#### Explain solution RD Sharma class 12 chapter Straight Line in Space exercise 24.1 question 9 sub question (iv) maths

Answer         : $\frac{49}{2} \text { sq.units }$

Hint               : To solve this we use area of parallelogram formula

Given           : $d_{1}=2 \hat{\imath}+3 \hat{\jmath}+6 \hat{k} \quad, d_{2}=3 \hat{\imath}-6 \hat{\jmath}+2 \hat{k}$

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

\begin{aligned} &d_{1} \times d_{2}=\left|\begin{array}{ccc} \hat{\imath} & \hat{\jmath} & \hat{k} \\ 2 & 3 & 6 \\ 3 & -6 & 2 \end{array}\right| \\\\ \quad &=\hat{\imath}(3 \times 2-(-6 \times 6))-\hat{\jmath}(2 \times 2-3 \times 6)+\hat{k}(2 \times-6-3 \times 3) \\\\ &\quad=42 \hat{\imath}+14 \hat{\jmath}-21 \hat{k} \end{aligned}

$\begin{gathered} A=\frac{1}{2}\left(d_{1} \times d_{2}\right) \\\\ =\frac{1}{2} \sqrt{42^{2}+14^{2}+-21^{2}} \\\\ =\frac{1}{2} \sqrt{144+196+9} \end{gathered}$

\begin{aligned} &=\frac{1}{2} \sqrt{349} \\\\ &=\frac{49}{2} \text { sq.units } \end{aligned}