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

Answer         : $\frac{\sqrt{26}}{2} \text { sq.units }$

Hint               : To solve this we use area of parallelogram formula

Given            : $3 \hat{\imath}+4 \hat{\jmath} \text { and } \hat{\imath}+\hat{\jmath}+\hat{k}$

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

$\begin{gathered} d_{1}=3 \hat{\imath}+4 \hat{\jmath} \\ d_{2}=\hat{\imath}+\hat{\jmath}+\hat{k} \end{gathered}$

\begin{aligned} &d_{1} \times d_{2}=\left|\begin{array}{lll} \hat{\imath} & \hat{\jmath} & \hat{k} \\ 3 & 4 & 0 \\ 1 & 1 & 1 \end{array}\right| \\\\ &=\hat{\imath}(4 \times 1-0 \times 1)-\hat{j}(3 \times 1-0 \times 1)+\hat{k}(3 \times 1-4 \times 1) \\\\ &=4 \hat{\imath}-3 \hat{\jmath}-1 \hat{k} \end{aligned}

\begin{aligned} &A \quad=\frac{1}{2}\left(d_{1} \times d_{2}\right) \\\\ &=\frac{1}{2} \sqrt{(-4)^{2}+(-3)^{2}+(-1)^{2}} \\\\ &=\frac{1}{2} \sqrt{16+9+1} \\\\ &=\frac{\sqrt{26}}{2} \text { sq.units } \end{aligned}