3.7 Perpendicular distance of a point from a 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

In order to find the perpendicular distance, if the equation of the line is given in vector form, convert it into cartesian form as it is easy to find the perpendicular distance in cartesian form.

Now consider a equation of line in cartesian form $$\frac{{x - {x_1}}}{a} = \frac{{y - {y_1}}}{b} = \frac{{z - {z_1}}}{c}$$ and a perpendicular is drawn from a point $P(\alpha ,\beta ,\gamma )$ to the line.

Let us assume the foot of perpendicular on the line to be $L$ and the co-ordinates of $L$ is calculated by putting the equation of line equals to $\lambda $ and calculate the value of $x$, $y$ and $z$ i.e., $$\frac{{x - {x_1}}}{a} = \frac{{y - {y_1}}}{b} = \frac{{z - {z_1}}}{c} = \lambda $$

where $\lambda $ is any constant. Therefore, the co-ordinates of $L$ is $$({x_1} + a\lambda ,{y_1} + b\lambda ,{z_1} + c\lambda )$$

Now, the vector $\overrightarrow {PL} $ can be written as $$\overrightarrow {PL} = \left( {{x_1} + a\lambda - \alpha } \right)\widehat i + \left( {{y_1} + b\lambda - \beta } \right)\widehat j + \left( {{z_1} + c\lambda - \gamma } \right)\widehat k$$

Since, $\overrightarrow {PL} $ is perpendicular to $\overrightarrow {AB} $, therefore $\overrightarrow {PL} $ is also perpendicular to the parallel vector $\overrightarrow z $ of the line i.e., $$\overrightarrow z = a\widehat i + b\widehat j + c\widehat k$$ Therefore,

$$\begin{equation} \begin{aligned} \overrightarrow {PL} .\overrightarrow z = 0 \\ \left[ {\left( {{x_1} + a\lambda - \alpha } \right)\widehat i + \left( {{y_1} + b\lambda - \beta } \right)\widehat j + \left( {{z_1} + c\lambda - \gamma } \right)\widehat k} \right].\left[ {a\widehat i + b\widehat j + c\widehat k} \right] = 0 \\ a\left( {{x_1} + a\lambda - \alpha } \right) + b\left( {{y_1} + b\lambda - \beta } \right) + c\left( {{z_1} + c\lambda - \gamma } \right) = 0 \\ \therefore \lambda = \frac{{a(\alpha - {x_1}) + b(\beta - {y_1}) + c(\gamma - {z_1})}}{{{a^2} + {b^2} + {c^2}}} \\\end{aligned} \end{equation} $$

Putting the value of $\lambda $ in $$({x_1} + a\lambda ,{y_1} + b\lambda ,{z_1} + c\lambda )$$ we get the co-ordinates of foot of perpendicular.

Now, the perpendicular distance of a point can find out using distance formulae between the given point and foot of perpendicular.