JEE Main 2015MathematicsThree Dimensional GeometryHardMCQ

JEE Main 2015 — Three Dimensional Geometry Question with Solution

JEE Main 2015 (04 Apr)

Question

The distance of the point 1, 0, 2 from the point of intersection of the line x-23=y+14=z-212 and the plane x-y+z=16, is 

Choose an option

Show full solutionCorrect option: A
Correct answer
A 1 3

Step-by-step explanation

Let x-23=y+14=z-212=t  

General points is (2+3t,- 1+4t,2+12t)

It lies on the plane x-y+z=16.

⇒2+3t+1-4t+2+12t=16

⇒t=1

Hence, the point of intersection will be

(2+3(1),-1+4(1),2+12(1))=5,3,14

Distance from (1,0,2) = 5 - 1 2 + 3 - 0 2 + 1 4 - 2 2

= 4 2 + 3 2 + 1 2 2

= 1 3

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About this question

This is a previous-year question from JEE Main 2015, covering the Three Dimensional Geometry chapter of Mathematics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.