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Scalar Triple Product

Mediummathematics

Three vectors a = i + 2j + 3k, b = 2i + j + k and c = 3i + j + 2k are given; compute their scalar triple product [a b c].

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About This Question

Subject
mathematics
Chapter
vector algebra
Topic
scalar triple product
Difficulty
Medium
Year
2025
Tags
advanced-calculus-drillscalar-triple-productdeterminantparallelepiped-volumelinear-independence

Solution

Correct Answer:

The scalar triple product [a b c] equals a . (b x c) and is evaluated as the determinant whose rows are the components of the three vectors. Geometrically it gives the signed volume of the parallelepiped spanned by the vectors, and its vanishing signals coplanarity, a frequently tested JEE Advanced criterion. Setting up the determinant with rows (1,2,3), (2,1,1), (3,1,2), expand along the first row: 1[(1)(2) - (1)(1)] - 2[(2)(2) - (1)(3)] + 3[(2)(1) - (1)(3)] = 1(2 - 1) - 2(4 - 3) + 3(2 - 3) = 1(1) - 2(1) + 3(-1) = 1 - 2 - 3 = -4. Option 6 takes the wrong sign and an inflated magnitude; option 0 would mean the vectors are coplanar, which the nonzero determinant refutes; option -12 triples a minor erroneously. Plausibility check: since the determinant is nonzero, the three vectors are linearly independent and span a genuine three-dimensional parallelepiped, consistent with a nonzero volume.

This medium difficulty mathematics question is from the chapter vector algebra, covering the topic of scalar triple product. It appeared in the 2025 exam.

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