each case we use the formula for the angle between two vectors given in Frame 10, letting **u** denote the first vector of the pair, **v** the second vector and *θ* the angle between the two vectors.**(a)** Here **u · v** = (1, 2, 0) · (3, −1, 2) = 3 − 2 + 0 = 1,andThenso**(b)**HereandThen so3.2 Post-audio exercisesExample 42Let **u** and **v** be the position vectors (6, 8) and (−12, 5), respectively.**(a)** Sketch **u** and **v** on a single diagram. On the same diagram, sketch the projection of **u** onto **v**, and the projection of **v** onto **u**.**(b)** Determine the angle between **u** and **v**.**(c)** Determine the projection of **u** onto **v**, and the projection of **v** onto **u**.**(a)****(b)** Let *θ* denote the angle between **u** and **v**. Now**(c)** The projection of **u** onto **v** is The projection of **v** onto **u** is Example 43Determine the angle between the vectors **u** = (3, 4, 5) and **v** = (1, 0, −1) in ^{3}.Let *θ* denote the angle between **u** and **v**. NowandThusExample 44Find a vector of length 2 that is perpendicular to both of the vectors **a** = (2, 1, 0) and **b** = (1, 0, −1).Let the vector we want be denoted by **v** = (*x*,*y*,*z*), for some real numbers *x*, *y* and *z*.Since ‖**v**‖ = 2,Since **v** is perpendicular to **a**,thus 2*x* + *y* = 0,that is,Since **v** is perpendicular to **b**,thus *x* − *z* = 0,that is,Substituting the expressions for *y* in equation (S.4) and for *z* in equation (S.5) into equation (S.3), we obtain *x*^{2} + (−2*x*)^{2} + *x*^{2} = 4;thus 6*x*^{2} = 4,soIt follows that the two possible vectors that satisfy the given conditions are3.3 Equation of a plane in three-dimensional Euclidean spaceWe stated in Section 1.7 that the general form of the equation of a plane in ^{3} iswhere *a*, *b* and *c* are not all zero. We can prove this now, using the dot product.How can we specify a plane uniquely in ^{3}? One possibility is to specify three points that lie in the plane. However, this does not enable us to prove equation (3.1) without a significant amount of algebra; so we adopt a different approach.For a given plane in ^{3}, there is a particular direction that can be used to specify the plane – the direction perpendicular to all the vectors in that plane.DefinitionA vector that is perpendicular to all the vectors in a given plane is called a **normal vector** to the plane.*Note:* A normal vector to a plane is also called a *normal* to the plane, and its direction is said to be *normal to* the plane.If **n** is a normal vector to a given plane, then so is *k***n**, for any non-zero real number *k*. If *k* > 0, then *k***n** is in the same direction as **n**, whereas if *k* < 0, then *k***n** is in the opposite direction to **n**.
A normal vector **n** does not determine a plane uniquely, as there are infinitely many planes that have **n** as a normal; these planes are parallel to one another. However, if we specify both a normal vector and a point that lies in the plane, then the plane is determined uniquely.Example 6Determine the equation of the plane in ^{3} that contains the point *P*(2, 3, 4) and has **n** = (1, 2, −1) as a normal.If the point *Q*(*x*, *y*, *z*) lies in the plane, then the vector must be perpendicular to the normal vector **n**; it follows that must be zero.
Hence
so (*x* − 2) × 1 + (*y* − 3) × 2 + (*z* − 4) × (−1) = 0.This equation can be rearranged in the form *x* + 2*y* − *z* = 4;this is the equation of the plane.In general, let **n** = (*a*, *b*, *c*) be normal to a given plane that contains the point *P*(*x*_{1}, *y*_{1}, *z*_{1}). Then an arbitrary point *Q*(*x*, *y*, *z*) lies in the plane if and only if the vectors and **n** are orthogonal; that is, if and only if .
