#### Please solve RD Sharma class 12 chapter Straight Line in Space exercise 24.1 question 3 sub question (i) maths textbook solution

Answer   :$\frac{1}{3}(-\hat{\imath}+2 \hat{\jmath}+2 \hat{k})$

Hint         : To solve this equation , use magnitude and $\vec{a} \times \vec{b}$

Given      :$4 \hat{\imath}-\hat{\jmath}+3 \hat{k} ;-2 \hat{\imath}+\hat{\jmath}-2 \hat{k}$

Solution :$\vec{a}=4 \hat{\imath}-\hat{\jmath}+3 \hat{k}$

$\vec{b}=-2 \hat{\imath}+\hat{\jmath}-2 \hat{k}$

\begin{aligned} \vec{a} \times \vec{b} &=\left|\begin{array}{ccc} \hat{\imath} & \hat{\jmath} & \hat{k} \\ 4 & -1 & 3 \\ -2 & 1 & -2 \end{array}\right| \\\\ &=\hat{i}(1 \times 2-3 \times 1)-\hat{j}(4 \times-2-3 \times-2)+\hat{k}(4 \times 1-(-1 \times-2)) \end{aligned}

$=-\hat{i}+2 \hat{j}+2 \hat{k}$

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

\begin{aligned} &=\sqrt{9} \\\\ &=3 \\\\ &\frac{\vec{a} \times \vec{b}}{|\vec{a}+\vec{b}|}=\frac{1}{3}(-\hat{\imath}+2 \hat{\jmath}+2 \hat{k}) \end{aligned}