3.4 Further exercises
Example 47

(a) Find the angle between each of the pairs of vectors:
(3, 1) and (1, −2); i + 2j and −3i + j − 2k.
(b) Determine the projection of (3, 1) onto (1, −2), and the projection of i + 2j onto −3i + j − 2k.
Answer
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 + 2j and q = −3i + j − 2k, we have
(b) When p = (3, 1) and q = (1, −2), the projection of p onto q is
When p = i + 2j and q = −3i + j − 2k, the projection of p onto q is
Example 48
Determine 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.
Answer
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
so
that 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} + 4y + 4 + y^{2} = 4,
that is,
2y^{2} + 4y = 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 49
Determine the equation of the plane that contains the point (−1, 3, 2) with (1, 2, −1) as a normal.
Answer
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 + 2y − z = (−1) × 1 + 3 = 2 + 2 = (−1),
that is,
x + 2y − z = 3.
Example 50
Determine the equation of the plane through (1, 0, 2), (0, 3, 4) and (0, −1, 0).
Answer
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 obtain
Substituting −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 − 2b = −b,
that is,
a = b.
It follows that the equation of the plane is
bx + by − bz = −b,
that is,
x + y − z = −1.