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 - 9 = 0?
Here, we have to find the equation of plane passing through the line of intersection of planes 2x - y = 0 and 3z - y = 0 which is perpendicular to plane 4x + 5y - 3z - 9 = 0 As we know that, equation of a plane passing through the intersection of these planes is given by: ((a1x+b1y+c1z+d)+λ(a2x+b2y+c2z+d)=0,where λ is a scalar ⇒(2x−y)+λ(3z−y)=0 ⇒2x+(−1−λ)y+3λz=0 ........(1) The direction ratios of the plane represented by (1) are: 2, (- 1 - λ), 3λ The direction ratios of the plane 4x + 5y - 3z - 9 = 0 are: 4, 5, - 3 ∵ plane represented by (1) is perpendicular to plane 4x + 5y - 3z - 9 = 0 ⇒ 8 - 5 - 5λ - 9λ = 0 ⇒ 3 - 14λ = 0 ⇒ λ = 3/14 By substituting λ = 3/14 in equation (1), we get ⇒ 28x - 17y + 9z = 0 Hence, equation of the required plane is 28x - 17y + 9z = 0