# Math Help - Normal to Plane

1. ## 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?

2. 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 $\pi$ which prep. to the vector $$ and contains $$ is:
$\pi: a(x-x_o)+b(y-y_o)+c(z-z_o)=0$

$(x-3)+2(y-2)+3(z-1)=0$
$x+2y+3z=10$.