Coplanarity Of Vectors

Question

For what value of the scalar lambda are the three vectors i + j + k, i + lambda j + 2k and 2i + j + 3k coplanar in space?

RankGuru · Accepted Answer

Three vectors are coplanar precisely when their scalar triple product vanishes, because a flattened parallelepiped encloses zero volume. Setting the determinant of their components to zero is the canonical JEE Advanced coplanarity test. Form the determinant with rows (1,1,1), (1,lambda,2), (2,1,3) and set it to zero. Expanding along the first row: 1[(lambda)(3) - (2)(1)] - 1[(1)(3) - (2)(2)] + 1[(1)(1) - (lambda)(2)] = (3 lambda - 2) - (3 - 4) + (1 - 2 lambda) = (3 lambda - 2) + 1 + (1 - 2 lambda) = lambda. Setting lambda = 0 makes the determinant vanish, so the vectors are coplanar exactly when lambda = 0. Option 2 mishandles the middle minor's sign; option 1 satisfies only a partial minor; option 3 overshoots. Plausibility check: with lambda = 0 the middle vector is i + 2k, and one verifies i + 2k = (2i + j + 3k) - (i + j + k), a genuine linear combination of the other two, confirming all three lie in one plane.

Coplanarity Of Vectors

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