Question

If $\overrightarrow{a}=\hat{i}+\hat{j}+\hat{k}$,

$\overrightarrow{b}=\hat{i}+\hat{j}-\hat{k}$,

$\overrightarrow{c}=\hat{i}-\hat{j}+\hat{k}$,

$\overrightarrow{d}=\hat{i}-\hat{j}-\hat{k}$
then $|(\overrightarrow{a}\times \overrightarrow{b}).(\overrightarrow{c}\times \overrightarrow{d})|=$

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

The correct option is A $0$

Consider, $|(\overline{a}\times \overline{b})\cdot (\overline{c}\times \overline{d})|$

$=\overrightarrow{a.}(\left(\overline{b.}\overline{d}\right)\overline{c}-\left(\overline{b.}\overline{c}\right)\overline{d})$
$=\left(\overline{a.}\overline{c}\right)\left(\overline{b.}\overline{d}\right)-\left(\overline{b.}\overline{c}\right)\overline{a}.\overline{d}$

$\therefore |(\overline{a}\times \overline{b})\cdot (\overline{c}\times \overline{d})|=(\overline{a}\cdot \overline{c})(\overline{b}\cdot \overline{d})-(\overline{b}\cdot \overline{c})(\overline{a}\cdot \overline{d})$ .... $\left(i\right)$

Now, $(\overline{a}\cdot \overline{c})\times (\overline{b}\cdot \overline{d})$

$=(1-1+1)\times (1-1+1)=1\times 1=1$

$\therefore (\overline{a}\cdot \overline{c})\times (\overline{b}\cdot \overline{d})=1$

$(\overline{b}\cdot \overline{c})\times (\overline{a}\cdot \overline{d})$

$=\left(\right(1,1,-1)\cdot (1,-1,1\left)\right)\times \left(\right(1,1,1)\cdot (1,-1,-1\left)\right)$

$=(1-1-1)\times (1-1-1)=(-1)\times (-1)=1$

$\therefore (\overline{b}\cdot \overline{c})\times (\overline{a}\cdot \overline{d})=1$

From $\left(i\right)$

$|(\overline{a}\times \overline{b})\cdot (\overline{c}\times \overline{d})|=1-1=0$

Hence, $|(\overline{a}\times \overline{b})\cdot (\overline{c}\times \overline{d})|=0$