Q. 105.0( 2 Votes )

Find the equation of the plane through the line of intersection of the planes and , which is at a unit distance from the origin?

Answer :

and


x + 3y + 6 = 0 and 3x – y – 4z = 0


x + 3y + 6 + k(3x – y – 4z) = 0


x(1 + 3k) + y(3 – k) – 4zk + 6 = 0


Distance from origin to plane =


36 = (1 + 3k)2 + (3 – k)2 + (4k)2


36 = 1 + 6k + 9k2 + 9 – 6k + k2 + 16k2


26 = 26k2


k2 = 1


k = 1


case :1 k = 1


x + 3y + 6 + 1(3x – y – 4z) = 0


4x + 2y – 4z + 6 = 0


Case :2 k = – 1


x + 3y + 6 – 1(3x – y – 4z) = 0


2x – 4y – 4z – 6 = 0


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