Question

If a ray of light through $A(1,2,3)$ strikes the plane $x+y+z=12$ at $B$ and on the reflection, passes through point $C(3,5,9),$ then

• A. coordinates of $B$ are $(-7,0,19)$
• B. equation of reflected ray is $\frac{x-5}{2}=\frac{y-6}{1}=\frac{z-7}{-2}$
• C. equation of plane containing the incident ray and the reflected ray is $3x-4y+z+2=0$
• D. equation of incident ray is $\frac{x-1}{6}=\frac{y-2}{-2}=\frac{z-3}{16}$

Answer 1

Answer: (A), (B), (C)

The correct options are
A coordinates of $B$ are $(-7,0,19)$
B equation of reflected ray is $\frac{x-5}{2}=\frac{y-6}{1}=\frac{z-7}{-2}$
C equation of plane containing the incident ray and the reflected ray is $3x-4y+z+2=0$

We know image of point $(x_1,y_1,z_1)$ in plane $ax+by+cz+d=0$ is
$\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}=-2\left(\frac{a{x}_1+b{y}_1+c{z}_1+d}{a^2+b^2+c^2}\right)$

So image of $A(1,2,3)$ in the plane $x+y+z=12$ is
$\frac{x-1}{1}=\frac{y-2}{1}=\frac{z-3}{1}=-2\left(\frac{1\cdot 1+1\cdot 2+1\cdot 3-12}{1^2+1^2+1^2}\right)\Rightarrow \frac{x-1}{1}=\frac{y-2}{1}=\frac{z-3}{1}=4\Rightarrow x=5,y=6,z=7$


Image of point $A(1,2,3)$ in plane is $(5,6,7),$ which lies on the reflected line $BC$
DR’s of line $BC$ is $(5-3,6-5,7-9)\equiv (2,1,-2)$
Equation of line $BC$ which is passing through $(5,6,7)$
$\frac{x-5}{2}=\frac{y-6}{1}=\frac{z-7}{-2}=\lambda$ (let)
$\Rightarrow x=2\lambda +5,y=\lambda +6,z=-2\lambda +7$
$\therefore B\equiv (2\lambda +5,\lambda +6,-2\lambda +7)\cdots \left(1\right)$

As point $B$ lies on plane $x+y+z=12$
$\therefore 2\lambda +5+\lambda +6-2\lambda +7-12=0\Rightarrow \lambda +6=0\Rightarrow \lambda =-6$
Putting value of $\lambda =-6$ in $\left(1\right),$ we get
$B\equiv (-7,0,19)$

Equation of reflected ray $BC$ is
$\frac{x-5}{2}=\frac{y-6}{1}=\frac{z-7}{-2}$

Equation of incident ray $AB$ is
$\frac{x-1}{-7-1}=\frac{y-2}{0-2}=\frac{z-3}{19-3}\Rightarrow \frac{x-1}{-8}=\frac{y-2}{-2}=\frac{z-3}{16}$

Equation of plane containing the incident and reflected ray is
$\left|\begin{array}{ccc}x-1& y-2& z-3\\ 3-1& 5-2& 9-3\\ -7-1& 0-2& 19-3\end{array}\right|=0$

$\Rightarrow (x-1)(48+12)-(y-2)(32+48)+(z-3)(-4+24)=0\Rightarrow 60(x-1)-80(y-2)+20(z-3)=0\Rightarrow 3(x-1)-4(y-2)+(z-3)=0\Rightarrow 3x-4y+z-3+8-3=0\Rightarrow 3x-4y+z+2=0$