JEE Mains · Maths · STD 12 - 11. three dimension geometry
Let the plane \(2 x+3 y+z+20=0\) be rotated through a right angle about its line of intersection with the plane \(x-3 y+5 z=8\). If the mirror image of the point \(\left(2,-\frac{1}{2}, 2\right)\) in the rotated plane is \(B ( a , b , c )\), then
- A \(\frac{a}{8}=\frac{b}{5}=\frac{c}{-4}\)
- B \(\frac{ a }{4}=\frac{ b }{5}=\frac{ c }{-2}\)
- C \(\frac{a}{8}=\frac{b}{-5}=\frac{c}{4}\)
- D \(\frac{a}{4}=\frac{b}{5}=\frac{c}{2}\)
Answer & Solution
Correct Answer
(A) \(\frac{a}{8}=\frac{b}{5}=\frac{c}{-4}\)
Step-by-step Solution
Detailed explanation
Let equation of rotated plane be: \((2 x +3 y + z +20)+\lambda( x -3 y +5 z -8)=0\) \((2+\lambda) x +(3-3 \lambda) y +(1+5 \lambda) z +20-8 \lambda=0\) Above plane is perpendicular to \(2 x +3 y + z +20=0\) So,…
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