#### Provide solution for RD Sharma maths class 12 chapter Straight Line in Space exercise 24.1 question 7 sub question (ii)

Answer         : $2 \hat{\imath}-2 \hat{\jmath}+\hat{k}$

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

Given            :$\vec{a}=3 \hat{\imath}+\hat{\jmath}-4 \hat{k} \vec{b}=6 \hat{\imath}+5 \hat{\jmath}-2 \hat{k}$      magnitude = 3

Solution       :$3 \times \frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|}$

$\vec{a} \times \vec{b}=\left|\begin{array}{ccc} \hat{\imath} & \hat{j} & \hat{k} \\ 3 & 1 & -4 \\ 6 & 5 & -2 \end{array}\right|$

\begin{aligned} &=\hat{\imath}(1 \times-2-(5 \times-4))-\hat{\jmath}(3 \times-2-(-4) \times 6)+\hat{k}(3 \times 5-1 \times 6) \\\\ &=18 \hat{\imath}-18 \hat{\jmath}+9 \hat{k} \\\\ &=9(2 \hat{\imath}-2 \hat{\jmath}+\hat{k}) \end{aligned}

\begin{aligned} &|\vec{a} \times \vec{b}|=9 \sqrt{2^{2}+(-2)^{2}+1^{2}} \\\\ &=9 \sqrt{4+4+1} \\\\ &=9 \sqrt{9} \\\\ &=9 \times 3 \end{aligned}

\begin{aligned} &=27 \\\\ &3 \times \frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|}=\frac{3 \times 9(2 \hat{i}-2 \hat{j}+\hat{k})}{27} \\\\ &=2 \hat{\imath}-2 \hat{j}+\hat{k} \end{aligned}