Chapter 8.4, Problem 27E

To determine

To show:For any vector as $\overrightarrow{a}\cdot a\times \left(-\overrightarrow{a}\right)=0$ .

Answer to Problem 27E

It is shown below for any vector $\overrightarrow{a}.a\times \left(-\overrightarrow{a}\right)=0$ .

Explanation of Solution

Given information:

The vector $\overrightarrow{a}.a\times \left(-\overrightarrow{a}\right)=0$ .

To show for any vector $\overrightarrow{a}.a\times \left(-\overrightarrow{a}\right)=0$ .

Let, $\overrightarrow{a}=\left({\overrightarrow{a}}_{1,}{\overrightarrow{a}}_2,{\overrightarrow{a}}_3\right)$ and $-\overrightarrow{a}=\left(-{\overrightarrow{a}}_1,-{\overrightarrow{a}}_2,-{\overrightarrow{a}}_3\right)$ .

The cross product of the vector $\overrightarrow{a}\times \left(-\overrightarrow{a}\right)$ is given as,

  $\overrightarrow{a}\times \left(-\overrightarrow{a}\right)=\left|\begin{array}{l}ijk\\ a_1{a}_{2\text{}}{a}_3\\ -a_1-a_2-a_3\end{array}\right|$

Expand it along with $i,j,k$ respectively.

  $\begin{array}{c}\overrightarrow{a}\times -\overrightarrow{a}=\overrightarrow{i}\left|\begin{array}{l}{a}_2{a}_3\\ -a_2-a_3\end{array}\right|-\overrightarrow{j}\left|\begin{array}{l}{a}_1{a}_3\\ -a_1-a_3\end{array}\right|+\overrightarrow{k}\left|\begin{array}{l}{a}_1{a}_2\\ -a_1-a_2\end{array}\right|\\ =0\overrightarrow{i}-0\overrightarrow{j}+0\overrightarrow{k}\\ =0\end{array}$

Therefore, it is shown above for any vector $\overrightarrow{a}.a\times \left(-\overrightarrow{a}\right)=0$ .