Since , this condition can be written in the formso (*x* − *x*_{1}) × *a* + (*y* − *y*_{1}) × *b* + (*z* − *z*_{1}) × *c* = 0.This equation can be rearranged in the form *ax* + *by* + *cz* = *ax*_{1} + *by*_{1} + *cz*_{1},that is, *ax* + *by* + *cz* = *d*,where *d* = *ax*_{1} + *by*_{1} + *cz*_{1}.Theorem 1The equation of the plane that contains the point (*x*_{1}, *y*_{1}, *z*_{1}) and has **n** = (*a*, *b*, *c*) as a normal is *ax* + *by* + *cz* = *d*,where *d* = *ax*_{1} + *by*_{1} + *cz*_{1}.Once we know the equation of a plane, we can ‘read off’ the components of a normal vector as they are the coefficients of *x*, *y* and *z* in the equation. For instance, one normal to the plane with equation *x* − 2*y* + 3*z* = 7 is **n** = (1, −2, 3).We shall not make use of the formula for *d* given in the above theorem. In practice, it is simpler to use the following corollary.Corollary to Theorem 1The equation of the plane that contains the point (*x*_{1}, *y*_{1}, *z*_{1}) and has **n** = (*a*, *b*, *c*) as a normal iswhere **x** = (*x*, *y*, *z*) and **p** = (*x*_{1}, *y*_{1}, *z*_{1}).**Proof**Since *ax* + *by* + *cz* = *d*, from Theorem 1, it follows that **x · n** = **p · n**.Example 7Determine the equation of the plane in ^{3} that contains the point (1, −1, 4) and has (2, −2, 3) as a normal.The equation of the plane is where **x** = (*x*, *y*, *z*), **p** = (1, −1, 4) and **n** = (2, −2, 3); in other words, the equation of the plane is that is, 2*x* − 2*y* + 3*z* = 1 × 2 + (−1) × (−2) + 4 × 3,or 2*x* − 2*y* + 3*z* = 16.Example 45Determine the equation of each of the following planes:**(a)** the plane that contains the point (1, 0, 2) and has (2, 3, 1) as a normal;**(b)** the plane that contains the point (−1, 1, 5) and has (4, −2, 1) as a normal.We use the formulafor the equation of a plane, where **x** = (*x*, *y*, *z*), **n** is a normal to the plane and **p** is a point in the plane.**(a)** Here **n** = (2, 3, 1) and **p** = (1, 0, 2), so the equation of the plane isThis can be expressed in the form 2*x* + 3*y* + *z* = 1 × 2 + 0 × 3 + 2 × 1,that is,2*x* + 3*y* + *z* = 4.**(b)** Here **n** = (4, −2, 1) and **p** = (−1, 1, 5), so the equation of the plane is This can be expressed in the form 4*x* − 2*y* + *z* = (−1) × 4 + 1 × (−2) + 5 × 1,that is, 4*x* − 2*y* + *z* = −1.We can use the general form of the equation of a plane to find the *equation of a plane that contains three given points*, as follows. We assume that the equation of the plane is *ax* + *by* + *cz* = *d*, and then substitute the coordinates of the three points in turn into this equation; this gives three simultaneous equations for *a*, *b* and *c*, which we solve.Example 46Determine the equation of the plane that contains the points (3, 0, 0), (0, 4, 0) and (3, 4, 5).Write down a vector that is normal to this plane.Let the equation of the plane bewhere *a*, *b*, *c* and *d* are real, and *a*, *b* and *c* are not all zero.Substituting the coordinates of each of the points in turn into equation (S.6), we obtain From the first two equations we obtain Substituting these expressions into the third equation, we obtain so *d* + *d* + 5*c* = *d*,henceSubstituting these expressions for *a*, *b* and *c* into equation (S.6), we see that the equation of the plane can be written as Dividing by *d*, we obtain It follows that one possible normal vector is .3.4 Further exercisesExample 47**(a)** Find the angle between each of the pairs of vectors:
(3, 1) and (1, −2); **i** + 2**j** and −3**i** + **j** − 2**k**.
**(b)** Determine the projection of (3, 1) onto (1, −2), and the projection of **i** + 2**j** onto −3**i** + **j** − 2**k**.Denote by **p** the first vector of each pair, by **q** the second vector and by *θ* the angle between the vectors.**(a)** When **p** = (3, 1) and **q** = (1, −2), we have It follows that When **p** = **i** + 2**j** and **q** = −3**i** + **j** − 2**k**, we have **(b)** When **p** = (3, 1) and **q** = (1, −2), the projection of **p** onto **q** isWhen **p** = **i** + 2**j** and **q** = −3**i** + **j** − 2**k**, the projection of **p** onto **q** is Example 48Determine the two vectors of length 2 which make an angle of /4 with the vector **p** = (2, −2). Verify that the two vectors that you have found are perpendicular to each other.Let such a vector be **r** = (*x*, *y*). Since **r** has length 2 and makes an angle /4 with **p** = (2, −2),In component form we have sothat is, *x* = *y* + 2.Also, **r** has length 2, so Substituting the expression for *x* into equation (S.7), we obtain (*y* + 2)^{2} + *y*^{2} = 4,thus *y*^{2} + 4*y* + 4 + *y*^{2} = 4,that is, 2*y*^{2} + 4*y* = 0.This has solutions *y* = 0 and y = −2;the corresponding values of *x* are *x* = 2 and *x* = 0.It follows that the two vectors of length 2 making an angle of /4 with (2, −2) are (2, 0) and (0, −2).The dot product of these two vectors is so the two vectors are