3D Geometry
Intersection of two lines; distance between two points
nta_pyq_2025_apr
Grade 12
Question:
Let the line passing through the points $(-1,2,1)$ and parallel to the line $\dfrac{x-1}{2}=\dfrac{y+1}{3}=\dfrac{z}{4}$ intersect the line $\dfrac{x+2}{3}=\dfrac{y-3}{2}=\dfrac{z-4}{1}$ at the point $P$. Then the distance of $P$ from the point $Q(4,-5,1)$ is:
$5$
$5\sqrt{5}$
$5\sqrt{6}$
$10$
Step-by-Step Solution
Key Concept: Write both lines in parametric form and solve for the intersection point $P$, then compute $|PQ|$.
Line 1: $(2\lambda-1,3\lambda+2,4\lambda+1)$.
Line 2: $(3\mu-2,2\mu+3,\mu+4)$.
$2\lambda-1=3\mu-2 \Rightarrow 2\lambda=3\mu-1$ ... (i)
$3\lambda+2=2\mu+3 \Rightarrow 3\lambda=2\mu+1$ ... (ii)
$4\lambda+1=\mu+4 \Rightarrow 4\lambda=\mu+3$ ... (iii)
From (ii) and (iii): $3(\mu+3)/4=(2\mu+1)/3$... Solving: $\lambda=1$, $\mu=1$. $P=(1,5,5)$.
$PQ=\sqrt{(4-1)^2+(-5-5)^2+(1-5)^2}=\sqrt{9+100+16}=\sqrt{125}=5\sqrt{5}$.
Correct Answer: 2