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  • Let the plane be rotated through a right angle about its line of i

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 \mathrm{B}(\mathrm{a}, \mathrm{b}, \mathrm{c}), then:

Option: 1

\begin{aligned} &\frac{a}{8}=\frac{b}{5}=\frac{c}{-4} \\ \end{aligned}


Option: 2

\frac{a}{4}=\frac{b}{5}=\frac{c}{-2} \\


Option: 3

\frac{a}{8}=\frac{b}{-5}=\frac{c}{4} \\


Option: 4

\frac{a}{4}=\frac{b}{5}=\frac{c}{2}


Answers (1)

best_answer

Let equation of rotated plane be :

\mathrm{(2 x+3 y+z+20)+\lambda(x-3 y+5 z-8)=0} \\

\mathrm{(2+\lambda) x+(3-3 \lambda) y+1(1+5 \lambda) z+20-8 \lambda=0}

Above plane is perpendicular to \mathrm{2x+3y+z+20=0}

So, \mathrm{(2+\lambda) \cdot 2+(3-3 \lambda) \cdot 3+(1+5 \lambda) \cdot 1=0 \Rightarrow \lambda=7}

\mathrm{\Rightarrow } Equation of rotated plane:

\mathrm{x-2y+4z-4=0}

Mirror image of   \mathrm{A\left ( 2,\frac{-1}{2},2 \right )} in rotated plane is  \mathrm{B\left (a,b,c \right )}

Equation of   \mathrm{A B: \frac{x-2}{1}=\frac{y+1 / 2}{-2}=\frac{z-2}{4}=k}

Let coordinate of   \mathrm{B\: b e\left(2+k,-\frac{1}{2}-2 k, 2+4 k\right)}

Midpoint of  \mathrm{A B \text { is }\left(2+\frac{k}{2},-\frac{1}{2}-k, 2+2 k\right)} which  will lie on the plane \mathrm{x-2 y+4 z-4=0}

\mathrm{Hence\: k=\frac{-2}{3}}

\mathrm{Therefore\: B\: is \left(\frac{4}{3}, \frac{5}{6},-\frac{2}{3}\right) \equiv\left(\frac{8}{6}, \frac{5}{6},\frac{-4}{6}\right) }

\mathrm{\text { So, } \frac{a}{8}=\frac{b}{5}=\frac{c}{-4}}

Hence the correct answer is option 1.

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