Question

If $\overrightarrow{a}=2\hat{i}+\hat{j}+3\hat{k},\overrightarrow{b}=3\hat{i}+2\hat{j}+\hat{k},\overrightarrow{c}=\hat{i}-\hat{j}-4\hat{k}$ and $\overrightarrow{d}=\hat{i}+2\hat{j}-\hat{k},$ then $(\overrightarrow{a}\times \overrightarrow{b})\times (\overrightarrow{c}\times \overrightarrow{d})=$

• A. $-8(\hat{i}+\hat{j}+\hat{k})$
• B. $16(\hat{i}-\hat{j}+2\hat{k})$
• C. $24(\hat{i}+\hat{j}-2\hat{k})$
• D. $-12(\hat{i}+\hat{j}+\hat{k})$

Answer 1

The correct option is C $24(\hat{i}+\hat{j}-2\hat{k})$

We know,
$(\overrightarrow{a}\times \overrightarrow{b})\times (\overrightarrow{c}\times \overrightarrow{d})=\left[\begin{array}{ccc}\overrightarrow{a}& \overrightarrow{b}& \overrightarrow{d}\end{array}\right]\overrightarrow{c}-\left[\begin{array}{ccc}\overrightarrow{a}& \overrightarrow{b}& \overrightarrow{c}\end{array}\right]\overrightarrow{d}=\left|\begin{array}{ccc}2& 1& 3\\ 3& 2& 1\\ 1& 2& -1\end{array}\right|\overrightarrow{c}-\left|\begin{array}{ccc}2& 1& 3\\ 3& 2& 1\\ 1& -1& -4\end{array}\right|\overrightarrow{d}=8(\hat{i}-\hat{j}-4\hat{k})+16(\hat{i}+2\hat{j}-\hat{k})=24\hat{i}+24\hat{j}-48\hat{k}=24(\hat{i}+\hat{j}-2\hat{k})$