Question

Find the image of :
$(-2,3,4)$ in the yz-plane.

Answer 1

Let the image of $(-2,3,4)$ be $(a,b,c)$Now the midpoint of $P(-2,3,4)$ & $Q(a,b,c)$ must lie on $y-z$ plane

On $y-z$ plane, $x=0$ ....... $\left(1\right)$

$\Rightarrow$ midpoint $M=(\frac{a-2}{2},\frac{b+3}{2},\frac{c+4}{2})$

For $M,x=0\Rightarrow \frac{a-2}{2}=0\Rightarrow a=2$ ........$\left(2\right)$

Now the line joining $P$ to $Q$ ( in general any point to its image) must be perpendicular to $yz$ plane (in generalthe mirror about which reflection is taken)

So, the line joining $P$ & $Q$ must be paralle to normal of $yz$ plane i.e $x$-axis be in proportion to that of $X$-axis hence d.r. of $PQ=(x_Q-x_P,y_Q-y_P,z_Q-z-[P\left]\right)$

d.r. of $x$-axis = $(1,0,0)$

$\Rightarrow x_Q-x_P=\lambda \times 1\Rightarrow a-(-2)=\lambda$

$y_Q-y_P=\lambda \times 0b-3=0$

$z_Q-z_P=\lambda \times 0c-4=0$

Hence, we have $b=3$......$\left(3\right)$ and $c=4$..........$\left(4\right)$

So, from eqn $\left(2\right),\left(3\right)$ & $\left(4\right)$ we get

image $=Q\equiv (2,3,4)$