3D Geometry
Intersection of lines; distance from a plane
nta_pyq_2023_jan
Grade 12

Question:

If the lines $\frac{x-1}{1} = \frac{y-2}{2} = \frac{z+3}{1}$ and $\frac{x-a}{2} = \frac{y+2}{3} = \frac{z-3}{1}$ intersect at the point P, then the distance of the point P from the plane $z = a$ is:
16
28
10
22

Step-by-Step Solution

Key Concept: Set parametric points on both lines equal to find intersection, determine $a$, then compute distance from plane.
$P=(23,46,19)$, $a=-9$. Distance from $z=-9$ is $|19-(-9)|=28$. Answer: (2)
Correct Answer: 28

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