3D Geometry
Intersection of Lines — Distance from a Point
nta_pyq_2024_apr
Grade 12
Question:
Let $d$ be the distance of the point of intersection of the lines $\dfrac{x+6}{3}=\dfrac{y}{2}=\dfrac{z+1}{1}$ and $\dfrac{x-7}{4}=\dfrac{y-9}{3}=\dfrac{z-4}{2}$ from the point $(7,8,9)$. Then $d^2+6$ is equal to:
Step-by-Step Solution
Key Concept: Parametrise: Line 1: $(3\lambda-6,2\lambda,-\lambda+(-1))=(3\lambda-6,2\lambda,\lambda-1)$. Line 2: $(4\mu+7,3\mu+9,2\mu+4)$. Equate and solve.
Intersection at $(3,6,2)$. $d^2=69$. $d^2+6=75$.
Correct Answer: 4