Need solution for rd sharma maths class 12 chapter 24 vector or cross product Exercises Multiple choice questions question 6 maths textbook solution

Given: Vector $\overrightarrow{a}$ & $\overrightarrow{b}$are inclined at angle$\theta =120^{\circ}$. If $\left | \overrightarrow{a} \right |=1 , \left | \overrightarrow{b} \right |=2$then $\left | \left ( \overrightarrow{a}+3\overrightarrow{b} \right )\times \left ( 3\overrightarrow{a}-\overrightarrow{b} \right ) \right |$is equal to

Hint: You must know about cross product.

Explanation:

\begin{aligned} |\vec{a}|=1,|\vec{b}|=2, \theta=120^{\circ} \\ \end{aligned}

$\qquad|(\vec{a}+3 \vec{b}) \times(3 \vec{a}-\vec{b})|^{2}=|3(\vec{a} \times \vec{a})-(\vec{a} \times \vec{b})+9(\vec{b} \times \vec{a})-3(\vec{b} \times \vec{b})|^{2} \\$

$=|3(0)+(\vec{b} \times \vec{a})+9(\vec{b} \times \vec{a})-3(0)|^{2} \quad[\because(\vec{a} \times \vec{a})=0 \text { and }(\vec{b} \times \vec{b})=0] \\$

$=|10(\vec{b} \times \vec{a})|^{2} \\$

$=|\vec{b} \times \vec{a}|=|\vec{b}| \vec{a} \mid \sin \theta$

Substitute,

\begin{aligned} |\vec{a}|=1,|\vec{b}|=2, \theta=120^{\circ} \\ \end{aligned}

${l} =|\vec{b}||\vec{a}| \sin 120^{\circ}$                                                         $\left[\begin{array}{l} \because \sin 120^{\circ}=\sin \left(180^{\circ}-60^{\circ}\right) \\ =\sin 60^{\circ}=\frac{\sqrt{3}}{2} \end{array}\right]$

$\\ =(2)(1)\left(\frac{\sqrt{3}}{2}\right) \\ =\sqrt{3}$

$\qquad|(\vec{a}+3 \vec{b}) \times(3 \vec{a}-\vec{b})|^{2}=\left | 10\left ( \sqrt{3} \right ) \right |^{2}=300$