Multivariate Calculus - Bruce E. Shapiro

32 LECTURE 4. LINES AND CURVES IN 3D
Since t must be the same in all three equations, we obtain the symmetric equations
of a line,
x − x 0
= y − y 0
= z − z 0
(4.13)
v x v y v z
Example 4.2 Find the symmetric equations of the line we found in example 4.1.
Solution. In the earlier example we calculated that
v = (v x , v y , v z ) = (5, −1, −2)
is parallel to the line, and that the line goes through the point
Hence
(x 0 , y 0 , z 0 ) = (2, −1, 5)
x 0 = 2, y 0 = −1, z 0 = 5
v x = 5, v y = −1, v z = −2
and thus the symmetric equations of the line are
x − 2
5
= y + 1
−1 = z − 5
−2
Example 4.3 Find the equation of the plane that contains the two lines
x = −2 + 2t, y = 1 + 4t, z = 2 + t
and
x = 2 + 4t, y = 3 + 2t, z = 1 − t
Solution. The equation of a plane through a point P that has a normal vector n is
n · (x, y, z) = P · n
Thus we need to find (a) a point on the plane; and (b) a vector that is perpendicular
to the plane.
One way to find the normal vector is to find two non-parallel vectors in the
plane and take their cross product. But we can read off two such vectors from the
equation of the line:
u = (2, 4, 1)
v = (4, 2, −1)
Hence a normal vectors is
⎛
⎞ ⎛ ⎞ ⎛ ⎞
0 −1 4 4 −6
n = u × v = ⎝ 1 0 −2⎠
⎝ 2 ⎠ = ⎝ 6 ⎠
−4 2 0 −1 −12
Revised December 6, 2006. Math 250, Fall 2006