Question

Prove that distance from the point P to the line given by the equation $r\left(t\right)=P_0+td$ is given by $\frac{\Vert d\times \overrightarrow{P_{0}P}\Vert }{\Vert d\Vert }$.

Short answer

The distance from the point P to the line given by the equation $r\left(t\right)=P_0+td$ is given by $\frac{\Vert d\times \overrightarrow{P_{0}P}\Vert }{\Vert d\Vert }$.

Step-by-step solution

Step 1. Prove

We have to prove that the distance from the point P to the line given by the equation $r\left(t\right)=P_0+td$ is given by $\frac{\Vert d\times \overrightarrow{P_{0}P}\Vert }{\Vert d\Vert }$.

Now, let the parametrization equation of the line is $r\left(t\right)=P_0+td$ and the point $P_0$ is on the line and direction vector parallel to the line.

Let's assume Q is the point on the line which is closest to the point P.

We will use the distance formula to calculate the length of $\overrightarrow{PQ}$.

So, $\Vert \overrightarrow{PQ}\Vert =\Vert \overrightarrow{P_{0}P}\Vert \sin \theta$ .......(i)

Step 2. Prove

The formula of the distance from a point to the perpendicular line is

$\begin{array}{rl}& \Vert d\times \overrightarrow{P_{0}P}\Vert =\Vert d\Vert \overrightarrow{P_{0}P}\Vert \sin \theta \\ & \frac{\Vert d\times \overrightarrow{P_{0}P}\Vert }{\Vert d\Vert }=\Vert \overrightarrow{P_{0}P}\Vert \sin \theta \\ & \Vert \overrightarrow{P_{0}P}\Vert \sin \theta =\frac{\Vert d\times \overrightarrow{P_{0}P}\Vert }{\Vert d\Vert }\end{array}$.

Now, put equation (i) so,

$\Vert \overrightarrow{PQ}\Vert =\Vert \overrightarrow{P_{0}P}\Vert \sin \theta$

$\Vert \overrightarrow{PQ}\Vert =\Vert \overrightarrow{P_{0}P}\Vert \sin \theta =\frac{\Vert d\times \overrightarrow{P_{0}P}\Vert }{\Vert d\Vert }$

$\Vert \overrightarrow{PQ}\Vert =\frac{\Vert d\times \overrightarrow{P_{0}P}\Vert }{\Vert d\Vert }$

Hence proved.