Question

Let $\pi$ be the plane parallel to $y-$axis and containing the points $(1,0,1)$ and $(3,2,-1).$ Also, $A\equiv (4,0,0)$ and $B\equiv (6,0,-2)$ are two points and $P\equiv (x_0,y_0,z_0)$ is a variable point on the plane $\pi .$ Then which of the following is/are CORRECT?

• A. The equation of the plane $\pi$ is $x+z=2$
• B. If $(PA+PB)$ is minimum, then $|4x_0+y_0+2z_0|$ is $12$
• C. If $|PA-PB|\in [0,\sqrt{N}],$ then $N$ is $8$
• D. If the reflection of the line $AB$ in the plane $\pi$ is $\frac{x-2}{1}=\frac{y-\alpha }{0}=\frac{z+\beta }{-1},$ then $({\alpha }^4+{\beta }^4)$ is $16$

Answer 1

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

The correct options are
A The equation of the plane $\pi$ is $x+z=2$
B If $(PA+PB)$ is minimum, then $|4x_0+y_0+2z_0|$ is $12$
C If $|PA-PB|\in [0,\sqrt{N}],$ then $N$ is $8$
D If the reflection of the line $AB$ in the plane $\pi$ is $\frac{x-2}{1}=\frac{y-\alpha }{0}=\frac{z+\beta }{-1},$ then $({\alpha }^4+{\beta }^4)$ is $16$

Let the plane be $\alpha x+\beta z+1=0$
It passes through $(1,0,1)$ and $(3,2,-1)$
$\therefore \alpha =-\frac{1}{2};\beta =-\frac{1}{2}$
Equation of plane $\pi$ is $x+z=2$

Both $A$ and $B$ are on same side of $\pi$.
Reflection of $A$ in plane $\pi$ is given by
$\frac{x-4}{1}=\frac{y-0}{0}=\frac{z-0}{1}=\frac{-2(4+0-2)}{1^2+1^2}$
So, coordinates of reflection point $A^{\prime }$ is $(2,0,-2)$
Equation of line $A^{\prime }B:$ $\overrightarrow{r}=6\overrightarrow{i}-2\hat{k}+\lambda \left(4\hat{i}\right)$
For $P:6+4\lambda +0-2=2$
$\Rightarrow \lambda =-\frac{1}{2}$
$\therefore$ Coordinates of $P$ are $(4,0,-2)$
$\therefore |4x_0+y_0+2z_0|=12$

Now, $|PA-PB{|}_{min}=0$
$|PA-PB{|}_{max}$ will approach $AB=\sqrt{4+0+4}=\sqrt{8}$
$\therefore |PA-PB|\in [0,\sqrt{8}]$

Also, $A^{\prime }$ will lie on $\frac{x-2}{1}=\frac{y-\alpha }{0}=\frac{z+\beta }{-1}$
$\Rightarrow \frac{2-2}{1}=\frac{0-\alpha }{0}=\frac{-2+\beta }{-1}$
$\Rightarrow \alpha =0,\beta =2$
$\therefore {\alpha }^4+{\beta }^4=16$