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 \(\frac{\left\|d\times \overrightarrow{P_{0}P} \right\|}{\left\|d \right\|}\).

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 \(\frac{\left\|d\times \overrightarrow{P_{0}P} \right\|}{\left\|d \right\|}\).

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 \(\frac{\left\|d\times \overrightarrow{P_{0}P} \right\|}{\left\|d \right\|}\).

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 \(\overrightarrow{P Q}\).

So, \(\left\|\overrightarrow{P Q} \right\|=\left\|\overrightarrow{P_{0}P} \right\|\sin\theta\) .......(i)

Step 2. Prove

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

\(\begin{aligned}&\left\|d \times \overrightarrow{P_0 P}\right\|=\|d\| \overrightarrow{P_0 P} \| \sin \theta \\&\frac{\left\|d\times \overrightarrow{P_0 P}\right\|}{\|d\|}=\| \overrightarrow{P_0 P} \| \sin \theta \\&\left\|\overrightarrow{P_0 P}\right\| \sin \theta=\frac{\left\|d \times \overrightarrow{P_0 P}\right\|}{\|d\|}\end{aligned}\).

Now, put equation (i) so,

\(\|\overrightarrow{P Q}\|=\left\|\overrightarrow{P_0 P}\right\| \sin \theta\)

\(\|\overrightarrow{P Q}\|=\left\|\overrightarrow{P_0 P}\right\| \sin \theta=\frac{\left\|d \times \overrightarrow{P_0 P}\right\|}{\|d\|}\)

\(\|\overrightarrow{P Q}\|=\frac{\left\|d \times \overrightarrow{P_0 P}\right\|}{\|d\|}\)

Hence proved.