Question

The equation of the plane through the line 2x-y=0=y-3z and perpendicular to 4x+5y-3z=8 is:

Answer 1

Let plane eqn. be (2x-y) + a(y-3z) = 0 ; where a = parameter to be found.

So plane eqn is 2x + (a-1)y -3az = 0.

As this plane is perpendicular to plane 4x+5y-3z=8 , their normals are also perpendicular.

So 2(4) + (a-1)(5) + (-3a)(-3) = 0 , ie , a = -3/14.

Substitute value of 'a' in plane eqn.

We get eqn. 2x - 17y/14 + 9z/14 = 0.

ie, Plane eqn will be 28x -17y + 9z = 0.
keshav yaduvanshi
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