Multivariate Calculus - Bruce E. Shapiro

LECTURE 2. VECTORS IN 3D 19
The Equation of a Plane
To find the equation of a plane through a point P = (x 1 , y 1 , z 1 ) that is perpendicular
to some vector n = (A, B, C),
1. Let Q = (x, y, z) be any other point in the plane.
2. Let v be a vector in the plane pointing from P to Q:
v = Q − P = (x − x 1 , y − y 1 , z − z 1 )
3. Since v must be perpendicular to n, the dot product v · n = 0, hence
4. Rearrange to give
A(x − x 1 ) + B(y − y 1 ) + (z − z 1 ) = 0
Ax + By + Cz = Ax 1 + By 1 + Cz 1 = n · P
Example 2.9 Find the equation of a plane passing through (5, 1, -7) that is perpendicular
to the vector (2,1,5)
Solution By the construction described above, the equation is
2x + y + 5z = (2)(5) + (1)(1) + (5)(−7) = −24
Example 2.10 Find the angle between the planes
3x − 4y + 7z = 5
and
2x − 3y + 4z = 0
Solution. We find the normal vectors to the planes by reading off the coefficients of
x, y, and z; the normal vectors are n = (3, −4, 7) and m = (2, −3, 4). Their angle
of intersection θ is the same as the angle between the planes. To get this angle, we
take the dot product, since m · n = ‖m‖n‖ cos θ. We calculate that
‖m‖ = √ 29
‖n‖ = √ 74
m · n = (2)(3) + (−3)(−4) + (4)(7) = 6 + 12 + 28 = 46
cos θ =
m · n
‖m‖‖n‖ = 46
(29)(74) = 46
2146 ≈ 0.0214
θ = arccos .214 ≈ 88.8 deg
Example 2.11 Find the equation of a plane through (−1, 2, −3) and parallel to the
plane 2x + 4y − z = 6.
Math 250, Fall 2006 Revised December 6, 2006.