Thread: Normal to Plane

Find the equation of the plane in R^3 that is perpendicular to (1,2,3) and contains the point (3,2,1).

So the normal to the plane is (1,2,3)

So: x+2y+3z=d

Sub the point in and solve for d.

(3) + 2(2) + 3(1) + d = 0
10 + d = 0
d = -10

So the equation is
x + 2y + 3z -10 = 0

Is this the correct method?

Originally Posted by millerst

Find the equation of the plane in R^3 that is perpendicular to (1,2,3) and contains the point (3,2,1).

So the normal to the plane is (1,2,3)

So: x+2y+3z=d

Sub the point in and solve for d.

(3) + 2(2) + 3(1) + d = 0
10 + d = 0
d = -10

So the equation is
x + 2y + 3z -10 = 0

Is this the correct method?

The equation of the plane which prep. to the vector and contains is:


Your plane:

or:
.
As you said.