Calc 3: Find the equation of the plane given 2 points on the plane: (3,4,1) and (3,1,-7)

[daniel_calc3]

Dear Readers,

I have a question lifted from an online video series. It goes like this.

Find the equation of the plane given 2 points on the plane: (3,4,1) and (3,1,-7).
The plane is perpendicular to: 8x + 9y + 3z = 13

The official answer given is
-63(x-3) + 64(y-4) -24(z-1) = 0

Can I say that the normal vector to the plane is <8,9,3>? shouldn't the answer be one of these:
8(x-3)+9(y-4)+3(z-1) = 0 or
8(x-3)+9(y-1)+3(z+7) = 0 ?

Thank you so much.

 

[blamocur]

Can I say that the normal vector to the plane is <8,9,3>? shouldn't the answer be one of these:

No. This vector is normal to the perpendicular plane.
It might be easier to discuss the problem if we start naming the planes. For example, your plane P contains points (3,4,1) and (3,1,-7) and is perpendicular to plane Q. The equation for plane Q is 8x + 9y + 3z = 13. You are asked to find the equation for plane P.
Hint: a vector normal to Q is parallel to P.

 

[daniel_calc3]

Thank you.

No. This vector is normal to the perpendicular plane. It might be easier to discuss the problem if we start naming the planes. For example, your plane P contains points (3,4,1) and (3,1,-7) and is perpendicular to plane Q. The equation for plane Q is 8x + 9y + 3z = 13. You are asked to find the equation for plane P. Hint: a vector normal to Q is parallel to P.