Q.242. Equation of a plane

Question 242.
Find the equation of the plane which passses through the point A (4,2,1) and is perpendicular to the planes with equation 5x-2y+6z+1 = 0 and 2x-y-z = 4.

Answer 242.
The normal to the plane 5x-2y+6z+1=0 is
n1 = (5, -2, 6).
The normal to the plane 2x-y-z=4 is
(2, -1, -1).

As the required plane is perpendicular to both these planes, the normal to the required plane is
n1 x n2
= (5, -2, 6) x (2, -1, -1)
= (8, 17, -1)
 => equation of the plane is of the form
8x + 17y - z = k
As it passes through the point A(4, 2, 1),
8(4) + 17(2) - (1) = k => k = 65
=> eqn. of the reqd. plane is
8x + 17y - z = 65.

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