Q. 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 $BC = 5\, units$. Then the area (in sq. units) of this triangle, given that the point $A(1, -1, 2)$, is :
Solution:
$\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} \times5 \times\frac{2\sqrt{34}}{5} = \sqrt{34} $