perpendicular.Example 49Determine the equation of the plane that contains the point (−1, 3, 2) with (1, 2, −1) as a normal.We use the corollary to Theorem 1. The equation of the plane is given by **x · n** = **p · n**, where **x** = (*x*, *y*, *z*), **p** = (−1, 3, 2) and **n** = (1, 2, −1).Thus the equation of the plane is *x* + 2*y* − *z* = (−1) × 1 + 3 = 2 + 2 = (−1),that is, *x* + 2*y* − *z* = 3.Example 50Determine the equation of the plane through (1, 0, 2), (0, 3, 4) and (0, −1, 0).Let the equation of the plane be for some real numbers *a*, *b*, *c* and *d*, where *a*, *b* and *c* are not all zero.Substituting the coordinates of the three points in turn into equation (S.8), we obtainSubstituting −*b* for *d* in equations (S.9) and (S.10), we obtain Equation (S.13) gives *c* = −*b*; if we then substitute −*b* for *c* in equation (S.12), we obtain *a* − 2*b* = −*b*,that is, *a* = *b*.It follows that the equation of the plane is *bx* + *by* − *bz* = −*b*,that is, *x* + *y* − *z* = −1.4 Conics4.1 Conic sections*Conic section* is the collective name given to the shapes that we obtain by taking different plane slices through a double cone. The shapes that we obtain from these cross-sections are drawn below. It is thought that the Greek mathematician Menaechmus discovered the conic sections around 350 bc.The circle in slice 7 can be regarded as a special case of an ellipse.We use the term *non-degenerate conic sections* to describe those conic sections that are parabolas, ellipses or hyperbolas; and the term *degenerate conic sections* to describe the single point, single line and pair of lines.*Note:* We usually use ‘conic’ rather than ‘conic section’ once we have described how conics arise.There are some interesting features of the parabola, ellipse and hyperbola that we note for use later. The ellipse and the hyperbola both have a *centre*; that is, there is a point about which rotation through an angle is a symmetry of the conic. For example, for the ellipse and hyperbola illustrated above, the centre is the origin. On the other hand, the parabola does not have a centre. The hyperbola has two lines, called *asymptotes*, which it approaches.4.2 CirclesRecall that a *circle* in ^{2} is the set of points (*x*, *y*) that lie at a fixed distance, called the *radius*, from a fixed point, called the *centre* of the circle. We can use the techniques of coordinate geometry to find the equation of a circle with a given centre and radius.
Let a circle have centre *C*(*a*, *b*) and radius *r*. Then, if *P*(*x*, *y*) is an arbitrary point on the circumference of the circle, the distance *CP* equals *r*. It follows from the formula for the distance between two points in the plane that Theorem 2The equation of a circle in ^{2} with centre (*a*, *b*) and radius *r* is (*x* − *a*)^{2} + (*y* − *b*)^{2} = *r*^{2}.For example, it follows from this formula that the circle with centre (−1, 2) and radius √3 has equationwhich can be simplified to give *x*^{2} + 2*x* + 1 + *y*_{2} − 4*y* + 4 = 3,or *x*^{2} + *y*^{2} + 2*x* − 4*y* + 2 = 0.Example 51Determine the equation of each of the following circles, given the centre and radius:**(a)** centre the origin, radius 1;**(b)** centre the origin, radius 4;**(c)** centre (3, 4), radius 2.We use the standard formula for the equation of a circle of given centre and radius given in Theorem 2.**(a)** This circle has equation (*x* − 0)^{2} + (*y* − 0)^{2} = 1^{2},which can be rewritten in the form *x*^{2} + *y*^{2} = 1.**(b)** This circle has equation (*x* − 0)^{2} + (*y* − 0)^{2} = 4^{2},which can be rewritten in the form *x*^{2} + *y*^{2} = 16.**(c)** This circle has equation (*x* − 3)^{2} + (*y* − 4)^{2} = 2^{2},which can be rewritten in the form *x*^{2} + *y*^{2} − 6*x* − 8*y* + 21 = 0.If we expand the brackets in equation (4.1) and collect the corresponding terms, we can rewrite equation (4.1) in the form *x*^{2} + *y*^{2} − 2*ax* − 2*by* + (*a*^{2} + *b*^{2} − *r*^{2}) = 0.Then, if we write *f* for −2*a*, *g* for −2*b* and *h* for *a*^{2} + *b*^{2} − *r*^{2}, this equation takes the form *Note:* The coefficients of *x*^{2} and *y*^{2} are equal.In many situations, equation (4.1) is more useful than equation (4.2) for determining the equation of a particular circle.We have seen that the equation of a circle can be written in the form On the other hand, given an equation of the form (4.2), can we determine whether it represents a circle? If it does represent a circle, can we determine its centre and radius?For example, consider the set of points (*x*, *y*) in the plane that satisfy the equation *Note:* In equation (4.3) the coefficients of *x*^{2} and *y*^{2} are both 1.In order to transform equation (4.3) into an equation of the form equation (4.1), we use a technique called ‘completing the square’. We rewrite the terms that involve only *x* and the terms that involve only *y*, as follows:*Note:* −2 is half the coefficient of *x* and +3 is half the coefficient of *y* in equation (4.3).Substituting the expressions above into equation (4.3), we obtain ((*x* − 2)^{2} − 4) + ((*y* + 3)^{2} − 9) + 9 = 0,that is, (*x* − 2)^{2} + (*y* + 3)^{2} = 4.