Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

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

The intercepts on the coordinate axes are $a=2$, $b=3$, and $c=4$.

The coordinates of the points where the plane cuts the axes are $A(2,0,0)$, $B(0,3,0)$, and $C(0,0,4)$.

The area of the triangle $ABC$ can be found using the areas of its projections on the coordinate planes, denoted as ${\Delta }_{xy}$, ${\Delta }_{yz}$, and ${\Delta }_{zx}$.

${\Delta }_{xy}=\frac{1}{2}ab=\frac{1}{2}(2)(3)=3$

${\Delta }_{yz}=\frac{1}{2}bc=\frac{1}{2}(3)(4)=6$

${\Delta }_{zx}=\frac{1}{2}ca=\frac{1}{2}(4)(2)=4$

The total area of the triangle is given by $\Delta =\sqrt{{\Delta }_{xy}^2+{\Delta }_{yz}^2+{\Delta }_{zx}^2}$.

$\Delta =\sqrt{3^2+6^2+4^2}$

$\Delta =\sqrt{9+36+16}=\sqrt{61}$ sq. units.

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