Vectors & 3D Geometry
Line through P parallel to L₁, intersecting two planes
MJAT_TS7_P2
Grade 12
Question:
A line from $P(1,3,2)$ parallel to $L_1: \frac{x-1}{2}=\frac{y-4}{2}=\frac{z-1}{6}$ intersects plane $L_1: x-y+3z=6$ at $Q$. Another line through $Q$ perpendicular to $L_1$ intersects plane $L_2: 2x-y+z=-4$ at $R$. Which is/are TRUE?
A) $|PQ|=6$
B) $R=(1,6,3)$
C) Centroid of $\triangle PQR=\left(\frac{4}{3},\frac{14}{3},\frac{5}{3}\right)$
D) Perimeter of $\triangle PQR=2+\sqrt{6}+\sqrt{11}$
Step-by-Step Solution
Key Concept: Line through $P(1,3,2)$ with direction $(2,2,6)$: parametric $(1+2t,3+2t,2+6t)$. Intersect $x-y+3z=6$: $(1+2t)-(3+2t)+3(2+6t)=6\Rightarrow 1+2t-3-2t+6+18t=6\Rightarrow 4+18t=6\Rightarrow t=1/9$. $Q=(1+2/9,3+2/9,2+6/9)=(11/9,29/9,24/9)$.
A ✓, C ✓. Answer: A, C.
Correct Answer: AC