Find the equation of the plane through the center of...

Question

Let Line L be the line of intersection of the planes x-2y+z=0 and 3x+y-2z=9.

Find the equation of the plane through the center of the sphere x² -8x+y²-2y+z²=8 and perpendicular to line L.

Answer 1

A normal vector for the first plane is (1, -2 , 1)

A normal vector for the second plane is (3, 1, -2)

The cross product of these two is ((3,5,7) This is a normal vector for the desired plane.

Thus the equation of the desired plane is 3x + 5y + 7 z = Q for some number Q

By completing the square, the center of the circle is found to be the point (4,2,0)

Substituting this point into the equation of the desired plane shows that Q = 22

So the equation of the desired plane is

3x + 5y + 7z= 22