Question

For any vector $\overrightarrow{p}$, the value of $\frac{3}{2}\{|\overrightarrow{p}\times \hat{i}{|}^2+|\overrightarrow{p}\times \hat{j}{|}^2+|\overrightarrow{p}\times \hat{k}{|}^2\}$ is

• A. ${\overrightarrow{p}}^2$
• B. $2{\overrightarrow{p}}^2$
• C. $3{\overrightarrow{p}}^2$
• D. $4{\overrightarrow{p}}^2$

Answer 1

Answer: (A)

${\overrightarrow{p}}^2$

$\begin{array}{c}\left[\because \left|a\times {\left|a\times b\right|}^2\right|=\left|\begin{array}{cc}a.a& a.b\\ a.b& b.b\end{array}\right|\right]\\ \frac{3}{2}\left\{\left|\begin{array}{cc}p.p& p.\hat{i}\\ p.\hat{i}& \hat{i}.\hat{i}\end{array}\right|+\left|\begin{array}{cc}p.p& p.\hat{j}\\ p.\hat{j}& \hat{j}.\hat{j}\end{array}\right|+\left|\begin{array}{cc}p.p& \hat{p}.\hat{k}\\ \hat{p}.\hat{k}& \hat{k}.\hat{k}\end{array}\right|\right\}\\ \frac{3}{2}\left\{p^2-p^2+p^2-p^2+p^2-p^2\right\}\left[\because \hat{i}.\hat{i}=\hat{j}.\hat{j}=\hat{k}.\hat{k}=1\right]\\ \therefore \frac{3}{2}\times 0=0\end{array}$