*Note:* We can ‘read off’ the centre and radius of the circle from this equation.It follows that equation (4.3) represents a circle with centre (2, −3) and radius 2.In general, we can use the same method, ‘completing the square’, to rewrite the equation *x*^{2} + *y*^{2} + *fx* + *gy* + ^{h} = 0in the form from which we can ‘read off’ the centre and radius.*Note:* Again, the coefficients of *x*^{2} and *y*^{2} are both 1.The constant in the first bracket is half the coefficient of *x* in the previous equation, and the constant in the second bracket is half the coefficient of *y*.Theorem 3An equation of the form *x*^{2} + *y*^{2} + *fx* + *gy* + ^{h} = 0represents a circle with if and only if .*Remark:* It follows from equation (4.4) that if , then there are no points (*x*, *y*) that satisfy the equation *x*^{2} + *y*^{2} + *fx* + *gy* + *h* = 0; and if , then the given equation simply represents the single point .Example 52Determine the condition on the numbers *f*, *g* and *h* in the equation *x*^{2} + *y*^{2} + *fx* + *gy* + *h* = 0for the circle with this equation to pass through the origin.Since the origin lies on the circle, its coordinates (0, 0) must satisfy the equation of the circle. Thus 0^{2} + 0^{2} + *f* × 0 + *g* × 0 + *h* = 0,which reduces to the condition *h* = 0.Example 53Determine the centre and radius of each of the circles given by the following equations.**(a)** *x*^{2} + *y*^{2} − 2*x* − 6*y* + 1 = 0**(b)** 3*x*^{2} + 3*y*^{2} − 12*x* − 48*y* = 0**(a)** We can complete the square in the equation *x*^{2} + *y*^{2} − 2*x* − 6*y* + 1 = 0to obtain (*x* − 1)^{2} − 1 + (*y* − 3)^{2} − 9 + 1 = 0,or (*x* − 1)^{2} + (*y* − 3)^{2} = 9.So the circle has centre (1, 3) and radius √9 = 3.(Alternatively, use the general formula for centre and radius in Theorem 3 with *f* = −2, *g* = −6 and *h* = 1.)**(b)** Here the coefficients of *x*^{2} and *y*^{2} are both 3, so we divide the equation by 3 to obtain Completing the square gives (*x* − 2)^{2} − 4 + (*y* − 8)^{2} − 64 = 0,or (*x* − 2)^{2} + (*y* − 8)^{2} + 68.So the circle has centre (2, 8) and radius (Alternatively, apply the general formula for centre and radius to the equation in the form of equation (S.14), with *f* = −4, *g* = −16 and *h* = 0.)Example 54Determine the set of points (*x*, *y*) in ^{2} that satisfy each of the following equations.**(a)** *x*^{2} + *y*^{2} + *x* + *y* + 1 = 0**(b)** *x*^{2} + *y*^{2} − 2*x* + 4*y* + 5 = 0**(c)** 2*x*^{2} + 2*y*^{2} + *x* − 3*y* − 5 = 0**(a)** If we complete the square in the equation we obtain the equation This equation represents the empty set, since its left-hand side is always non-negative whereas its right-hand side is negative.(Alternatively, we can use information about the quantity . In equation (S.15), we have *f* = 1, *g* = 1 and *h* = 1. Thusso the set must be empty.)**(b)** If we complete the square in the equation we obtain the equation (*x* − 1)^{2} + (*y* + 2)^{2} = 0.Thus the set in the plane represented by equation (S.16) is the single point (1, −2).(Alternatively, we can use information about the quantity . In equation (S.16), we have *f* = −2, *g* = 4 and *h* = 5. Thusso the set is the single point **(c)** Here the coefficients of *x*^{2} and *y*^{2} are both 2, so we divide the equation by 2 to give If we complete the square in equation (S.17), we obtain the equation that is,Thus the set in the plane represented by equation (S.17) is a circle with centre and radius .(Alternatively, we can use information about the quantity . In equation (S.17), we have and . Thus so the set is the circle with centre and radius 4.3 Focus–directrix definitions of the non-degenerate conicsEarlier, we defined the conic sections as the curves of intersection of planes with a double cone. One of these conic sections, the circle, can be defined as the set of points a fixed distance from a fixed point.Here we define the other non-degenerate conics, the parabola, ellipse and hyperbola, as sets of points that satisfy a somewhat similar condition.These three *non-degenerate conics* (the parabola, ellipse and hyperbola) can be defined as the set of points *P* in the plane that satisfy the following condition: the distance of *P* from a fixed point is a constant multiple *e* of the distance of *P* from a fixed line. The fixed point is called the *focus* of the conic, the fixed line is called its *directrix*, and the constant multiple *e* is called its *eccentricity*.The different conics arise according to the value *e* of the eccentricity, as follows.EccentricityA non-degenerate conic isan ellipse if 0 ≤ *e* < 1,

a parabola if *e* = 1,

a hyperbola if *e* > 1.

