Multivariate Calculus - Bruce E. Shapiro

LECTURE 3. THE CROSS PRODUCT 27
5. In fact, every point X in the plane can be represented by some vector −−→ PX
pointing from P to X, and hence we can say the plane is the locus of all points
satisfying n · −−→ PX = 0 and hence
n · (x − p x , y − p y , z − p z ) = 0
or equivalently,
since P = (p x , p y , p z ).
n · (x, y, z) = n · P
It is usually easier to reconstruct this procedure each time than it is to memorize
a set of formulas, as in the following example.
Example 3.5 Find the equation of the plane that contains the three points P =
(1, 3, 0), Q = (3, 4, −3) and R = (3, 6, 2).
Solution. We start by finding two vectors in the plane, for example,
−−→
QP = P − Q = (1, 3, 0) − (3, 4, −3) = (−2, −1, 3)
−−→
QR = R − Q = (3, 6, 2) − (3, 4, −3) = (0, 2, 5)
Then a normal vector to the plane is given by
n = −−→ QP × −−→ QR
= (−2, −1, −3) × (0, 2, 5)
i j k
=
−2 −1 −3
∣ 0 2 5 ∣
= i
∣ 3 −1
∣ ∣ ∣∣∣ −5 6 ∣ − j −4 −1
∣∣∣ 2 6 ∣ + k 4 3
2 −5∣
= −11i + 10j − 4k
= (−11, 10, −4)
Then for any point (x, y, z) in the plane, the vector from R (where R was chosen
completely randomly from the three points) to (x, y, z) is given by
v = (x − 3, y − 6, z − 2)
Since v and n must be perpendicular, then their dot product is zero:
0 = n · v = (−11, 10, −4) · (x − 3, y − 6, z − 2)
= −11(x − 3) + 10(y − 6) − 4(z − 2)
= −11x + 33 + 10y − 60 − 4z + 8
= −11x + 10y − 4z − 19
Hence the equation of the plane that contains the three points is
11x − 10y + 4z = −19
Math 250, Fall 2006 Revised December 6, 2006.