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  • If the equation of plane passing through the mirror image of a point with respect to line <img alt="\frac{x+1}{2}=\frac{y-3}{1}=\frac{z+2}{-1}" src="/la

If the equation of plane passing through the mirror image of a point \left ( 2,3,1 \right ) with respect to line \frac{x+1}{2}=\frac{y-3}{1}=\frac{z+2}{-1} and containing the line \frac{x-2}{3}=\frac{1-y}{2}=\frac{z+1}{1}is \alpha x+\beta y+\gamma z=24, then \alpha +\beta +\gamma is equal to :
 
Option: 1 19
Option: 2 21
Option: 3 18
Option: 4 20

Answers (1)

best_answer

Let point A be (2, 3, 1)

\\L_{1}: \frac{x+1}{2}=\frac{y-3}{1}=\frac{z+2}{-1}=\lambda\\\text{let any point on line L}_1\text{ is B}\;(2\lambda-1,\lambda+3,-\lambda-2)

\text { Now if } \mathrm{B} \text { is foot of perpendicular of } \mathrm{A} \text { in } \mathrm{L}_{1} \text { , then } \mathrm{AB} \perp \mathrm{L}_{1}

\\ 2(2 \lambda-3)+1(\lambda)-(-\lambda-3)=0 \\ 6 \lambda-3=0 \Rightarrow \lambda=\frac{1}{2} \\ \text {Hence } B\left(0, \frac{7}{2},-\frac{5}{2}\right)

\text {Now image } A^{\prime}(-2,4,-6)

Now equation of plane containing A'(-2,4,-6) and line 

\mathrm{L}_{2}: \frac{\mathrm{x}-2}{3}=\frac{\mathrm{y}-1}{-2}=\frac{\mathrm{z}+1}{1} \text { is }

\\\left|\begin{array}{ccc} \mathrm{x}-2 & \mathrm{y}-1 & \mathrm{z}+1 \\ 3 & -2 & 1 \\ 4 & -3 & 5 \end{array}\right|=0 \\ \Rightarrow 7 \mathrm{x}+11 \mathrm{y}+\mathrm{z}=24 \\ \text { Hence } \alpha=7, \beta=11, \mathrm{y}=1

Posted by

himanshu.meshram

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