#### Need solution for RD Sharma maths class 12 chapter Straight Line in Space exercise 24.1 question 8 sub question (iii)

Answer         :  $10 \sqrt{3}\; s q . \text { units }$

Hint               : To solve this equation we use area of parallelogram

Given             : $3 \hat{\imath}+\hat{\jmath}-2 \hat{k} \quad \text { and } \hat{\imath}-3 \hat{\jmath}+4 \hat{k}$

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

\begin{aligned} &\vec{a}=3 \hat{\imath}+\hat{\jmath}-2 \hat{k} \\\\ &\vec{b}=\hat{\imath}-3 \hat{j}+4 \hat{k} \end{aligned}

\begin{aligned} &\vec{a} \times \vec{b}=\left|\begin{array}{ccc} \hat{\imath} & \hat{\jmath} & \hat{k} \\ 3 & 1 & -2 \\ 1 & -3 & 4 \end{array}\right| \\\\ &=\hat{\imath}(1 \times 4-(-2)(-3))-\hat{\jmath}(3 \times 4-2 \times-1)+\hat{k}(3 \times-3-1 \times 1) \\\\ &=-2 \hat{\imath}-14 \hat{\jmath}-10 \hat{k} \end{aligned}

\begin{aligned} &|\vec{a} \times \vec{b}|=\sqrt{(-2)^{2}+(-14)^{2}+(-10)^{2}} \\\\ &=\sqrt{4+196+100} \\\\ &=\sqrt{300} \\\\ &=10 \sqrt{3} \text { sq.units } \end{aligned}