The point of intersection of the plane 3 x - 5 y + 2 z = 6 with the straight line passing through the origin and perpendicular to the plane 2 x - y - z = 4 is

Question

The point of intersection of the plane 3 x - 5 y + 2 z = 6 with the straight line passing through the origin and perpendicular to the plane 2 x - y - z = 4 is, 1 , - 1 , - 1, - 1 , - 1,2, 4,2 , 2, 4 3 , - 2 3 , - 2 3

MARKS App · Accepted Answer

A vector parallel to line is 2 i ^ - j ^ - k ^ So equation of line is x - 0 2 = y - 0 - 1 = z - 0 - 1 Any point on the line is 2 λ , - λ , - λ It must satisfy the given plane 6 λ + 5 λ - 2 λ = 6 9 λ = 6 ⇒ λ = 2 3 Point is 4 3 , - 2 3 , - 2 3

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