*Note:* When *e* = 0, the ellipse is a circle; the focus is the centre of the circle, and the directrix is ‘at infinity’.4.4 Parabola (<i>e</i> = 1)A *parabola* is defined to be the set of points *P* in the plane whose distances from a fixed point *F* are equal to their distances from a fixed line *d*. We obtain a parabola *in standard form* ifthe focus *F* lies on the *x*-axis, and has coordinates (*a*, 0), where *a* > 0;the directrix *d* is the line with equation *x* = −*a*.Thus the origin lies on the parabola, since it is equidistant from *F* and *d*.Let *P* (*x*, *y*) be an arbitrary point on the parabola, and let *M* be the foot of the perpendicular from *P* to the directrix. Since *FP* = *PM*, by the definition of the parabola, it follows that *FP*^{2} = *PM*^{2}; that is,
(*x* − *a*)^{2} + (*y* − 0)^{2} = (*x* + *a*)^{2}.Multiplying out the brackets, we obtain
*x*^{2} − 2*ax* + *a*^{2} + *y*^{2} = *x*^{2} + 2*ax* + *a*^{2},
which simplifies to the equation
*y*^{2} = 4*ax*.
Each point with coordinates of the form (*at*^{2}, 2*at*), where *t* ∈ , lies on the parabola, since (2*at*)^{2} = 4*a* × *at*^{2}. Conversely, we can write the coordinates of each point on the parabola in the form (*at*^{2}, 2*at*). If we choose *t* = *y*/(2*a*), then *y* = 2*at* andas required. Thus there is a one-one correspondence between the real numbers *t* and the points of the parabola.We call the *x*-axis the *axis* of a parabola in standard form, since the parabola is symmetric with respect to this line, and we call the origin the *vertex* of a parabola in standard form, since it is the point of intersection of the axis with the parabola. A parabola has no centre.We summarise these facts as follows:Parabola in standard formA parabola in standard form has equationIt can also be described by the parametric equations It has focus (*a*, 0) and directrix *x* = −*a*; its axis is the *x*-axis and its vertex is the origin.Example 8Consider the parabola with equation *y*^{2} = 2*x* and parametric equations
, *y* = *t* (*t* ∈ ).**(a)** Write down the focus, vertex, axis and directrix of the parabola.**(b)** Determine the equation of the chord that joins the two distinct points *P* and *Q* with parameters *t*_{1} and *t*_{2}, respectively. Determine the condition on *t*_{1} and *t*_{2} such that the chord *PQ* passes through the focus of the parabola. Such a chord is called a *focal chord*.**(a)** The parabola is in standard form, where 4*a* = 2, that is, . It follows that its focus is , its vertex is (0, 0), its axis is the *x*-axis and the equation of its directrix is .**(b)** The coordinates of *P* and *Q* are and , respectively.
We must consider the cases and separately.We suppose first that . Then the gradient of *PQ* isSince lies on the line *PQ*, the equation of *PQ* isMultiplying both sides by *t*_{1} + *t*_{2}, we obtain so and hence If , then *t*_{1} = −*t*_{2} (as t_{1} ≠ *t*_{2}, since *P* and *Q* are distinct). When *t*_{1} = −*t*_{2}, *PQ* is parallel to the *y*-axis and the equation of *PQ* isthat is, 0 = 2*x* + *t*^{1}*t*^{2}.Thus equation (4.5) is the equation of *PQ* in the case *t*_{1} = −*t*_{2} also, for then *t*_{1} + *t*_{2} = 0.
Thus *PQ* passes through the focus if and only if satisfies equation (4.5), which is the case if and only if (*t*_{1} + *t*_{2})0 = 1 + *t*_{1}*t*_{2};that is, if and only if *t*_{1}*t*_{2} = −1.Example 55Consider the parabola with equation *y*^{2} = *x* and parametric equations
.**(a)** Write down the focus, vertex, axis and directrix.**(b)** Determine the equation of the chord that joins the two distinct points *P* and *Q* with parameters *t*_{1} and *t*_{2}, respectively.**(c)** Determine the condition on *t*_{1} and *t*_{2}, and so on *P* and *Q*, such that the focus of the parabola is the midpoint of the chord *PQ*.
**(a)** The parabola is in standard form, where 4*a* = 1, that is, . It follows that the focus is , the vertex is (0, 0), the axis is the *x*-axis, and the equation of the directrix is .**(b)** The coordinates of *P* and *Q* are and , respectively. It follows that if , then the gradient of *PQ* is Since lies on the line *PQ*, it follows that the equation of *PQ* is Multiplying both sides by 2(*t*_{1} + *t*_{2}), we obtain so and hence If , then *t*_{1} = −*t*_{2} (as *t*_{1} ≠ −*t*_{2}, since *P* and *Q* are distinct). If *t*_{1} = −*t*_{2}, then *PQ* is parallel to the *y*-axis and the equation of *PQ* isthat is,Thus equation (S.18) is the equation of *PQ* in the case *t*_{1} = −*t*_{2} also, for then *t*_{1} + *t*_{2} = 0.**(c)** The midpoint of *PQ* is This is the focus if and only if which is the case if and only if *t*_{2} = −*t*_{1} and , that is, and so *t*_{1} = ± 1. Thus the midpoint of *PQ* is the focus if and only if *P* and *Q* are the points and .*Remark:* The focus is the midpoint of *PQ* if and only if *PQ* is the chord through the focus parallel to the *y*-axis.4.5 Ellipse (0 < <i>e</i> < 1)An *ellipse* with eccentricity *e* (where 0 < *e* < 1) is the set of points *P* in the plane whose distances from a fixed point *F* are *e* times their distances from a fixed line *d*. We obtain such an ellipse *in standard form* ifthe focus *F* lies on the *x*-axis, and has coordinates (*ae*, 0), where *a* > 0;the directrix *d* is the line with equation *x* = *a*/*e*.
