Question

If $|\overline{a}|=|\overline{b}|=|\overline{c}|=2$ and $\overline{a}\cdot \overline{b}=\overline{b}\cdot \overline{c}=\overline{c}\cdot \overline{a}=2,then\left[\overline{a}\overline{b}\overline{c}\right]cos{45}^{\circ }$ is equal to

• A. $1$
• B. $2$
• C. $3\sqrt{2}$
• D. $4$

Answer 1

The correct option is C $3\sqrt{2}$

$\overrightarrow{a}.\overrightarrow{b}=2$

$\mid a\mid \mid b\mid \cos {\theta }_1=2$

$\cos {\theta }_1=\frac{1}{2},\sin {\theta }_1=\frac{\sqrt{3}}{2}$

${\theta }_1={60}^{\circ }$

Similarly,

Angle between $\overrightarrow{b}.\overrightarrow{c}$ and $\overrightarrow{c}.\overrightarrow{b}$ is also ${60}^{\circ }$

$[\overrightarrow{a}.\overrightarrow{b}.\overrightarrow{c}]=\mid (\overrightarrow{a}\times \overrightarrow{b}).\overrightarrow{c}\mid$ $=\mid a\mid \mid b\mid \left(\sin {\theta }_1\right)\hat{n}.\overrightarrow{c}$ $=\mid \frac{4\sqrt{3}}{2}\hat{n}.\overrightarrow{c}\mid$

$\Rightarrow [\overrightarrow{a}.\overrightarrow{b}.\overrightarrow{c}]$ $=\mid \frac{4\sqrt{3}}{2}\hat{n}.\cos {30}^{\circ }\mid =\mid 4\sqrt{3}\times \frac{\sqrt{3}}{2}\mid =\mid 6\mid$

$\Rightarrow [\overrightarrow{a}.\overrightarrow{b}.\overrightarrow{c}]\cos {45}^{\circ }$$=6\cos {45}^{\circ }=\frac{6}{\sqrt{2}}=3\sqrt{2}$