Question

Let $\alpha ,\beta ,\gamma ,\delta$ be real numbers such that ${\alpha }^2+{\beta }^2+{\gamma }^2\ne 0$ and $\alpha +\gamma =1.$ Suppose the point $(3,2,-1)$ is the mirror image of the point $(1,0,-1)$ with respect to the plane $\alpha x+\beta y+\gamma z=\delta .$ Then which of the following statements is/are TRUE?

• A. $\alpha +\beta =2$
• B. $\delta -\gamma =3$
• C. $\delta +\beta =4$
• D. $\alpha +\beta +\gamma =\delta$

Answer 1

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

$\delta +\beta =4$


Given equation of the plane is $\alpha x+\beta y+\gamma z=\delta .$
$Q$ is mid point of $P{P}^{\prime }$
$Q=(\frac{1+3}{2},\frac{0+2}{2},\frac{-1-1}{2})=(2,1,-1)$
Since point $Q$ lie on the plane $\alpha x+\beta y+\gamma z=\delta ,$
$\therefore \alpha \left(2\right)+\beta \left(1\right)+\gamma (-1)=\delta .$
$\Rightarrow 2\alpha +\beta -\gamma =\delta \cdots \left(1\right)$

$P{P}^{\prime }$ is normal to given plane.
So, $\frac{\alpha }{2}=\frac{\beta }{2}=\frac{\gamma }{0}=\lambda \text{(let)}$
$\Rightarrow \alpha =2\lambda ,\beta =2\lambda ,\gamma =0$
Since $\alpha +\gamma =1,$
$\therefore 2\lambda +0=1$
$\Rightarrow \lambda =\frac{1}{2}$
$\alpha =2\lambda =2\times \frac{1}{2}=1,\beta =2\lambda =2\times \frac{1}{2}=1,\gamma =0$

Now putting values of $\alpha ,\beta ,\gamma$ in eq. $\left(1\right),$ we get
$2\left(1\right)+1-0=\delta$
$\Rightarrow \delta =3$
$\therefore \delta -\gamma =3-0=3$
$\delta +\beta =3+1=4$
$\alpha +\beta +\gamma =1+1+0=2(\ne \delta )$
$\alpha +\beta =1+1=2$