Q.396. Equation of a plane through a line and perpendicular to a plane

Question 396.
Find the equation of a plane passing through the line
2x-y= 0 = 3z-y and perpendicular to the plane 4x+5y+3z = 8.

Answer 396.
Eqns. of all the planes through the line of intersection of the planes
2x - y = 0 and 3z - y = 0 is
2x - y + k (y - 3z) = 0
=> 2x + (k - 1)y - 3kz = 0
 Let k = k' represent a plane which is perpendicular to the given plane 4x + 5y + 3z = 8
Its normal is n1 = [2, (k'-1), - 3k']
Normal of the given plane 4x + 5y + 3z = 8 is n2 = (4, 5, 3)
=> n1 . n2 = 0
=> 2*4 + (k' - 1)*5 - (3k')*3 = 0
=> 8 + 5k' - 5 - 9k' = 0
=> k' = 3/4
 => eqn. of the reqd. plane is
2x + (3/4 - 1)y - 9z/4 = 0
=> 8x - y - 9z = 0.

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