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. $5\sqrt{17}$
• B. $\sqrt{34}$
• C. $6$
• D. $2\sqrt{34}$

Answer 1

The correct option is B $\sqrt{34}$


Let $AD$ be the perpendicular drawn from point $A$ on $BC$.
The vertices $B$ and $C$ of $△ABC$ lie on the line $\frac{x+2}{3}=\frac{y-1}{0}=\frac{z}{4}$
Coordinates of $D$ can be expressed as $(3\lambda -2,1,4\lambda )$.
DR's of $AD$ are $(3\lambda -3,2,4\lambda -2)$
$AD$ is perpendicular to $BC$.
$\therefore 3(3\lambda -3)+0\left(2\right)+4(4\lambda -2)=0$
$\Rightarrow 9\lambda -9+0+16\lambda -8=0$
$\Rightarrow \lambda =\frac{17}{25}$

Coordinates of $D$ are $(\frac{1}{25},1,\frac{68}{25})$
Length of $AD=\sqrt{{(\frac{1}{25}-1)}^2+(1-(-1){)}^2+{(\frac{68}{25}-2)}^2}$
$=\sqrt{\frac{576}{625}+4+\frac{324}{625}}$
$=\sqrt{\frac{3400}{625}}=\frac{2}{5}\sqrt{34}$

Area of triangle$=\frac{1}{2}\times 5\times \frac{2}{5}\sqrt{34}=\sqrt{34}$ sq. units