Using the scalar triple product to prove that vectors are coplanar

Formula for the scalar triple product

The scalar triple product $|a\cdot(b\times c)|$ of three vectors $a\langle{a_1},a_2,a_3\rangle$, $b\langle{b_1},b_2,b_3\rangle$ and $c\langle{c_1},c_2,c_3\rangle$ will be equal to $0$ when the vectors are coplanar, which means that the vectors all lie in the same plane.

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$a$, $b$, and $c$ are coplanar if $|a\cdot(b\times c)|=0$

$b\times c$ is the cross product of $b$ and $c$, and we’ll find it using the $3\times 3$ matrix

$\begin{vmatrix}\bold i&\bold j&\bold k\\b_1&b_2&b_3\\c_1&c_2&c_3\end{vmatrix}=\bold i\begin{vmatrix}b_2&b_3\\c_2&c_3\end{vmatrix}-\bold j\begin{vmatrix}b_1&b_3\\c_1&c_3\end{vmatrix}+\bold k\begin{vmatrix}b_1&b_2\\c_1&c_2\end{vmatrix}$

$=\bold i(b_2c_3-b_3c_2)-\bold j(b_1c_3-b_3c_1)+\bold k(b_1c_2-b_2c_1)$

We’ll convert the result of the cross product into standard vector form, and then take the dot product of $a\langle{a_1},a_2,a_3\rangle$ and the vector result of $b\times c$.

$|a\cdot(b\times c)|$

The final answer is the scalar triple product. If it’s equal to $0$, then we’ve proven that the vectors are coplanar.

How to use the scalar triple product to verify that three vectors are coplanar (that they lie in the same plane)

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Proving that the vectors are coplanar

Example

Prove that the vectors are coplanar.

$a\langle3,3,-3\rangle$

$b\langle1,0,-2\rangle$

$c\langle2,3,-1\rangle$

We’ll use the scalar triple product, and we’ll start by calculating the cross product of $b$ and $c$, $b\times c$.

$b\times c=\begin{vmatrix}\bold i&\bold j&\bold k\\1&0&-2\\2&3&-1\end{vmatrix}$

$b\times c=\bold i\begin{vmatrix}0&-2\\3&-1\end{vmatrix}-\bold j\begin{vmatrix}1&-2\\2&-1\end{vmatrix}+\bold k\begin{vmatrix}1&0\\2&3\end{vmatrix}$

$b\times c=\bold i[(0)(-1)-(-2)(3)]-\bold j[(1)(-1)-(-2)(2)]+\bold k[(1)(3)-(0)(2)]$

$b\times c=\bold i(0+6)-\bold j(-1+4)+\bold k(3-0)$

$b\times c=6\bold i-3\bold j+3\bold k$

$b\times c=\langle6,-3,3\rangle$

The final answer is the scalar triple product. If it’s equal to 0, then we’ve proven that the vectors are coplanar.

Next we’ll take the dot product of $a\langle3,3,-3\rangle$ and $b\times c=\langle6,-3,3\rangle$.

$|a\cdot(b\times c)|=(3)(6)+(3)(-3)+(-3)(3)$

$|a\cdot(b\times c)|=18-9-9$

$|a\cdot(b\times c)|=0$

Since the scalar triple product of the vectors $a\langle3,3,-3\rangle$, $b\langle1,0,-2\rangle$ and $c\langle2,3,-1\rangle$ is equal to $0$,

$|a\cdot(b\times c)|=0$

the vectors $a$, $b$, and $c$ are coplanar.

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