Question

For any vector $\overrightarrow{a}$, the value of $(\overrightarrow{a}\times \hat{i}{)}^2+(\overrightarrow{a}\times \hat{j}{)}^2+(\overrightarrow{a}\times \hat{k}{)}^2$ is equal to

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

Answer 1

The correct option is D $2{\overrightarrow{a}}^2$

Let $\overrightarrow{a}=x\hat{i}+y\hat{j}+z\hat{k}$

$\therefore {\overrightarrow{a}}^2=x^2+y^2+z^2$

$\therefore \overrightarrow{a}\times \hat{i}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ x& y& z\\ 1& 0& 0\end{array}\right|$

$=\hat{i}\left[0\right]-\hat{j}[-z]+\hat{k}[-y]$

$=z\hat{j}-y\hat{k}$

$\therefore (\overrightarrow{a}\times \hat{i}{)}^2=(z\hat{j}-y\hat{k}\left)\right(z\hat{j}-y\hat{k})$

$=y^2+z^2$

Similarly, $(\overrightarrow{a}\times \hat{j}{)}^2=x^2+z^2$

And $(\overrightarrow{a}\times \hat{k}{)}^2=x^2+y^2$

$\therefore (\overrightarrow{a}\times \hat{i}{)}^2+(\overrightarrow{a}\times \hat{j}{)}^2+(\overrightarrow{a}\times \hat{k}{)}^2=y^2+z^2+x^2+z^2+x^2+y^2$

$=2(x^2+y^2+z^2)=2{\overrightarrow{a}}^2$