#### Explain solution RD Sharma class 12 chapter Straight Line in Space exercise 24.1 question 8 sub question (iv) maths

Answer          :  $4 \sqrt{2} \text { sq. units }$

Hint                : To solve this equation we use area of parallelogram

Given              $\hat{\imath}-3 \hat{\jmath}+\hat{k} \text { and } \hat{\imath}+\hat{\jmath}+\hat{k}$

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

Area of parallelogram $=|\vec{a} \times \vec{b}|$

\begin{aligned} &\vec{a} \times \vec{b}=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & -3 & 1 \\ 1 & 1 & 1 \end{array}\right| \\\\ &\vec{a} \times \vec{b}=\hat{i}[-3.1-1.1]-\hat{j}[1.1-1.1]+\hat{k}[1.1-1 \cdot(-3)] \end{aligned}

\begin{aligned} &=\hat{\mathrm{i}}[-3-1]-\hat{\mathrm{j}}[1-1]+\hat{\mathrm{k}}[1+3] \\\\ &\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}=-4 \hat{\mathbf{i}}+4 \hat{\mathrm{k}} \end{aligned}

Area of parallelogram

\begin{aligned} &=\sqrt{-4^{2}+4^{2}} \\\\ &=4 \sqrt{2} \end{aligned}