Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

The equation of the plane is $\frac{x}{3}+\frac{y}{2}-\frac{z}{4}=1$.

The plane intersects the coordinate axes at $A(3,0,0)$, $B(0,2,0)$, and $C(0,0,-4)$.

The position vectors of the points are $\vec{a}=3\hat{i}$, $\vec{b}=2\hat{j}$, and $\vec{c}=-4\hat{k}$.

The vectors representing the sides of the triangle are:

$\overrightarrow{AB}=\vec{b}-\vec{a}=-3\hat{i}+2\hat{j}$

$\overrightarrow{AC}=\vec{c}-\vec{a}=-3\hat{i}-4\hat{k}$

The cross product of these vectors is:

$\overrightarrow{AB}\times \overrightarrow{AC}=(-3\hat{i}+2\hat{j})\times (-3\hat{i}-4\hat{k})=-8\hat{i}-12\hat{j}+6\hat{k}$

The area of $△ABC$ is given by $\frac{1}{2}|\overrightarrow{AB}\times \overrightarrow{AC}|$.

Area $=\frac{1}{2}\sqrt{(-8{)}^2+(-12{)}^2+6^2}$

Area $=\frac{1}{2}\sqrt{64+144+36}$

Area $=\frac{1}{2}\sqrt{244}$

Area $=\frac{1}{2}\times 2\sqrt{61}=\sqrt{61}$ sq. units.

Answer: $\sqrt{61}$ sq. units