#### Provide solution for RD Sharma maths class 12 chapter Vector or Cross Product exercise 24.1 question 34

$\sqrt{61}$ Square units

Hint:

To solve this we use area of triangle

Given:

$A(1,1,2) ; B(2,3,5) ; C(1,5,5)$

Solution:

\begin{aligned} &D=\frac{1}{2}(\overrightarrow{A B} \times \overrightarrow{B C}) \\\\ &\Rightarrow \overrightarrow{A B}=\overrightarrow{O B}-\overrightarrow{O A} \\\\ &=(2 \hat{i}+3 \hat{j}+5 \hat{k})-(\hat{i}+\hat{j}+2 \hat{k}) \\\\ &=\hat{i}+2 \hat{j}+3 \hat{k} \end{aligned}

\begin{aligned} &\Rightarrow \overrightarrow{B C}=\overrightarrow{O C}-\overrightarrow{O B} \\\\ &=(\hat{i}+5 \hat{j}+5 \hat{k})-(2 \hat{i}+3 \hat{j}+5 \hat{k}) \\\\ &=-\hat{i}+2 \hat{j} \end{aligned}

\begin{aligned} &D=\frac{1}{2}(\overrightarrow{A B} \times \overrightarrow{B C}) \\\\ &=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ -1 & 2 & 0 \end{array}\right| \end{aligned}

\begin{aligned} &=\hat{i}(0-6)+\hat{j}(0-(-3))+\hat{k}(2-(-2)) \\\\ &=-6 \hat{i}-3 \hat{j}+4 \hat{k} \end{aligned}

\begin{aligned} &=(\overrightarrow{A B} \times \overrightarrow{B C})=\sqrt{(-6)^{2}+(-3)^{2}+(4)^{2}} \\\\ &=\sqrt{36+9+6} \\\\ &=\sqrt{61} \text { square units } \end{aligned}