3D Geometry
Distance Along a Given Line Direction
nta_pyq_2024_jan
Grade 12
Question:
The distance, of the point $(7,-2,11)$ from the line $\dfrac{x-6}{1}=\dfrac{y-4}{0}=\dfrac{z-8}{-3}$ along the line $\dfrac{x-5}{2}=\dfrac{y-1}{-3}=\dfrac{z-5}{6}$, is:
Step-by-Step Solution
Key Concept: From $(7,-2,11)$, move along direction $(2,-3,6)$: $(7+2t,-2-3t,11+6t)$. Find $t$ such that this point lies on the first line $\frac{x-6}{1}=\frac{y-4}{0}=\frac{z-8}{-3}$. Then distance $=|t|\sqrt{4+9+36}=7|t|$.
$t=-2$. Distance $=7\times2=14$.
Correct Answer: 2