Question

If $\overrightarrow{a}=\hat{i}+\hat{j},\overrightarrow{b}=\hat{i}+\hat{k}$, then $\left\{\right\{(\overrightarrow{a}\times \overrightarrow{b})\times \overrightarrow{a}\}\times \overrightarrow{a}\}\times \overrightarrow{b}=$

• A. $\hat{i}+\hat{j}+\hat{k}$
• B. $2(\hat{i}+2\hat{j}-\hat{k})$
• C. $-\hat{i}+\hat{j}+\hat{k}$
• D. $\hat{i}-\hat{j}+\hat{k}$

Answer 1

Answer: (B)

$2(\hat{i}+2\hat{j}-\hat{k})$

Firstly we have

$(\overline{a}\times \overline{b})\times \overline{a}$

$=a^2\overline{b}-(\overline{a}.\overline{b})\overline{a}$

$=2\overline{i}+2\overline{k}-\overline{i}-\overline{j}$

$=\overline{i}-\overline{j}+2\overline{k}$

Now, the result obtained above is cross producted with $\overline{a}$ to get

$(\overline{i}-\overline{j}+2\overline{k})\times (\overline{i}+\overline{j})$

$=0+\overline{k}+\overline{k}+0+2\overline{j}-2\overline{i}$

$=-2\overline{i}+2\overline{j}+2\overline{k}$

The result obtained is cross producted with $\overline{b},$ to get $(-2\overline{i}+2\overline{j}+2\overline{k})\times (\overline{i}+\overline{k})$

$=0+2\overline{j}-2\overline{k}+2\overline{i}+2\overline{j}+0$

$=2\overline{i}+4\overline{j}-2\overline{k}$

So answer is option $C$