Provide solution for RD Sharma maths class 12 chapter Straight Line in Space exercise 24.1 question 3 sub question (ii)

Answer         : $\frac{1}{11}(\hat{\imath}+\hat{\jmath}-3 \hat{k})$

Hint               : To solve this equation we use $\frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|}$

Given            : $\vec{a}=2 \hat{\imath}+\hat{\jmath}+\hat{k}$
$\vec{b}=\hat{\imath}+2 \hat{\jmath}+\hat{k}$

Solution       : $\frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|}$

$\vec{a} \times \vec{b}=\left|\begin{array}{ccc} \hat{\imath} & \hat{\jmath} & \hat{k} \\ 2 & 1 & 1 \\ 1 & 2 & 1 \end{array}\right|$

\begin{aligned} &=\hat{\imath}(1 \times 1-2 \times 1)-\hat{\jmath}(2 \times 1-1 \times 1)+\hat{k}(2 \times 2-1 \times 1) \\\\ &=\hat{\imath}(-1)-\hat{\jmath}(1)+\hat{k}(3) \\\\ &=-\hat{\imath}-\hat{\jmath}+3 \hat{k} \end{aligned}

$|\vec{a} \times \vec{b}|=\sqrt{(-1)^{2}+(-1)^{2}+3^{2}}$

\begin{aligned} &=\sqrt{11} \\\\ &\frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|}=\frac{1}{\sqrt{11}}(-\hat{\imath}-\hat{\jmath}+3 \hat{k}) \end{aligned}