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Two lines in 3D space are given: M1 passes through the origin with direction ratios (1,1,1), i.e. x = y = z, and M2 passes through the origin with direction ratios (1,2,3), i.e. x/1 = y/2 = z/3. A third line M3 passes through the point A = (1,1,1) and forms a triangle together with M1 and M2, where A = (1,1,1) is one vertex of the triangle, the origin O = (0,0,0) is a second vertex (lying on both M1 and M2), and the third vertex B lies on M2. If the area of this triangle is sqrt(6) square units, find the coordinates of B (i.e., the point where M3 meets M2).

1. (1, 2, 3)
2. (2, 4, 6)
3. (4/3, 8/3, 4)
4. (1, 5, 7)

Correct answer: (2, 4, 6)

Solution

The cross product OA x OB = (1,1,1) x (s,2s,3s) = (s,-2s,s) has magnitude s*sqrt(6). Setting (1/2)*s*sqrt(6) = sqrt(6) gives s = 2, so B = (2,4,6).

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