devarshi-dt-logo

Question:

If the distance of the point P(1, -2, 1) from the plane x + 2y - z = α, where α > 0 is 5, then the foot of the perpendicular from P is

(8/3, 4/3, -1/3)

(4/3, -1/3, 1/3)

(1/3, 2/3, 10/3)

(2/3, -5/3, 5/2)

Solution:

Distance from plane
15 = |α + 5|
Let (x1, y1, z1) be the foot of perpendicular.
Then, x1 - 1 = y1 + 2 = z1 - 1 = -λ(α + 5)/9
(x1, y1, z1) = (8/3, 4/3, -1/3)