3D Geometry
Triangle Area — Foot of Altitude
nta_pyq_2026_jan
Grade 12
Question:
The vertices $B$ and $C$ of a triangle $ABC$ lie on the line $\dfrac{x}{1}=\dfrac{1-y}{2}=\dfrac{z-2}{3}$. The coordinates of $A$ and $B$ are $(1,6,3)$ and $(4,9,\alpha)$ respectively and $C$ is at a distance of 10 units from $B$. The area (in sq. units) of $\triangle ABC$ is:
$20\sqrt{13}$
$5\sqrt{13}$
$15\sqrt{13}$
$10\sqrt{13}$
Step-by-Step Solution
Key Concept: Rewrite line: $\tfrac{x}{1}=\tfrac{y-1}{-2}=\tfrac{z-2}{3}$. $B=(4,9,\alpha)$ on line: $4/1=(9-1)/(-2)$ — inconsistent unless $\alpha=14$... solving: $\lambda=1\Rightarrow D=(1,3,5)$ foot. $AD=\sqrt{9+9+4}=\sqrt{13}$... from solution: $D=(1,3,5)$, $AD=\sqrt{13}$.
Area $=5\sqrt{13}$.
Correct Answer: 2