#### Need solution for RD Sharma maths class 12 chapter Vector or Cross Product exercise 24.1 question 35

$\frac{1}{2} \sqrt{274}$  Square units

Hint:

To solve this we use area of triangle

Given:

$A(1,2,3) ; B(2,-1,4) ; C(4,5,-1)$

Solution:   $D=\frac{1}{2}(\overrightarrow{A B} \times \overrightarrow{A C})$

\begin{aligned} &\Rightarrow \overrightarrow{A B}=\overrightarrow{O B}-\overrightarrow{O A} \\\\ &=(2 \hat{i}-\hat{j}+4 \hat{k})-(\hat{i}+2 \hat{j}+3 \hat{k}) \\\\ &=\hat{i}-3 \hat{j}+\hat{k} \end{aligned}

\begin{aligned} &\Rightarrow \overrightarrow{B C}=\overrightarrow{O C}-\overrightarrow{O B} \\\\ &=(\hat{i}+2 \hat{j}+3 \hat{k})-(2 \hat{i}-\hat{j}+4 \hat{k}) \\\\ &=3 \hat{i}+3 \hat{j}-4 \hat{k} \end{aligned}

\begin{aligned} &D=\frac{1}{2}(\overrightarrow{A B} \times \overrightarrow{B C}) \end{aligned} \mid

$=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & -3 & 1 \\ 3 & 3 & -4 \end{array}\right|$

\begin{aligned} &=\hat{i}(12-3)+\hat{j}(-4-3)+\hat{k}(3+9) \\\\ &=9 \hat{i}+7 \hat{j}+12 \hat{k} \\\\ &=(\overrightarrow{A B} \times \overrightarrow{A C})=\sqrt{81+49+144} \\\\ &=\frac{1}{2} \sqrt{274} \text { square units } \end{aligned}