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JEE Main 20249 AprilEvening Shift
Exam
JEE Main
Year
2024
Date
9 April
Shift
Evening Shift
Subject
Mathematics
Chapter
Three Dimensional Geometry

JEE Main 2024 · 9 April · Evening Shift · Mathematics · Three Dimensional Geometry

The square of the distance of the image of the point (6,1,5) in the...

The square of the distance of the image of the point (6,1,5)(6,1,5) in the line x−13=y2=z−24\frac{x-1}{3}=\frac{y}{2}=\frac{z-2}{4}, from the origin is _________The square of the distance of the image of the point (6,1,5) in the line (x-1)/(3)=(y)/(2)=(z-2)/(4) , from the origin

Correct answer

0

Step-by-step solution

The square of the distance of the image of the point (6,1,5) in the line (x-1)/(3)=(y)/(2)=(z-2)/(4) , from the origin
Let M(3λ+1,2λ,4λ+2)\mathrm{M}(3 \lambda+1,2 \lambda, 4 \lambda+2) AM→⋅b→=0\overrightarrow{\mathrm{AM}} \cdot \overrightarrow{\mathrm{b}}=0 ⇒9λ−15+4λ−2+16λ−12=0\Rightarrow \quad 9 \lambda-15+4 \lambda-2+16 \lambda-12=0 ⇒29λ=29\Rightarrow \quad 29 \lambda=29 ⇒λ=1\Rightarrow \quad \lambda=1 M(4,2,6),\mathrm{M}(4,2,6),\mathrm{I}=(2,3,7)=(2,3,7) Required Distance =4+9+49=62=\sqrt{4+9+49}=\sqrt{62} Ans. 62
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