Let *P* (*x*, *y*) be an arbitrary point on the ellipse, and let *M* be the foot of the perpendicular from *P* to the directrix. Since *FP* = *e* × *PM*, by the definition of the ellipse, it follows that *FP*^{2} = *e*^{2} × *PM*^{2}; that is,Multiplying out the brackets, we obtain *x*^{2} − 2*aex* + *a*_{2}*e*_{2} + *y*_{2} = *e*_{2}*x*_{2} − 2*aex* + *a*_{2},which simplifies to the equation *x*_{2}(1 − *e*_{2}) + *y*_{2} = *a*_{2}(1 − *e*_{2}),that is,Substituting *b* for , so that *b*^{2} = *a*^{2}(1 − *e*^{2}), we obtain the standard form of the equation of the ellipseThis equation is symmetric in *x* and in *y*, so that the ellipse also has a second focus *F*′ at (−*ae*, 0), and a second directrix *d*′ with equation *x* = −*a*/*e*.The ellipse intersects the axes at the points (±*a*, 0) and (0, ±*b*). We call the line segment joining the points (±*a*, 0) the *major axis* of the ellipse, and the line segment joining the points (0, ±*b*) the *minor axis* of the ellipse. Since *b* < *a*, the minor axis is shorter than the major axis. The origin is the centre of this ellipse.*Note:* Since 0 < *e* < 1, we have 0 < *b* < *a*.
Each point with coordinates (*a* cos *t*, *b* sin *t*) lies on the ellipse, since Then, just as for the parabola, we can check that gives a parametric representation of the ellipse.An ellipse with eccentricity *e* = 0 is a circle. In this case, *a* = *b* and the circle *x*^{2} + *y*^{2} = *a*^{2} can be parametrised by We summarise these facts about ellipses (including circles) as follows:Ellipse in standard formAn ellipse in standard form has equationIt can also be described by the parametric equations If *e* > 0, it has foci (±*ae*, 0) and directrices *x* = ±*a*/*e*; its major axis is the line segment joining the points (±*a*, 0), and its minor axis is the line segment joining the points (0, ±*b*).If *e* = 0, the ellipse is a circle.Example 56Let *P* be a point , *t* ∈ , on the ellipse with equation *x*^{2} + 2*y*^{2} = 1.**(a)** Determine the foci *F* and *F*′ of the ellipse.**(b)** Determine the gradients of *FP* and *F*′*P*, when these lines are not parallel to the *y*-axis.**(c)** Find those points *P* on the ellipse for which *FP* is perpendicular to *F*′*P*.
**(a)** This ellipse is of the form with *a* = 1 and , so If *e* denotes the eccentricity of the ellipse, so that *b*^{2} = *a*^{2}(1 − *e*^{2}), then we haveit follows that , so .In the general case, the foci are (±*ae*, 0); it follows that here the foci are .**(b)** Let *F* and *F*′ be and , respectively. (It does not matter which way round these are chosen.)Then the gradient of *FP* is where we know that , since *FP* is not parallel to the *y*-axis.Similarly, the gradient of *F*′*P* is where we know that , since *F*′*P* is not parallel to the *y*-axis.**(c)** When *FP* is perpendicular to *F*′*P*, we have We may rewrite this in the form so 2 cos^{2}*t* − 1 + sin^{2}*t* = 0.Since cos^{2}*t* + sin^{2}*t* = 1, it follows that cos^{2}*t* = 0 and so cos*t* = 0. This occurs only when *t* = ±/2; that is, at the points .4.6 Hyperbola (<i>e</i> > 1)A *hyperbola* is the set of points *P* in the plane whose distances from a fixed point *F* are *e* times their distances from a fixed line *d*, where *e* > 1. We obtain a hyperbola *in standard form* ifthe focus *F* lies on the *x*-axis, and has coordinates (*ae*, 0), where *a* > 0;the directrix *d* is the line with equation *x* = *a*/*e*.
