Find the equation of the plane through the line of intersection of the planes 2x – y = 0 and 3z – y = 0 which is perpendicular to the plane 4x + 5y – 3z = 8 ?
we know that, equation of a plane passing through the line of intersection of two planes
a1x + b1y + c1z + d1 = 0 and a2x + b2y + c2z + d2 = 0 is given by
(a1x + b1y + c1z + d1) + k(a2x + b2y + c2z + d2) = 0
So equation of plane passing through the line of intersection of given two planes
2x – y = 0 and 3z – y = 0 is
(2x – y) + k(3z – y) = 0
2x – y + 3kz – ky = 0
x(2) + y( – 1 – k) + z(3k) = 0 …… (1)
we know that, two planes are perpendicular if
a1a2 + b1b2 + c1c2 = 0 …… (2)
given, plane (1) is perpendicular to plane
4x + 5y – 3z = 8 …… (3)
Using (1) and (3) in equation (2)
(2)(4) + ( – 1 – k)(5) + (3k)( – 3) = 0
8 – 5 – 5k – 9k = 0
3 – 14k = 0
– 14k = – 3
put the value of k in equation (1)
x(2) + y( – 1 – k) + z(3k) = 0
x(2) + y( – 1 – 
) + z(3
) = 0
x(2) + y(
) + z(
) = 0
x(2) + y(
) + z(
) = 0
multiplying with 14 we get
28x – 17y + 9z = 0
Equation of required plane is, 28x – 17y + 9z = 0