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

Answer 1

Answer: (C)

$2|\overrightarrow{a}{|}^2$

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

And we know that, $\hat{i}\times \hat{j}=\hat{k},\hat{j}\times \hat{k}=\hat{i},\hat{k}\times \hat{i}=\hat{j}$

So,

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

Here,

${(-y\hat{k}+z\hat{j})}^2={\left(\sqrt{{(-y)}^2+z^2}\right)}^2=y^2+z^2{(x\hat{k}-z\hat{i})}^2={\left(\sqrt{x^2+{(-z)}^2}\right)}^2=x^2+z^2{(-x\hat{j}+y\hat{i})}^2={\left(\sqrt{{(-x)}^2+y^2}\right)}^2=x^2+y^2$

Putting these values we get from,

${(-y\hat{k}+z\hat{j})}^2+{(x\hat{k}-z\hat{i})}^2+{(-x\hat{j}+y\hat{i})}^2=y^2+z^2+x^2+z^2+x^2+y^2=2(x^2+y^2+z^2)$

Now, for the vector $\overrightarrow{a}=x\hat{i}+y\hat{j}+z\hat{k}$,

$x^2+y^2+z^2={\left|\overrightarrow{a}\right|}^2\therefore 2(x^2+y^2+z^2)=2{\left|\overrightarrow{a}\right|}^2$

Hence, the required value is $2{\left|\overrightarrow{a}\right|}^2.$