Question

The vertices $B$ and $C$ of a $\Delta$ABC lie on the line, $\frac{x+2}{3}=\frac{y-1}{0}=\frac{z}{4}$ such that $BC=5units$. Then the area (in sq. units) of this triangle, given that the point $A(1,-1,2)$, is :

• A. $2\sqrt{34}$
• B. $\sqrt{34}$
• C. 6
• D. $\sqrt{17}$

Answer 1

Answer: (B)

$\overrightarrow{AD}.\left(3\hat{i}+4\hat{k}\right)=0$
$3\left(3\lambda -3\right)+0+4\left(4\lambda -2\right)=0$
$\left(9\lambda -9\right)+\left(16\lambda -8\right)=0$
$25\lambda =17\Rightarrow \lambda =\frac{17}{25}$
$\therefore \overrightarrow{AD}=\left(\frac{51}{25}-3\right)\hat{i}+2\hat{j}+\left(\frac{68}{25}-2\right)\hat{k}$
$=\frac{24}{25}\hat{i}+2\hat{j}+\frac{18}{25}\hat{k}$
$\left|\overrightarrow{AD}\right|=\sqrt{\frac{576}{625}+4+\frac{324}{625}}$
$=\sqrt{\frac{900}{625}+4}=\sqrt{\frac{3400}{625}}$
$=\sqrt{34}.\frac{10}{25}=\frac{2}{5}\sqrt{34}$
Area of $\Delta =\frac{1}{2}\times 5\times \frac{2\sqrt{34}}{5}=\sqrt{34}$