Let *P* (*x*, *y*) be an arbitrary point on the hyperbola, and let *M* be the foot of the perpendicular from *P* to the directrix. Since *FP* = *e* × *PM*, by the definition of the hyperbola, it follows that *FP*^{2} = *e*^{2} × *PM*^{2}; that is,Multiplying out the brackets, we obtain *x*^{2} − 2*aex* + *a*_{2}*e*_{2} + *y*_{2} = *e*_{2}*x*_{2} − 2*aex* + *a*_{2},which simplifies to the equation *x*_{2}(*e*_{2} − 1) − *y*_{2} = *a*_{2}(*e*_{2} − 1),that is,Substituting *b* for , so that *b*^{2} = *a*^{2}(*e*^{2} − 1), we obtain the standard form of the equation of the hyperbola This equation is symmetric in *x* and in *y*, so that the hyperbola also has a second focus *F*′ at (−*ae*, 0), and a second directrix *d*′ with equation *x* = −*a*/*e*.The hyperbola intersects the *x*-axis at the points (±*a*, 0). We call the line segment joining the points (±*a*, 0) the *major axis* or *transverse axis* of the hyperbola, and the line segment joining the points (0, ±*b*) the *minor axis* or *conjugate axis* of the hyperbola (this is not a chord of the hyperbola). The origin is the centre of this hyperbola.
Each point with coordinates (*a* sec *t*, *b* tan *t*) lies on the hyperbola, since *Note: * In general, sec^{2}*t* = 1 + tan^{2}*t*.Then, just as for the parabola, we can check that gives a parametric representation of the hyperbola.*Note:* An alternative parametrisation, using hyperbolic functions, is *x* = *a*cosh*t*, y = *b*sinh*t* (*t* ∈ ).Two other features of the shape of the hyperbola stand out.First, the hyperbola consists of two separate curves or *branches*.Secondly, the lines with equations *y* = ±*bx*/*a* divide the plane into two pairs of opposite sectors; the branches of the hyperbola lie in one pair. As *x* → ±∞, the branches of the hyperbola get closer and closer to these two lines. We call the lines *y* = ±*bx*/*a* the *asymptotes* of the hyperbola.
We summarise these facts as follows:Hyperbola in standard formA hyperbola in standard form has equationIt can also be described by the parametric equationsIt has foci (±*ae*, 0) and directrices *x* = ±*a*/*e*; its major axis is the line segment joining the points (±*a*, 0), and its minor axis is the line segment joining the points (0, ±*b*).Example 57Let *P* be a point , *t* ∈ , on the hyperbola with equation *x*^{2} − 2*y*^{2} = 1.**(a)** Determine the foci *F* and *F*′ of the hyperbola.**(b)** Determine the gradients of *FP* and *F*′*P*, when these lines are not parallel to the *y*-axis.**(c)** Find the point *P* on the hyperbola, in the first quadrant, for which *FP* is perpendicular to *F*′*P*.
**(a)** This hyperbola is of the form with *a* = 1 and , so . If *e* denotes the eccentricity of the hyperbola, so that *b*^{2} = *a*^{2}(*e*^{2} − 1), we haveit follows that , so In the general case, the foci are (±*ae*, 0); it follows that here the foci are .**(b)** Let *F* and *F*′ be and , respectively. (It does not matter which way round these are chosen.)Then the gradient of *FP* is where we know that , since *FP* is not parallel to the *y*-axis.Similarly, the gradient of *F*′*P* is where we know that , since *F*′*P* is not parallel to the *y*-axis.**(c)** When *FP* is perpendicular to *F*′*P*, we have We may rewrite this in the form so 2sec^{2}*t* − 3 + tan^{2}*t* = 0.Since sec^{2}*t* = 1 + tan^{2}*t*, it follows that 3 tan^{2}*t* = 1.Since we are looking for a point *P* in the first quadrant, we choose .When , we have . Since we are looking for a point *P* in the first quadrant, we choose .It follows that the required point *P* has coordinates .4.7 Rectangular hyperbola (<i>e</i> = √2)If the eccentricity *e* of a hyperbola is equal to √2, then *e*^{2} = 2 and *b* = *a*. Then the asymptotes of the hyperbola have equations *y* = ±*x*, so they are at right angles. A hyperbola whose asymptotes are at right angles is called a *rectangular hyperbola*.
Then, if we use the asymptotes as new *x*- and *y*-axes (instead of the original *x*- and *y*-axes), the equation of the hyperbola can be written in the form *xy* = *c*^{2}, for some positive number *c*. (We omit the details.)The rectangular hyperbola with equation *xy* = *c*^{2} has the origin as its centre, and the *x*- and *y*-axes as its asymptotes. It can be described by the parametric equations
4.8 General equation of a conicYou have already met the parabola, ellipse and hyperbola. So far, you have considered the equation of a conic only when it is in standard form; that is, when the centre of the conic (if it has a centre) is at the origin, and the axes of the conic are parallel to the *x*- and *y*-axes. However, most of the conics that arise in calculations are not in standard form.We have seen that any circle can be described by an equation of the formMore generally, it can be shown that any conic can be described by an equation of the form where *A*, *B* and *C* are not all zero.Equation (4.6) is of the form (4.7) with *A* = *C* = 1 and *B* = 0.We know that the equations of the non-degenerate conics in standard form are Each of these equations is of the form (4.7).The equations of degenerate conics can also be expressed in the form (4.7). For example:However, an equation of the form (4.7) can also describe the empty set; an example of this is the equation *x*^{2} + *y*^{2} + 1 = 0, as there is no point (*x*, *y*) in ^{2} for which *x*^{2} + *y*^{2} = −1. (This is an unexpected possibility!) For simplicity in the statement of the theorem below, therefore, we add the empty set to our existing list of degenerate conics.Theorem 4Any conic has an equation of the form where *A*, *B*, *C*, *F*, *G* and *H* are real numbers, and *A*, *B* and *C* are not all zero. Conversely, the set of all points in ^{2} whose coordinates (*x*, *y*) satisfy an equation of the form (4.7) is a conic.*Note:* We omit a proof of this result.4.9 Further exercisesExample 58Determine the equation of the circle with centre (2, 1) and radius 3.The equation of the circle is (*x* − 2)^{2} + (*y* − 1)^{2} = 3^{2},which can be rewritten in the form *x*^{2} + *y*^{2} − 4*x* − 2*y* − 4 = 0.Example 59Determine the points of intersection of the line with equation *y* = *x* + 2 and the circle in Exercise 58.
