Question

The vertices $B$ and $C$ of a $△ABC$ lie on the line, $\frac{x+2}{3}=\frac{y-1}{0}=\frac{z}{4}$ such that $BC=5$ units. 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. $5\sqrt{17}$

Answer 1

Answer: (B)

$\sqrt{34}$

$\overline{AD}.(3\hat{i}+4\hat{k})=0$

$3(3\lambda -3)+0+4(4\lambda -2)=0$

$(9\lambda -9)+(16\lambda -8)=0$

$25\lambda =17\Rightarrow \lambda =\frac{17}{25}$

$\therefore \overline{AD}=(\frac{51}{25}-3)\hat{i}+2\hat{j}+(\frac{68}{25}-2)\hat{k}=\frac{24}{25}\hat{i}+2\hat{j}+\frac{18}{25}\hat{k}$

$\left|\overline{AD}\right|=\sqrt{\frac{576}{625}+4+\frac{324}{625}}=\sqrt{\frac{3400}{625}}=\frac{2}{5}\sqrt{34}$

Area of triangle=$\Delta =\frac{1}{2}\times 5\times \frac{2\sqrt{34}}{5}=\sqrt{34}$