If the line x - 2 3 = y + 1 2 = z - 1 - 1 intersects the plane 2 x + 3 y - z + 13 = 0 at a point P and the plane 3 x + y + 4 z = 16 at a point Q , then P Q is equal to

Question

If the line x - 2 3 = y + 1 2 = z - 1 - 1 intersects the plane 2 x + 3 y - z + 13 = 0 at a point P and the plane 3 x + y + 4 z = 16 at a point Q , then P Q is equal to, 2 7, 14, 2 14, 14

MARKS App · Accepted Answer

x - 2 3 = y + 1 2 = z - 1 - 1 = λ x = 3 λ + 2 , y = 2 λ - 1 , z = - λ + 1 Intersection with plane 2 x + 3 y - z + 13 = 0 Hence, 2 3 λ + 2 + 3 2 λ - 1 - - λ + 1 + 13 = 0 13 λ + 13 = 0 λ = - 1 ∴ P - 1 , - 3,2 Intersection with plane 3 x + y + 4 z = 16 Hence, 3 3 λ + 2 + 2 λ - 1 + 4 - λ + 1 = 16 λ = 1 Q 5,1 , 0 P Q = 6 2 + 4 2 + 2 2 = 56 = 2 14

Search any question & find its solution

Looking for more such questions to practice?