If the distance of point P 3,2 , 6 from the line x - 1 2 = y - 2 3 = z - 3 4 measured parallel to the plane 3 x - 5 y + 2 z = 5 is k , then the value of k 2 is equal to

Question

If the distance of point P 3,2 , 6 from the line x - 1 2 = y - 2 3 = z - 3 4 measured parallel to the plane 3 x - 5 y + 2 z = 5 is k , then the value of k 2 is equal to,

MARKS App · Accepted Answer

Coordinates of the point Q are 2 λ + 1,3 λ + 2,4 λ + 3 P Q → is perpendicular to the normal vector of the plane, P Q → = 2 λ - 2 i ^ + 3 λ j ^ + 4 λ - 3 k ^ Normal vector = 3 i ^ - 5 j ^ + 2 k ^ Hence, 3 2 λ - 2 - 5 3 λ + 2 4 λ - 3 = 0 ⇒ 6 λ - 6 - 15 λ + 8 λ - 6 = 0 ⇒ - λ - 12 = 0 ⇒ λ = - 1 2 So, the coordinates of the point Q are - 23 , - 34 , - 45 Hence, P Q = 26 2 + 36 2 + 51 2 = 4573

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