#### Provide solution for RD Sharma maths class 12 chapter Vector or Cross Product exercise 24.1 question 6

Answer         :  $-25 \hat{\imath}+35 \hat{\jmath}-55 \hat{k}$

Hint               : To solve this equation we put value in $(\vec{a}+2 \vec{b}) \text { and }(2 \vec{a}-\vec{b})$

Given            :$\vec{a}=3 \hat{\imath}-\hat{\jmath}-2 \hat{k}$

$\vec{b}=2 \hat{i}+3 \hat{j}+\hat{k}$

Find $(\vec{a}+2 \vec{b}) \times(2 \vec{a}-\vec{b})$

Solution      :

\begin{aligned} &(\vec{a}+2 \vec{b})=(3 \hat{\imath}-\hat{\jmath}-2 \hat{k}+2(2 \hat{\imath}+3 \hat{j}+\hat{k})) \\\\ &=3 \hat{i}+4 \hat{i}-\hat{j}+6 \hat{j}-2 \hat{k}+2 \hat{k} \\\\ &=7 \hat{\imath}+5 \hat{\jmath} \end{aligned}...............(1)

\begin{aligned} &(2 \vec{a}-\vec{b})=2(3 \hat{\imath}-\hat{\jmath}-2 \hat{k})-(2 \hat{\imath}+3 \hat{\jmath}+\hat{k}) \\\\ &=6 \hat{\imath}-2 \hat{\imath}-2 \hat{\jmath}-3 \hat{\jmath}-4 \hat{k}-\hat{k} \\\\ &=4 \hat{\imath}-5 \hat{\jmath}-5 \hat{k} \end{aligned}...............(2)

\begin{aligned} &(\vec{a}+2 \vec{b}) \times(2 \vec{a}-\vec{b})=\left|\begin{array}{ccc} \hat{\imath} & \hat{\jmath} & \hat{k} \\ 7 & 5 & 0 \\ 4 & -5 & -5 \end{array}\right| \\\\ &=\hat{\imath}(-25-0)-\hat{\jmath}(-35-0)+\hat{k}(-35-20) \\\\ &=-25 \hat{\imath}+35 \hat{\jmath}-55 \hat{k} \end{aligned}