3.8 Image of a point in a straight line

2.1 Vectorial form of a line passing through a given point and parallel to a given vector
2.2 Vectorial form of a line passing through two given points
2.3 Cartesian form of a line passing through a given point and parallel to a given vector
2.4 Cartesian form of a line passing through two given points

3.1 Condition for perpendicularity
3.2 Condition for parallelism
3.3 Co-planarity of two lines
3.4 Shortest distance between two lines
3.5 Distance between two skew lines
3.6 Distance between parallel lines
3.7 Perpendicular distance of a point from a line
3.8 Image of a point in a straight line

4.1 Equation of plane in normal form
4.2 Equation of a plane perpendicular to a given vector and passing through a given point
4.3 Equation of plane passing through a given point and parallel to the two given vectors
4.4 Equation of plane passing through three non-collinear points
4.5 Intercept form of plane
4.6 Equation of plane passing through the intersection of two given planes

5.1 Angle between two planes
5.2 Angle between a line and a plane
5.3 Distance of a point from a plane
5.4 Distance between two parallel planes

To find the image of a point $Q$ in a straight line, follow the same procedure explained above and find the foot of perpendicular first.

Now, let us assume the co-ordinates of image of point $Q$ be $({\alpha '},{\beta '},{\gamma '})$ and foot of perpendicular $L$ is the mid-point of given point $P$ and image $Q$.

Therefore, apply mid-point formulae between them we get,

$$\begin{equation} \begin{aligned} \frac{{\alpha + \alpha '}}{2} = {x_1} + a\lambda \\ \frac{{\beta + \beta '}}{2} = {y_1} + b\lambda \\ \frac{{\gamma + \gamma '}}{2} = {z_1} + c\lambda \\\end{aligned} \end{equation} $$

Therefore,

$$\begin{equation} \begin{aligned} \alpha ' = 2({x_1} + a\lambda ) - \alpha \\ \beta ' = 2\left( {{y_1} + b\lambda } \right) - \beta \\ \gamma ' = 2\left( {{z_1} + c\lambda } \right) - \gamma \\\end{aligned} \end{equation} $$

Question 9. Find the image of the point $(1,6,3)$ in the line $$\frac{x}{1} = \frac{{y - 1}}{2} = \frac{{z - 2}}{3}$$

Solution: Let $P(1,6,3)$ be the given point and $L$ be the foot of perpendicular from $P$ to the given line. The co-ordinates of any general point on line can be find out by $$\frac{{x - 0}}{1} = \frac{{y - 1}}{2} = \frac{{z - 2}}{3} = \lambda $$ i.e., $$x = \lambda ,\;\;y = 2\lambda + 1,\;\;z = 3\lambda + 2$$ Therefore, the co-ordinates of $L$ be $$(\lambda ,2\lambda + 1,3\lambda + 2)$$ Now, the vector $PL$ can be written as $$\overrightarrow {PL} = (\lambda - 1)\widehat i + (2\lambda - 5)\widehat j + (3\lambda - 1)\widehat k$$

Since $\overrightarrow {PL} $ is perpendicular to $\overrightarrow {AB} $, therefore, it is also perpendicular to the parallel vector $\overrightarrow b = \widehat i + 2\widehat j + 3\widehat k$ of the given line. Therefore, $$\begin{equation} \begin{aligned} \overrightarrow {PL} .\overrightarrow b = 0 \\ \left[ {(\lambda - 1)\widehat i + (2\lambda - 5)\widehat j + (3\lambda - 1)\widehat k} \right].\left( {\widehat i + 2\widehat j + 3\widehat k} \right) = 0 \\ (\lambda - 1).1 + (2\lambda - 5).2 + (3\lambda - 1).3 = 0 \\ \lambda = 1 \\\end{aligned} \end{equation} $$

So, co-ordinates of $L$ are $(1,3,5)$.

Let $Q({x_1},{y_1},{z_1})$ be the image of point $P(1,6,3)$ and $L$ is the mid-point of $PQ$. Apply mid-point formulae, we get $$\begin{equation} \begin{aligned} 1 = \frac{{{x_1} + 1}}{2},3 = \frac{{{y_1} + 6}}{2},5 = \frac{{{z_1} + 3}}{2} \\ {x_1} = 1,{y_1} = 0,{z_1} = 7 \\\end{aligned} \end{equation} $$

Therefore, the image of point $P(1,6,3)$ in the given line is $(1,0,7)$.