At the points of intersection of the line and the circle, the following equations must be satisfied: *y* = *x* + 2and *x*^{2} + *y*^{2} − 4*x* − 2*y* − 4 = 0.Substituting the expression for *y* from the first equation into the second equation, we obtain *x*^{2} + (*x* + 2)^{2} − 4*x* − 2(*x* + 2) − 4 = 0.Multiplying this out and collecting terms, we obtain 2*x*^{2} − 2*x* − 4 = 0,or *x*^{2} − *x* − 2 = 0.This can be factorised as (*x* − 2)(*x* + 1) = 0,so *x* = 2 or *x* = −1.Substituting these values of *x* into the equation *y* = *x* + 2 of the line, we find that the corresponding values of *y* are *y* = 4 and *y* = 1.Hence the points of intersection of the line and the circle are (2, 4) and (−1, 1).Example 60This question concerns the parabola *y*^{2} = 4*ax* (*a* > 0) with parametric equations *x* = *at*^{2}, *y* = 2*at* and focus *F*. Let *P* and *Q* be points on the parabola with parameters *t*_{1} and *t*_{2}, respectively. Prove that:**(a)** if *PQ* subtends a right angle at the vertex *O* of the parabola, then *t*_{1}*t*_{2} = −4;
**(b)** if *t*_{1} = 2 and *PQ* is perpendicular to *FP*, then
.**(a)** The gradient of *OP* isand the gradient of *OQ* is *OP* and *OQ* are perpendicular (in other words, *PQ* subtends a right angle at *O*) if *m*_{1}*m*_{2} = −1, that is, if which reduces to *t*_{1}*t*_{2} = −4, as required.**(b)** Here *P* is the point (4*a*, 4*a*). Also, *F* is (*a*, 0), so the gradient of *FP* is and the gradient of *PQ* is *FP* and *PQ* are perpendicular if *m*_{1}*m*_{2} = −1, that is, if which reduces to , so , as required.Example 61This question concerns the rectangular hyperbola *xy* = *c*^{2} (*c* > 0) with parametric equations *x* = *ct*, *y* = *c*/*t*. Let *P* and *Q* be points on the hyperbola with parameters *t*_{1} and *t*_{2}, respectively.**(a)** Determine the equation of the chord *PQ*.
**(b)** Determine the coordinates of the point *N* where *PQ* meets the *x*-axis.**(c)** Determine the midpoint *M* of *PQ*.**(d)** Prove that *OM* = *MN*, where *O* is the origin.**(a)** The gradient of the chord *PQ* isso the equation of *PQ* is that is,**(b)** At the point *N* where *PQ* meets the *x*-axis, *y* = 0 and equation (S.19) also holds, so *x* = *c*(*t*_{1} + *t*_{2}). Hence *N* is the point with coordinates (*c*(*t*_{1} + *t*_{2}), 0).**(c)** The midpoint *M* of *PQ* is **(d)** By the Distance Formula, the distance *OM* is Also,Thus *OM* = *MN*.*Remark:* An alternative way of proving part (d) is as follows: Since the point *M* is vertically above the midpoint of *ON*, the triangle *OMN* must be an isosceles triangle, so *OM* = *MN*.ConclusionThis free course provided an introduction to studying Mathematics and Statistics. It took you through a series of exercises designed to develop your approach to study and learning at a distance and helped to improve your confidence as an independent learner.Keep on learning Study another free courseThere are more than **800 courses on OpenLearn** for you to choose from on a range of subjects. Find out more about all our free courses. Take your studies furtherFind out more about studying with The Open University by visiting our online prospectus. If you are new to university study, you may be interested in our Access Courses or Certificates. What’s new from OpenLearn?
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Except for third party materials and otherwise stated (see terms and conditions). This content is made available under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 LicenceCourse image: Alpha du centaure in Flickr made available under Creative Commons Attribution 2.0 Licence.The audio extracts are taken from M208 *Pure Mathematics*.**Don't miss out:**If reading this text has inspired you to learn more, you may be interested in joining the millions of people who discover our free learning resources and qualifications by visiting The Open University - www.open.edu/openlearn/free-courses