3D Geometry
Intersection of Lines in 3D — Distance from Another Line
nta_pyq_2024_jan
Grade 12
Question:
The lines $\dfrac{x-2}{2}=\dfrac{y}{-2}=\dfrac{z-7}{16}$ and $\dfrac{x+3}{4}=\dfrac{y+2}{3}=\dfrac{z+2}{1}$ intersect at the point $P$. If the distance of $P$ from the line $\dfrac{x+1}{2}=\dfrac{y-1}{3}=\dfrac{z-1}{1}$ is $l$, then $14l^2$ is equal to
Step-by-Step Solution
Key Concept: Find intersection $P$ of the two lines, then compute distance from $P$ to the third line using the formula for point-to-line distance in 3D.
$P=(1,1,-1)$. $14l^2=108$.
Correct Answer: 108