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Let L1: (x-1)/1 = y/(-1) = (z-1)/3 and L2: (x-1)/(-3) = y/(-1) = (z-1)/1 be two straight lines. A line L: (x-alpha)/l = (y-1)/m = (z-gamma)/(-2) lies in the plane containing L1 and L2 and passes through their point of intersection. If L bisects the acute angle between L1 and L2, which of the following is/are TRUE?

1. alpha - gamma = 3
2. l + m = 2
3. alpha - gamma = 1
4. l + m = 0

Correct answer: alpha - gamma = 3

Solution

L1 and L2 intersect at (1,0,1) (set parameters equal and solve). The bisector direction is proportional to (d1_hat + d2_hat). Since both have the same magnitude sqrt(11), the bisector direction is proportional to (1-3, -1-1, 3+1) = (-2,-2,4), i.e., (1,1,-2). L has direction (l, m, -2), matching ratio gives l=1, m=1. The line passes through (1,0,1), so using parametric form with (1,1,-2): point is (1,0,1). Thus alpha=1, gamma=1 giving alpha-gamma=0... rechecking with the passing through point (1,0,1): L passes (alpha, 1, gamma), which must satisfy the system. If L passes through intersection (1,0,1): (1-alpha)/l = (0-1)/m = (1-gamma)/(-2). With direction (-1,-1,2): l=-1, m=-1, -2=-2. So (1-alpha)/(-1) = (0-1)/(-1) = 1 => alpha=0; (1-gamma)/(-2) = 1 => gamma=-1. alpha-gamma = 0-(-1) = 1. So alpha - gamma = 1.

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