Let $\overrightarrow{\mathbf{a}} = 2\widehat{\dot{\mathbf{i}}} + \widehat{\dot{\mathbf{j}}} - 2\widehat{\mathbf{k}},\ \ \overrightarrow{\mathbf{b}} = \widehat{\dot{\mathbf{i}}} + \widehat{\dot{\mathbf{j}}}$ If $\ \overrightarrow{\mathbf{c}}$ is a vector such that $\overrightarrow{\mathbf{a}} \bullet \overrightarrow{\mathbf{b}}\ \left| \overrightarrow{\mathbf{c}} ight|\$and$\ \left| \overrightarrow{\mathbf{c}} - \overrightarrow{\mathbf{a}} ight| = 2\sqrt{2}$ and angle between $\overrightarrow{\mathbf{a}} imes \ \overrightarrow{\mathbf{b}}\ and\ \overrightarrow{\mathbf{c}}\ is\ 30{^\circ},$ then $|(\overrightarrow{\mathbf{a}} imes \ \overrightarrow{\mathbf{b}}\ ) imes \ \overrightarrow{\mathbf{c}}|$ is

Question

Let $\overrightarrow{\mathbf{a}} = 2\widehat{\dot{\mathbf{i}}} + \widehat{\dot{\mathbf{j}}} - 2\widehat{\mathbf{k}},\ \ \overrightarrow{\mathbf{b}} = \widehat{\dot{\mathbf{i}}} + \widehat{\dot{\mathbf{j}}}$ If $\ \overrightarrow{\mathbf{c}}$ is a vector such that $\overrightarrow{\mathbf{a}} \bullet \overrightarrow{\mathbf{b}}\ \left| \overrightarrow{\mathbf{c}} ight|\$and$\ \left| \overrightarrow{\mathbf{c}} - \overrightarrow{\mathbf{a}} ight| = 2\sqrt{2}$ and angle between $\overrightarrow{\mathbf{a}} 	imes \ \overrightarrow{\mathbf{b}}\ and\ \overrightarrow{\mathbf{c}}\ is\ 30{^\circ},$ then $|(\overrightarrow{\mathbf{a}} 	imes \ \overrightarrow{\mathbf{b}}\ ) 	imes \ \overrightarrow{\mathbf{c}}|$ is

Dr. Abubacker · Accepted Answer

(a)
Given, $\overrightarrow{\mathbf{a}} = 2\widehat{\dot{\mathbf{i}}} + \widehat{\dot{\mathbf{j}}} - 2\widehat{\mathbf{k}},\ \ \overrightarrow{\mathbf{b}} = \widehat{\dot{\mathbf{i}}} + \widehat{\dot{\mathbf{j}}}$,
Now, $\left| \overrightarrow{\mathbf{a}} ight| = \sqrt{4 + 1 + 4} = 3$
Since, $\left| \overrightarrow{\mathbf{c}} - \overrightarrow{\mathbf{a}} ight| = 2\sqrt{2}$
$\Longrightarrow \left| \overrightarrow{\mathbf{c}} ight|^{2} + |\overrightarrow{\mathbf{a}}|^{2} - 2\overrightarrow{\mathbf{c}} \bullet \overrightarrow{\mathbf{a}} = 8$
$\Longrightarrow \ \left| \overrightarrow{\mathbf{c}} ight|^{2} + 9 - 2\left| \overrightarrow{\mathbf{c}} ight| = 8$
$\Longrightarrow \ \left| \overrightarrow{\mathbf{c}} ight|^{2} - 2\left| \overrightarrow{\mathbf{c}} ight| + 1 = 0$
$\Longrightarrow \left| \overrightarrow{\mathbf{c}} ight| = 1$
Now, $\left| \left( \overrightarrow{\mathbf{a}} 	imes \overrightarrow{\mathbf{b}} ight) 	imes \overrightarrow{\mathbf{c}} ight| = \left| \overrightarrow{\mathbf{a}} 	imes \overrightarrow{\mathbf{b}} ight|\left| \overrightarrow{\mathbf{c}} ight|\sin{30{^\circ}}$ ….(i)
$Now,\ \overrightarrow{\mathbf{a}} 	imes \overrightarrow{\mathbf{b}} = \left| \begin{matrix}
\widehat{\dot{\mathbf{i}}} & \widehat{\dot{\mathbf{j}}} & \widehat{\mathbf{k}} \
2 & 1 & - 2 \
1 & 1 & 0
\end{matrix} ight|$
$= \widehat{\dot{\mathbf{i}}}(0 + 2) - \widehat{\dot{\mathbf{j}}}(0 + 2) + \widehat{\mathbf{k}}(2 - 1)$
$= 2\widehat{\dot{\mathbf{i}}} - 2\widehat{\dot{\mathbf{j}}} + \widehat{\mathbf{k}}$
$\Longrightarrow \left| \overrightarrow{\mathbf{a}} 	imes \overrightarrow{\mathbf{b}} ight| = \sqrt{4 + 4 + 1} = 3$
$	herefore\$From Eq. (i),
$\left| \left( \overrightarrow{\mathbf{a}} 	imes \overrightarrow{\mathbf{b}} ight) 	imes \overrightarrow{\mathbf{c}} ight| = 3 \bullet 1 \bullet \frac{1}{2} = \frac{3}{2}$

Explanation

Reviewed by Dr. Abubacker

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