Scalar Triple Product & Coplanarity

MATScalar Triple Product

The scalar triple product $[\vec{a}\,\vec{b}\,\vec{c}] = \vec{a}\cdot(\vec{b}\times\vec{c})$ is a single number equal (in absolute value) to the volume of the parallelepiped with edges $\vec{a},\vec{b},\vec{c}$ . It vanishes exactly when the three vectors are coplanar, making it the standard coplanarity test.

ISC examines it via direct determinant evaluation, solving for an unknown that forces coplanarity, finding box volumes, and vector proofs of geometric facts.

Scalar triple product (determinant)

[\vec{a}\,\vec{b}\,\vec{c}] = \begin{vmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{vmatrix}

\vec{a},\vec{b},\vec{c}

have the listed components; the rows are the three vectors.

Definition via dot and cross

[\vec{a}\,\vec{b}\,\vec{c}] = \vec{a}\cdot(\vec{b}\times\vec{c})

A scalar; equals

\vec{b}\cdot(\vec{c}\times\vec{a})=\vec{c}\cdot(\vec{a}\times\vec{b})

by cyclic symmetry.

Coplanarity condition

[\vec{a}\,\vec{b}\,\vec{c}] = 0

Holds iff

\vec{a},\vec{b},\vec{c}

are coplanar (linearly dependent).

Volume of a parallelepiped

V = \left|\,[\vec{a}\,\vec{b}\,\vec{c}]\,\right|

\vec{a},\vec{b},\vec{c}

are the three edges meeting at a corner;

V

in cubic units.

The scalar triple product is unchanged by cyclic permutation: $[\vec{a}\,\vec{b}\,\vec{c}]=[\vec{b}\,\vec{c}\,\vec{a}]=[\vec{c}\,\vec{a}\,\vec{b}]$ .
Swapping any two vectors changes only the sign: $[\vec{a}\,\vec{b}\,\vec{c}]=-[\vec{b}\,\vec{a}\,\vec{c}]$ .
If any two of the three vectors are equal or parallel, the product is $0$ .
For coplanarity with an unknown, set the determinant to $0$ and solve the resulting equation in $\lambda$ .
Volume is the absolute value of the triple product; never report a negative volume.
Expanding the determinant by the first row reproduces $\vec{a}\cdot(\vec{b}\times\vec{c})$ exactly.
Four points $A,B,C,D$ are coplanar iff $[\vec{AB}\,\vec{AC}\,\vec{AD}]=0$ .
For a vector proof such as 'the diagonals of a rhombus are perpendicular', express the diagonals as $\vec{a}+\vec{b}$ and $\vec{a}-\vec{b}$ and show $(\vec{a}+\vec{b})\cdot(\vec{a}-\vec{b})=|\vec{a}|^2-|\vec{b}|^2=0$ since the sides are equal.

Where the marks go

Reporting volume as a signed (possibly negative) value instead of $\left|[\vec{a}\,\vec{b}\,\vec{c}]\right|$ .
Sign errors when expanding the $3\times3$ determinant, especially on the middle column.
Treating the triple product as a vector; it is a scalar (the cross product is computed first, then the dot).
Mis-setting the coplanarity test, e.g. equating the product to a nonzero value or confusing it with the perpendicularity test $\vec{a}\cdot\vec{b}=0$ .

How the board asks it

Predict the productexpanding the 3x3 determinant of components
Evaluate $[\vec{a}\,\vec{b}\,\vec{c}]$ for $\vec{a} = 2\hat{i} + 3\hat{j} - \hat{k}$ , $\vec{b} = \hat{i} - \hat{j} + 2\hat{k}$ and $\vec{c} = 3\hat{i} + \hat{j} + \hat{k}$ .
Numericalvectors coplanar iff the scalar triple product is $0$
Find $\lambda$ if the vectors $\vec{a} = \hat{i} - \hat{j} + \hat{k}$ , $\vec{b} = 2\hat{i} + \hat{j} - \hat{k}$ and $\vec{c} = \lambda\hat{i} - \hat{j} + \lambda\hat{k}$ are coplanar.
Derive / prove $A,B,C,D$ coplanar iff $[\vec{AB}\,\vec{AC}\,\vec{AD}]=0$
Show that the points $A(1, 2, 3)$ , $B(-1, -2, -1)$ , $C(2, 3, 2)$ and $D(4, 7, 6)$ are coplanar.
Numericalvolume equals $\lvert[\vec{a}\,\vec{b}\,\vec{c}]\rvert$
Find the volume of the parallelepiped whose coterminous edges are $\vec{a} = \hat{i} + 2\hat{j} + 3\hat{k}$ , $\vec{b} = -\hat{i} + \hat{j} + 2\hat{k}$ and $\vec{c} = 2\hat{i} + \hat{j} - \hat{k}$ .
Derive / provemultilinearity of the scalar triple product
Prove that $[\vec{a} + \vec{b}\ \ \vec{b} + \vec{c}\ \ \vec{c} + \vec{a}] = 2[\vec{a}\,\vec{b}\,\vec{c}]$ .

Practise this topic

PreviousVector (Cross) Product & Area

Written for Sublevo. Question text quoted anywhere in these notes is the Council’s and carries its year and paper; the board’s own diagrams are not reproduced.