Recipe

Compute the scalar triple product

Use when: you have three vectors in \mathbb{R}^3 and need either the volume of the parallelepiped they span, or a yes/no test for whether the three are coplanar

Goal: the signed number \det[\mathbf u\ \mathbf v\ \mathbf w], read as a volume (its absolute value) or as a coplanarity test (zero or not)

Taught in Area, volume, and the scalar triple product

1. 1

   Write \mathbf u, \mathbf v, \mathbf w into the columns of one 3\times3 matrix, in the order the problem states them — swapping two columns only flips the final sign, never changes whether the result is zero.

   Example

   For \mathbf u=(1,2,0), \mathbf v=(3,0,-1), \mathbf w=(-2,1,4): \det\begin{bmatrix} 1 & 3 & -2 \\ 2 & 0 & 1 \\ 0 & -1 & 4 \end{bmatrix}

   Why: the theorem is stated for exactly this matrix — \mathbf u,\mathbf v,\mathbf w as columns, in that order — so this is the object every later step operates on.

2. 2

   Cofactor expansion works along any row or column; look at all three rows and columns first and pick whichever one has the most zeros, since a zero entry's whole cofactor never needs to be computed.

   Example

   The bottom row, (0,-1,4), has a zero in its first slot — expanding along that row means one of the three 2\times2 minors can be skipped entirely.

   Why: this is the same determinant either way; picking the row or column with a zero is free savings, not a different method.

3. 3

   Cross out the row and column of each entry in turn, take the determinant of the leftover 2\times2 block, and attach the alternating +,-,+ sign that goes with that entry's position.

   Example

   Expanding along the bottom row (0,-1,4) (positions -,+ after the leading 0 term drops out): \det = -(-1)\det\begin{pmatrix}1&-2\\2&1\end{pmatrix} + 4\det\begin{pmatrix}1&3\\2&0\end{pmatrix} = 1(1+4) + 4(0-6) = 5 - 24 = -19

   ⚡ Shortcut

   For a 3\times3 determinant specifically, Sarrus's rule skips the cofactor bookkeeping entirely: copy the first two columns again to the right of the matrix, multiply along the three full diagonals running down-and-right, add those, then multiply along the three diagonals running down-and-left and subtract that sum. \begin{array}{ccc|cc}1&3&-2&1&3\\2&0&1&2&0\\0&-1&4&0&-1\end{array} Down-right diagonals: (1)(0)(4)+(3)(1)(0)+(-2)(2)(-1)=0+0+4=4. Down-left diagonals: (-2)(0)(0)+(1)(1)(-1)+(4)(2)(3)=0-1+24=23. Result: 4-23=-19 — the same number, no minors at all. (Sarrus's rule works ONLY at 3\times3; it doesn't generalize to 4\times4 and larger.)

   Why: cofactor expansion is what the theorem's proof actually relies on, so it's the method that always works at any size — Sarrus's rule is a 3\times3-only mental shortcut riding on top of it.

4. 4

   For a volume, take the absolute value — the sign only ever records orientation, never size. For a coplanarity test, check whether the determinant is exactly zero.

   Example

   \det=-19, so the volume is |-19|=19 cubic units; since -19\neq0, \mathbf u,\mathbf v,\mathbf w are not coplanar — they genuinely span a solid box.

   Why: absolute value strips out the sign that reordering the vectors would flip, leaving pure size for a volume; a zero determinant is exactly the algebraic signature of three vectors collapsing into a shared plane, with no box left to measure.