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

Given:$\left | \overrightarrow{a}\times \overrightarrow{b} \right |=4\left | \overrightarrow{a}\overrightarrow{b} \right |=2$

Hint: Using

$\left | \overrightarrow{a}\times \overrightarrow{b} \right |=\left | \overrightarrow{a} \right |\left | \overrightarrow{b} \right |\sin \theta ;$

$\left | \overrightarrow{a}. \overrightarrow{b} \right |=\left | \overrightarrow{a} \right |\left | \overrightarrow{b} \right |\cos \theta$

Explanation:

$\left | \overrightarrow{a}\times \overrightarrow{b} \right |=4$

$\left | \overrightarrow{a} \right |\left | \overrightarrow{b} \right |\sin \theta=4$                              ……………………. (1)

Also      $\left | \overrightarrow{a}.\overrightarrow{b} \right |=2$

$\left | \overrightarrow{a} \right |\left | \overrightarrow{b} \right |\cos \theta=2$                              ……………………. (2)

Squaring and adding (1) and (2)

\begin{aligned} &|\vec{a}|^{2}|\vec{b}|^{2} \sin ^{2} \theta+|\vec{a}|^{2}|\vec{b}|^{2} \cos ^{2} \theta=(4)^{2}+(2)^{2} \\ &\end{aligned}

$|\vec{a}|^{2}|\vec{b}|^{2}\left(\sin ^{2} \theta+\cos ^{2} \theta\right)=16+4$                                 $\quad\left[\sin ^{2} \theta+\cos ^{2} \theta=1\right] \\$

$|\vec{a}|^{2}|\vec{b}|^{2}(1)=20 \\$

$|\vec{a}|^{2}|\vec{b}|^{2}=20$