3D Geometry
Area of triangle with two vertices on a line
nta_pyq_2025_apr
Grade 12
Question:
Let in a $\triangle ABC$, the length of the side $AC$ be 6, the vertex $B$ be $(1,2,3)$ and the vertices $A$, $C$ lie on the line $\dfrac{x-6}{3}=\dfrac{y-7}{2}=\dfrac{z-7}{-2}$. Then the area (in sq. units) of $\triangle ABC$ is:
Step-by-Step Solution
Key Concept: Find the midpoint $M$ of $AC$ as the foot of perpendicular from $B$ to the given line, compute $|BM|$, then area $=\tfrac{1}{2}\times AC\times BM$.
Let $M=(3\lambda+6,2\lambda+7,-2\lambda+7)$ be general point on the line. $\overrightarrow{BM}=(3\lambda+5,2\lambda+5,-2\lambda+4)$.
$\overrightarrow{BM}\cdot(3,2,-2)=0$: $3(3\lambda+5)+2(2\lambda+5)-2(-2\lambda+4)=0$
$9\lambda+15+4\lambda+10+4\lambda-8=0 \Rightarrow 17\lambda=-17 \Rightarrow \lambda=-1$.
$\overrightarrow{BM}=(2,3,6)$, $|BM|=\sqrt{4+9+36}=7$.
Area $=\tfrac{1}{2}\times6\times7=21$.
Correct Answer: 2