Question

The plane 4x+7y+4z+81=0 is rotated through a right angle about its line of intersection with the plane 5x+3y+10z=25. The equation of the plane in its new position

Answer 1

hey mate.
here's the solution

Answer 2

The plane 4x+7y+4z+81=0 is rotated through a right angle about its line of intersection with the plane 5x+3y+10z=25. The equation of the plane in its new position is x−4y+6z=106

• The equation of the plane through the line of intersection of the planes

                 4x+ 7y + 4z + 81 = 0 and(5x+ 3y + 10z) = 25

• If (4x + 7y + 4z + 81) + λ(5+ 3y + 10z— 25) = 0

          (4+5λ)x + (7 + 3λ)y + (4 + 10λ)z + 81 - 25λ = 0

Which is parallel to

       x - 4y + 6z =k

       $\frac{4+5λ}{1} =\frac{7+3λ}{-4} =\frac{4+10λ}{6}$

Then,

•     We get λ = -1

Then from Eq. (i)

                  -x+ 4y - 6z + 106=0

                     x - 4y + 6z = 106

Therefore, the equation of the plane in its new position is x−4y+6z=106