Q.404. Distance of a point from a plane along a direction.

Question 404.
Find the distance of point (1,-2,3) from the plane 2x + 3y - 4z = 16 measured along the straight line with direction ratios 1,-2, 3.

Answer 404.
Point (1, -2, 3) lies on the line
(x, y, z) = (1, -2, 3) + k (1, -2, 3) whose direction ratios are 1, -2, 3 for k = 0.
=> x = 1 + k, y = -2 - 2k and z = 3 + 3k
Plugging in the equation of the plane
2 (1 + k) + 3 (-2 - 2k) - 4 (3 + 3k) = 16
=> - 16 - 16k = 16
=> k = - 2
=> the line intersects the plane in the point (1+k, -2-2k, 3+3k)
= (-1, 2, -3)
=> the required distance is the distance between
(-1, 2, -3) and (1, -2, 3)
= √[(-1 - 1)^2 + (2 + 2)^2 + (-3 -3)^2]
= 2√(14).

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