Multivariate Calculus - Bruce E. Shapiro

28 LECTURE 3. THE CROSS PRODUCT
Example 3.6 Find the equation of the plane through (2, −3, 2) and parallel to the
plane containing the vectors v = 4i + 3j − k, w = 2i − 5j + 6k.
Solution.The normal to the desired plane will also be perpendicular to the plane
containing v and w. Since the cross product of any two vectors is by definition
perpendicular to the plane containing both vectors, then one such normal vector is
i j k
n = v × w =
4 3 −1
∣2 −5 6 ∣
= i
∣ 3 −1
∣ ∣ ∣∣∣ −5 6 ∣ − j 4 −1
∣∣∣ 2 6 ∣ + k 4 3
2 −5∣
= 13i − 26j − 26k
Since a point on the plane is (2,-3,2), the equation of the plane is
n · (x, y, z) = n · P
(13, −26, −26) · (x, y, z) = (13, −26, −26) · (2, −3, 2)
13x − 26y − 26z = 52
Dividing the last equation through by 13 gives x − 2y − 2z = 4 as the equation of
the desired plane.
Revised December 6, 2006. Math 250, Fall 2006