Find an equation of the plane with the given characteristics. The plane passes through the points (4, 2, 1) and (4, 1, -6) and is perpendicular to the plane 8x + 9y+ 3z = 14.

Question

my answer was 57x-56y+8z-124=0 but it was wrong

Answer 1

To find the equation of a plane, you need to know its normal vector and a point on the plane.

We are given that the plane is perpendicular to the plane 8x + 9y + 3z = 14. So the normal vector of our plane is parallel to the normal vector of the given plane, which is <8,9,3>.

To find a point on the plane, we can use either of the given points. Let's use (4,2,1):

So now we have the normal vector and a point on the plane, and we can use the point-normal form of the equation of a plane:

<8,9,3> dot (x-4,y-2,z-1) = 0

Expanding the dot product:

8(x-4) + 9(y-2) + 3(z-1) = 0

Simplifying:

8x + 9y + 3z = 47

So the equation of the plane is 8x + 9y + 3z = 47.

Answer 2

AAAaannndd the bot gets it wrong yet again!

surely it is obvious that
8x + 9y + 3z = 47
is not perpendicular to
8x + 9y+ 3z = 14