JEE Main 2019 — Three Dimensional Geometry Question with Solution

The points $B$ and $C$ lie on the line $\frac{x+2}{3}=\frac{y-1}{0}=\frac{z}{4}.$

Draw perpendicular $AD$ on the line $BC.$

Clearly area of $\Delta ABC=\frac{1}{2}\cdot AD\cdot BC$

To find a point on the line, let $\frac{x+2}{3}=\frac{y-1}{0}=\frac{z}{4}=r$

$\Rightarrow x+2=3r, y-1=0, z=4r$

$\Rightarrow x=3r-2, y=1, z=4r$

Thus, the point $D\equiv \left(3r-2, 1, 4r\right)$

The direction ratios of a line joining two points $\left(x_1, y_1, z_1\right)$ and $\left(x_2, y_2, z_2\right)$ are $<x_2-x_1, y_2-y_1, z_2-z_1>$

Thus, the direction ratios of $AD$ are $<3r-2-1, 1+1, 4r-2>=<3r-3, 2, 4r-2>$

Since $AD$ is perpendicular to given line, hence the dot product of their direction ratios is zero.

$\Rightarrow 3·\left(3r-3\right)+0·2+4·\left(4r-2\right)=0$

$\Rightarrow 9r-9+16r-8=0$

$\Rightarrow 25r-17=0$

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

$\Rightarrow D\equiv \left(3\times \frac{17}{25}-2, 1, 4\times \frac{17}{25}\right)=\left(\frac{1}{25}, 1, \frac{68}{25}\right)$

The distance between the points $\left(x_1, y_1, z_1\right) \& \left(x_2, y_2, z_2\right)$ is $\sqrt{{\left(x_1-x_2\right)}^2+{\left(y_1-y_2\right)}^2+{\left(z_1-z_2\right)}^2}$

$\Rightarrow AD=\sqrt{{\left(\frac{1}{25}-1\right)}^2+{\left(1+1\right)}^2+{\left(\frac{68}{25}-2\right)}^2}$

$\Rightarrow AD=\sqrt{\frac{576}{625}+4+\frac{324}{625}}$

$\Rightarrow AD=\sqrt{\frac{136}{25}}$

$\Rightarrow AD=\frac{2}{5}\sqrt{34}$ units

Hence, the area of $\Delta ABC=\frac{1}{2}\cdot \left(\frac{2}{5}\sqrt{34}\right)\cdot 5=\sqrt{34}$ sq units.