Lines in 3D — Foot of Perpendicular, Section Formula
DAILY_CHALLENGE
Grade None
Question:
Let $L$ be the straight line joining the points $P(1,2,-1)$ and $Q(2,3,1)$. Let $S$ be the foot of the perpendicular drawn from the point $R(4,-1,5)$ to the line $L$. Another line passing through $R$ intersects $L$ at a point $T$ such that the point $S$ divides the line segment $PT$ internally in the ratio $|PS|:|ST|=1:2$, where $|PS|$ and $|ST|$ are the lengths of the line segments $PS$ and $ST$, respectively.
Then which of the following statements is (are) TRUE?
The orthocentre of the triangle $PRT$ is $\left(\dfrac{23}{5},-4,\dfrac{31}{5}\right)$
The orthocentre of the triangle $PRT$ is $(4,3,5)$
The area of the triangle $PRT$ is $6\sqrt{5}$
The area of the triangle $PRT$ is $18\sqrt{5}$
Step-by-Step Solution
Key Concept: In 3D, the orthocentre of a triangle lies on the plane of the triangle. Use the plane equation as a third constraint alongside two altitude perpendicularity conditions to uniquely determine H.
**Step 1: Find foot of perpendicular S**
Direction of $L$: $\vec{PQ}=(1,1,2)$. Parametric: $(1+t,2+t,-1+2t)$. $\vec{RS}\cdot\vec{PQ}=0$: $(t-3)+(t+3)+(4t-12)=0 \Rightarrow 6t=12 \Rightarrow t=2$. $S=(3,4,3)$.
**Step 2: Find T using section formula**
$S$ divides $PT$ in ratio $1:2$ internally: $S=\frac{2P+T}{3} \Rightarrow T=3S-2P=3(3,4,3)-2(1,2,-1)=(7,8,11)$. Verify on $L$: $t=6$ gives $(7,8,11)$. ✓
**Step 3: Area of triangle PRT**
$\vec{PR}=(3,-3,6)$, $\vec{PT}=(6,6,12)$. $\vec{PR}\times\vec{PT}=(-72,0,36)$. Area $=\frac{1}{2}\sqrt{72^2+36^2}=\frac{1}{2}\cdot36\sqrt{5}=18\sqrt{5}$. (D) TRUE ✓
**Step 4: Find orthocentre H**
$H$ lies on plane $2x-z=3$ (plane of $\triangle PRT$) and satisfies $\vec{RH}\perp\vec{PT}$ and $\vec{TH}\perp\vec{PR}$. Solving: $y=-4$, $x+2z=17$, $z=2x-3$. Get $x=23/5$, $z=31/5$. $H=(23/5,-4,31/5)$. (A) TRUE ✓
Correct Answer: A, D