3D Geometry
Area of right-angled triangle with foot of perpendicular
nta_pyq_2025_apr
Grade 12
Question:
Let $P$ be the foot of the perpendicular from the point $Q(10,-3,-1)$ on the line $\dfrac{x-3}{7}=\dfrac{y-2}{-1}=\dfrac{z+1}{-2}$. Then the area of the right angled triangle $PQR$, where $R$ is the point $(3,-2,1)$, is:
$9\sqrt{15}$
$\sqrt{30}$
$8\sqrt{15}$
$3\sqrt{30}$
Step-by-Step Solution
Key Concept: Find $P$ from the perpendicularity condition, then compute $\overrightarrow{PQ}$ and $\overrightarrow{PR}$ and use $\text{Area}=\tfrac{1}{2}|\overrightarrow{PQ}\times\overrightarrow{PR}|$.
General point $P=(7\lambda+3,-\lambda+2,-2\lambda-1)$.
$\overrightarrow{QP}\cdot(7,-1,-2)=0$: $(7\lambda-7)\cdot7+(-\lambda+5)\cdot(-1)+(-2\lambda)\cdot(-2)=0$
$49\lambda-49+\lambda-5+4\lambda=0 \Rightarrow 54\lambda=54 \Rightarrow \lambda=1$.
$P=(10,1,-3)$.
$\overrightarrow{PQ}=(0,-4,2)$, $\overrightarrow{PR}=(-7,-3,4)$.
$\overrightarrow{PQ}\times\overrightarrow{PR}=\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\0&-4&2\\-7&-3&4\end{vmatrix}=(-16+6)\hat{i}-(0+14)\hat{j}+(0-28)\hat{k}=-10\hat{i}-14\hat{j}-28\hat{k}$.
Wait — checking: $=\hat{i}(-16+6)-\hat{j}(0+14)+\hat{k}(0-28)$, Area $=\tfrac{1}{2}\sqrt{100+196+784}=\tfrac{1}{2}\sqrt{1080}=3\sqrt{30}$.
Correct Answer: 4