Scalar Triple ProductJEE Main

[a b c] — volume in a determinant — interactive Mathematics simulation for IIT-JEE.

Concept

The scalar triple product [a b c] = a · (b × c) equals the 3×3 determinant of components — and its absolute value is the volume of the parallelepiped on a, b, c. It vanishes exactly when the three vectors are coplanar: the single most-used test in 3D geometry. Cyclic shifts preserve it; swapping any two vectors flips the sign.

Key formula

[\vec a\, \vec b\, \vec c] = \vec a \cdot (\vec b \times \vec c) = \begin{vmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{vmatrix}; \quad V_{tet} = \tfrac{1}{6}|[\vec a\, \vec b\, \vec c]|

Derivation

b × c has magnitude = base parallelogram area and direction ⊥ base. Dotting with a projects a onto that normal — i.e. multiplies base area by height: volume.

Coplanarity: if a lies in the plane of b, c, it is ⊥ to b × c, so the dot product is zero. Cyclic symmetry [a b c] = [b c a] = [c a b] follows from determinant row-swap parity.

Scenarios to explore

Scalar Triple Product — [a b c] — the box volume determinant & coplanarity test.

Real-world applications

Coplanarity of four points A, B, C, D: [AB, AC, AD] = 0.
Volume of tetrahedra in 3D geometry problems.
Shortest distance between skew lines uses the box product in the numerator.

JEE exam tips

Dot and cross are interchangeable in the STP: a · (b × c) = (a × b) · c.
[a+b, b+c, c+a] = 2[a b c] — classic identity, drops out of linearity.
Skew-line distance: d = |[d₁ d₂ (r₂−r₁)]| / |d₁ × d₂|.

Common mistakes

Forgetting the 1/6 factor for tetrahedron volume.
Thinking [a b c] can't be negative — sign encodes orientation (handedness).
Writing a · b × c ambiguously — the cross MUST bind first (a·b) × c is meaningless.

Exam traps to avoid

[a b c] = 0 does NOT force any pair parallel — coplanar is weaker than collinear.
Repeated vector kills it instantly: [a a b] = 0 — spot before computing.