#### Provide solution for RD Sharma maths class 12 chapter Straight Line in Space exercise 24.1 question 9 sub question (ii)

Answer          : $\frac{\sqrt{6}}{2} \text { sq.units }$

Hint                : To solve this we use area of parallelogram formula

Given             : $2 \hat{\imath}+\hat{k} \text { and } \hat{\imath}+\hat{\jmath}+\hat{k}$

Solution        : Area of parallelogram   $=\frac{1}{2}\left(d_{1} \times d_{2}\right)$

$\begin{gathered} d_{1}=2 \hat{\imath}+\hat{k} \\\\ d_{2}=\hat{\imath}+\hat{\jmath}+\hat{k} \end{gathered}$

\begin{aligned} &d_{1} \times d_{2}\left|\begin{array}{ccc} \hat{\imath} & \hat{\jmath} & \hat{k} \\ 2 & 0 & 1 \\ 1 & 1 & 1 \end{array}\right| \\\\ &=\hat{\imath}(0 \times 1-1 \times 1)-\hat{\jmath}(2 \times 1-1 \times 1)+\hat{k}(2) \\\\ &=-\hat{\imath}-\hat{\jmath}+2 \hat{k} \end{aligned}

\begin{aligned} &\frac{1}{2}\left|d_{1} \times d_{2}\right|=\frac{1}{2} \sqrt{(-1)^{2}+(-1)^{2}+2^{2}} \\\\ &=\frac{1}{2} \sqrt{1+1+4} \\\\ &=\frac{\sqrt{6}}{2} \text { sq.units } \end{aligned}

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