Multivariate Calculus - Bruce E. Shapiro

LECTURE 4. LINES AND CURVES IN 3D 33
Note: alternatively, we could have calculated
i j k
n = u × v =
2 4 1
= −6i + 6j − 12k
∣4 2 −1∣ We also need to find a point on the plane. We do this by picking any time, say
t = 0, and substituting that into either of the lines. The first line, for example, gives
P = (−2, 1, 2) at t = 0. The equation of the plane through P with normal vector n
is then
(x, y, z) · (−6, 6, −12) = (−2, 1, 2) · (−6, 6, −12)
or
−6x + 6y + 12z = 12 + 6 − 24 = −6
Suppose we had picked a different point P, say, for example, the second line at t = 2,
P = (10, 7, −1). Then we would calculate
(x, y, z) · (−6, 6, −12) = (−6, 6, −12) · (10, 7, −1)
−6x + 6y + 12z = −60 + 42 + 12 = −6
In other words we get the same equation regardless of which point we pick
on the plane.
Example 4.4 Find the equation of the plane containing the line
x = 1 + 2t, y = −1 + 3t, z = 4 + t
and the point P = (1, −1, 5) (text section 14.4, exercise 19).
Solution. We already know one point on the plane; we also need a normal vector
to get the equation of the plane. We can get a second point on the plane from the
equation of a line; at t = 0 we have
The vector
Q = (x(0), y(0), z(0)) = (1, −1, 4)
u = −−→ QP = P − Q = (1, −1, 5) − (1, −1, 4) = (0, 0, 1)
We can get a second vector that lies in the plane from the equation of the line,
v = (2, 3, 1) which is the velocity vector of the line. Hence a normal vector to the
plane is
⎛ ⎞ ⎛ ⎞ ⎛ ⎞
0 −1 0 2 −3
u = v × v = (0, 0, 1) × (2, 3, 1) = ⎝1 0 0⎠
⎝3⎠ = ⎝ 2 ⎠
0 0 0 1 0
The equation of the plane is
(x, y, z) · (−3, 2, 0) = (1, −1, 5) · (−3, 2, 0)
−3x + 2y = −5.
Math 250, Fall 2006 Revised December 6, 2006.