Question

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

• A. $3\overrightarrow{a^2}$
• B. $\overrightarrow{a^2}$
• C. $2\overrightarrow{a^2}$
• D. None of these

Answer 1

The correct option is A $3\overrightarrow{a^2}$

Let any vector such that

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

Given that

$(a\times \hat{ı}{)}^2+(a\times \hat{ȷ}{)}^2+(a\times \hat{k}{)}^2$

$\Rightarrow \left[\right(x\hat{ı}+yj+z\hat{k})\times \hat{ı}{]}^2+[(x\hat{ı}+y\hat{j}+z\hat{k})x\hat{ȷ}{]}^2+\left[\right(x\hat{ı}+y\hat{j}+z\hat{k})\times \hat{k}{]}^2$

$\Rightarrow [-y\hat{k}+z\hat{ȷ}{]}^2+[x\hat{k}-z\hat{ı}{]}^2+[-x\hat{ȷ}+y\hat{ı}{]}^2$

$\text{As we know that}[-y\hat{k}+z\hat{ȷ}{]}^2={\left[\sqrt{(-y{)}^2+z^2}\right]}^2=y^2+z^2$

Similarly

$\Rightarrow y^2+z^2+x^2+z^2+x^2+y^2$

$\Rightarrow 2[x^2+y^2+z^2]$

also $(x^2+y^2+z^2)=|a{|}^2=a^2$

hence, $2[x^2+y^2+z^2]=2a^2$

$\therefore$ (c) is the correct option.