question_answer

The image of the line \[\frac{x-1}{9}=\frac{y-2}{-1}=\frac{z+3}{-3}\] in the plane \[3x-3y+10z-26=0,\] is

A) \[\frac{x-5/2}{9}=\frac{y-1/2}{-1}=\frac{z-2}{-3}\]

B) \[\frac{x+5/2}{9}=\frac{y+1/2}{-1}=\frac{z+2}{-3}\]

C) \[\frac{x-5/2}{9}=\frac{y+1/2}{-1}=\frac{z+2}{3}\]

D) None of these

Correct Answer: A

Solution :

[a] \[\frac{x-1}{9}=\frac{y-2}{-1}=\frac{z+3}{-3}=1\]                    ??.(1) \[3x-3y+10z-26=0\]        ........(2) Point M is the image of point A in plane (2) (i.e. foot of the perpendicular) \[\therefore \] (DR'S of the normal of the plane (2)) - (DR'S of the line AM) = 3,- 3,  10 Then equation of line \[AM\Rightarrow \frac{x-1}{3}=\frac{y-2}{-3}=\frac{z+3}{10}=\lambda \] \[M\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)=\left( 3\lambda +1,-3\lambda +2,10\lambda -3 \right)\]             \[\because \]point M lies in the plane (2) Then \[3{{x}_{1}}-3{{y}_{1}}+10{{z}_{1}}-26=0\] \[3\left( 3\lambda +1 \right)-3\left( -3\lambda +2 \right)+10\left( 10\lambda -3 \right)-26=0\] \[118\lambda =59\Rightarrow \lambda =1/2\] \[\therefore M\left\{ \frac{5}{2},\frac{1}{2},2 \right\}\] \[\therefore \] Equation of line BM \[\Rightarrow \frac{\left( x-\frac{5}{2} \right)}{9}=\frac{\left( y-\frac{1}{2} \right)}{-1}=\frac{(z-2)}{-3}\]