#### Please solve RD Sharma class 12 chapter Straight Line in Space exercise 24.1 question 9 sub question (i) maths textbook solution

Answer          : $\frac{15}{2} \text { sq.units }$

Hint                : To solve this we use area of parallelogram formula

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

Solution         : $d_{1}=-2 \hat{\imath}+\hat{\jmath}-2 \hat{k}$

\begin{aligned} &d_{2}=4 \hat{\imath}-\hat{\jmath}-3 \hat{k} \\\\ &d_{1} \times d_{2}=\left|\begin{array}{ccc} \hat{\imath} & \hat{\jmath} & \hat{k} \\ -2 & 1 & -2 \\ 4 & -1 & -3 \end{array}\right| \end{aligned}

\begin{aligned} &=\hat{\imath}(1 \times-3-(-2)(-1))-\hat{j}(-2 \times-3-4 \times 2)+\hat{k}(-2 \times-1-4 \times-1) \\\\ &=-5 \mathrm{i}-14 \mathrm{j}-2 \mathrm{k} \end{aligned}

\begin{aligned} &A=\frac{1}{2}\left(d_{1} \times d_{2}\right) \\\\ &=\frac{1}{2} \sqrt{(-5)^{2}+14^{2}+(-2)^{2}} \\\\ &=\frac{1}{2} \sqrt{25+196+4} \end{aligned}

\begin{aligned} &=\frac{1}{2} \sqrt{225} \\\\ &=\frac{15}{2} \text { sq.units } \end{aligned}

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