#### Please solve RD Sharma class 12 chapter Vector or Cross Product exercise 24.1 question 29 maths textbook solution

$\frac{\sqrt{61}}{2} \text { square units }$

Hint:

To solve this we use area of triangle ABC

Given:

\begin{aligned} &A(2,3,5) \\ &B(3,5,8) \\ &C(2,7,8) \end{aligned}

$\text { Area of } \Delta \mathrm{ABC}=\frac{1}{2}|\overrightarrow{A B} \times \overrightarrow{B C}|$

Solution:

\begin{aligned} &\overrightarrow{A B}=\hat{i}+2 \hat{j}+3 \hat{k} \\\\ &\overrightarrow{B C}=-\hat{i}+2 \hat{j} \\\\ &\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} \\\\ &|\overrightarrow{A B} \times \overrightarrow{B C}|=\sqrt{(-6)^{2}+(-3)^{2}+(4)^{2}} \end{aligned}

\begin{aligned} &|\overrightarrow{A B} \times \overrightarrow{B C}|=\sqrt{36+9+16} \\\\ &|\overrightarrow{A B} \times \overrightarrow{B C}|=\sqrt{61}\\\\ \end{aligned}

$Area \; of \; \Delta \mathrm{ABC}=\frac{1}{2}|\sqrt{61}|$

$=\frac{\sqrt{61}}{2} square\; units$