Multivariate Calculus - Bruce E. Shapiro

26 LECTURE 3. THE CROSS PRODUCT
Example 3.4 Let a = 〈3, 3, 1〉 , b = 〈−2, −1, 0〉 , c = 〈−2, −3, −1〉. Find a×(b×c)
Solution. By property (11) of theorem 3.2,
a × (b × c) = b(a · c) − c(a · b)
= (−2, −1, 0) ((3, 3, 1) · (−2, −3, −1)) − (−2, −3, −1) ((3, 3, 1) · (−2, −1, 0))
= −16(−2, −1, 0) + 9(−2, −3, −1)
= (32, 16, 0) + (−18, −27, −9)
= (14, −11, −9)
Finding the equation of a plane through three points P, Q, R
Suppose we are given the coordinates of three (non-collinear) points, P, Q, and R,
and want to find the equation of the plain to contains all three points. The following
procedure will give you this equation.
1. Use the points to define two vectors
−−→
PQ = Q − P
−−→
QR = R − Q
It does not matter which pair you use, as long as you take any two non-parallel
vectors from among the six possible vectors −−→ PQ, −−→ QR, −−→ RP, −−→ QP, −−→ RQ and −−→ PR.
(Note that if you chose −−→ PQ as your first vector, the second can be any of the
other vectors except for −−→ QP.
2. The vectors you chose in step (1) define the plane. Both vectors lie in the
plane that contains the three points. Calculate the cross product
n = −−→ PQ × −−→ QR
Since n is perpendicular to both vectors, it must be normal to the plane they
are contained in.
3. Pick any one of the three points P, Q, R, say P = (p x , p y , p z ), and let X =
(x, y, z) be any point in the plane. Then the vector
lies in the plane.
−−→
PX = X − P = (x − p x , y − p y , z − p z )
4. Since −−→ PX lies in the plane, and the vector n is normal to the plane, they must
be perpendicular to one another, i.e.,
n · −−→ PX = 0
Revised December 6, 2006. Math 250, Fall 2006