Question

The image of the point $(1,-3,5)$ w.r.t the plane containing the two lines $\frac{x+1}{-3}=\frac{y-3}{2}=\frac{z+2}{1}$ and $\frac{x}{1}=\frac{y-7}{-3}=\frac{z+7}{2}$ is:

• A. $(-1,-5,3)$
• B. $(1,5,3)$
• C. $(-2,7,3)$
• D. $(-2,7,-3)$

Answer 1

The correct option is A $(-1,-5,3)$

Equation of plane containing $L_1$ and $L_2$ is given by:
$\left|\begin{array}{ccc}x+1& y-3& z+2\\ -3& 2& 1\\ 1& -3& 2\end{array}\right|=0$
$(x+1)7-(y-3)(-7)+(z+2)7=0$
$\Rightarrow x+1+y-3+z+2=0$
$\Rightarrow x+y+z=0$
Required equation of plane is $x+y+z=0$
Let the image of $(1,-3,5)$ be $(\alpha ,\beta ,\gamma )$
$\frac{\alpha -x_1}{a}=\frac{\beta -y_1}{b}=\frac{\gamma -z_1}{c}=-2\frac{a{x}_1+b{y}_1+c{z}_1+d}{a^2+b^2+c^2}$
$\frac{\alpha -1}{1}=\frac{\beta +3}{1}=\frac{\gamma -5}{1}=-2\times \frac{1-3+5}{1+1+1}$
$\frac{\alpha -1}{1}=\frac{\beta +3}{1}=\frac{\gamma -5}{1}=-2\times 1$
$\alpha =-1,\beta =-5,\gamma =3$
Hence, image is $(-1,-5,3)$