 Vectors and conics

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2.7 Further exercises

Example 31

Let p = 2i − 3j + k and q = −i −2j −4k be two vectors in 3. Determine p + q, pq and 2p − 3q.

Since p = 2i − 3j + k and q = −i −2j −4k, we have and Example 32

Let u and v be the position vectors (1, 1) and (2, 1), respectively.

• (a) Determine the position vectors u + 2v, −u, −u + 3v and u − 3v.

• (b) On a single diagram, sketch u, v and the position vectors that you found in part (a).

• (a) Here,

• • and

• • (b)

• Example 33

Let 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 write

• • Thus

• • Then, 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 give

• • Thus we can write

• • (b)

• Example 34

Let 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 is Hence 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 is

• • that is,

• • where λ ∈ .

• (b) The point (7, 2) lies on the line ℓ if there is a value of λ such that

• • Equating 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 that

• • Equating components in turn, we obtain

• • The 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 36

The 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 as

• • Now b = a + c by the Parallelogram Law, since OB is the diagonal of the parallelogram with adjacent sides a and c. Thus

• • Next, 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 is

• • Multiplying 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 form

• • But 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 that

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