7.0 Image of a point in a plane

Let us assume the image of point $P(\alpha ,\beta ,\gamma )$ in the plane $ax+by+cz+d=0$ be $Q({x_1},{y_1},{z_1})$ and the line $PQ$ meet the plane at $R$.

Since $PQ$ is perpendicular to the plane means it is parallel to the normal vector of the plane.

Therefore, the direction ratios of the line $PQ$ are $a$, $b$ and $c$.

The equation of line $PQ$ passing through a point $P(\alpha ,\beta ,\gamma )$ and having direction ratios $a$, $b$ and $c$ is $$\frac{{x - \alpha }}{a} = \frac{{y - \beta }}{b} = \frac{{z - \gamma }}{c} = r(say)$$

Therefore, co-ordinate of any point on the line which also lie on the plane is considered as $$x = ar + \alpha ;\quad y = br + \beta ;\quad z = cr + \gamma $$ This must satisfy the equation of plane.

From there we can find the value of $r$. Since $R$ is the mid-point of $PQ$, using mid-point formulae, we can find the co-ordinates of $Q$ i.e., the image of point $P$ in the plane.

Question 14. Find the image of point $P(3,5,7)$ in the plane $2x+y+z=0$.

Solution: The given equation of plane is $2x+y+z=0$.

Therefore, the direction ratios of normal vector to the plane is $2$, $1$, $1$.

Co-ordinates of point $P$ is $(3,5,7)$.

Let us assume the co-ordinates of image of point $P$ be $Q({x_1},{y_1},{z_1})$ and the line $PQ$ meet the plane at $R$. Equation of line $PQ$ can be written as $$\frac{{x - 3}}{2} = \frac{{y - 5}}{1} = \frac{{z - 7}}{1} = r(say)$$ Therefore, co-ordinate of any point $R$ on the line which also lie on the plane is written as $$x = 2r + 3;\quad y = r + 5;\quad z = r + 7$$ Since $R$ lies on the plane,

$$\begin{equation} \begin{aligned} 2\left( {2r + 3} \right) + \left( {r + 5} \right) + \left( {r + 7} \right) = 0 \\ 6r + 18 = 0 \\ r = - 3 \\\end{aligned} \end{equation} $$Therefore, co-ordinates of point $R$ is $(-3,2,4)$. Since $R$ is the mid-point of $PQ$, using mid-point formulae we get, $$\begin{equation} \begin{aligned} - 3 = \frac{{{x_1} + 3}}{2};\quad 2 = \frac{{{y_1} + 5}}{2};\quad 4 = \frac{{{z_1} + 7}}{2} \\ {x_1} = - 9;\quad {y_1} = - 1;\quad {z_1} = 1 \\\end{aligned} \end{equation} $$ Therefore, the co-ordinates of image of point $P(3,5,7)$ is $Q(-9,-1,1)$.