### Example 1 of Triple Product (Dot Product and Cross Product)

Given three vectors $\begin{array}{l}\mathop u\limits_{}^ \to = \left\langle {\begin{array}{*{20}{c}}0\\0\\1\end{array}} \right\rangle \\\mathop v\limits^ \to = \left\langle {\begin{array}{*{20}{c}}2\\2\\1\end{array}} \right\rangle \\\mathop w\limits^ \to = \left\langle {\begin{array}{*{20}{c}}1\\1\\0\end{array}} \right\rangle \end{array}$ find the triple product, $$\mathop u\limits_{}^ \to \cdot \left( {\mathop v\limits^ \to \times \mathop w\limits^ \to } \right)$$.

Find the cross product first. To do this easily, write both vectors horizontally, and find the 2x2 determinant while covering your hand over the column you want to find. Basically, you have two 1x3 vectors (row x column), the result will also be 1x3. I'll show this below. Note that "i", "j" and "k" just represent the first, second, and third dimensions/components.

$\begin{array}{*{20}{c}}i&j&k\\2&2&1\\1&1&0\end{array}$

I want v x w, so I write v first and then w. To get the first component, put your hand over the first column, then find the 2x2 determinant, which is just a product of a diagonal minus the product of the other diagonal.

The first component is (2)(0) - (1)(1) = -1

Now put your hand over the second column, and find the 2x2 determinant, BUT, you need to add a negative sign to your answer. This is because of the cofactor matrix which you can write out, but basically the middle determinant needs to be negative. The second component is -[(2)(0) - (1)(1)] = 1

And the final component, put your hand to cover the third column and find the determinant, which is (2)(1) - (1)(2) = 0

Now just put the three components into a vector. $\left( {\mathop v\limits^ \to \times \mathop w\limits^ \to } \right) = \left\langle {\begin{array}{*{20}{c}}{ - 1}\\1\\0\end{array}} \right\rangle$

Then we find $$\mathop u\limits_{}^ \to \cdot \left( {\mathop v\limits^ \to \times \mathop w\limits^ \to } \right)$$. To find the dot product, multiply the components by each other and sum them up.

$$\begin{array}{l}\mathop u\limits_{}^ \to \cdot \left( {\mathop v\limits^ \to \times \mathop w\limits^ \to } \right) = \left\langle {\begin{array}{*{20}{c}}0\\0\\1\end{array}} \right\rangle \cdot \left\langle {\begin{array}{*{20}{c}}{ - 1}\\1\\0\end{array}} \right\rangle \\\mathop u\limits_{}^ \to \cdot \left( {\mathop v\limits^ \to \times \mathop w\limits^ \to } \right) = (0)( - 1) + (0)(1) + (1)(0)\\\mathop u\limits_{}^ \to \cdot \left( {\mathop v\limits^ \to \times \mathop w\limits^ \to } \right) = 0\end{array}$$

So the triple product is 0.

Please see our other linear algebra examples below:

Triple product examples (dot product of a cross product), including finding the volume of a parallelpiped:

Triple Product (Dot Product and Cross Product) Example 2

Volume of Parallelpiped Triple Product Example 3

Volume of Parallelpiped Triple Product Example 4

Also see our guide on determinants for 2x2, 3x3 and even a 5x5 matrix:

Determinants (Ex1) 