Let a plane P contain two lines r i lambda i j lambda R and

Question

Let a plane P contain two lines -> r = hat i + lambda ( hat i + hat j ), lambda in R and -> r = - hat j + mu ( hat j - hat k ), mu in R If Q(alpha , beta , gamma ) is the foot of the perpendicular drawn from the point M(1, 0, 1) to P, then 3(alpha + beta + gamma ) equals _______.

DivineJEE · Accepted Answer

Given lines, \ vec r = hat i + lambda (hat i + hat j) parallel to (hat i + hat j) \ Let, vec n_1 = (hat i + hat j) \ and vec r = -hat j + mu (hat j - hat k) parallel to (hat j - hat k) \ Let, vec n_2 = (hat j - hat k) \ Therefore Normal of plane, vec n = vec n_1 x vec n_2 \ vec n = hat i hat j hat k \ 1 1 0 \ 0 1 -1 \ = -hat i + hat j + hat k \ Line vec r = hat i + lambda (hat i + hat j) is on the plane so, \ point on the line (1, 0, 0) will be also on the plane. \ Therefore Equation of the plane, \ -1(x - 1) + 1(y - 0) + 1(z - 0) = 0 \ => x - y - z - 1 = 0 \ Foot of perpendicular from (x_1, y_1, z_1) on the plane, \ (x - x_1 / a) = (y - y_1 / b) = (z - z_1 / c) = -((ax_1 + by_1 + cz_1 + d) / a^2 + b^2 + c^2) \ Here foot of perpendicular is drawn from M(1, 0, 1), \ Therefore (x - 1 / 1) = (y - 0 / -1) = (z - 1 / -1) = -((1 - 0 - 1 - 1) / 3) \ Therefore x - 1 = (1 / 3) => x = (4 / 3) \ (y / -1) = (1 / 3) => y = -(1 / 3) \ (z - 1 / -1) = (1 / 3) => z = (2 / 3) \ According to the question, \ x = alpha, y = beta, z = gamma \ Therefore alpha = (4 / 3), beta = -(1 / 3), gamma = (2 / 3) \ Therefore 3(alpha + beta + gamma) = 3( (4 / 3) - (1 / 3) + (2 / 3) ) = 5

Explanation

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