91影视

Consider a point $P$ to the line $\mathcal{L}$

The goal is to calculate the distance between the points $P$and $Q$in the diagram.

The triangle $P{P}_{0}Q$is a right-angle triangle.

It can be written as, using trigonometric ratios.

$$\begin{gathered} \tan 胃=\frac{PQ}{P_{0}Q} \\ \tan 胃=\frac{PQ}{d} \end{gathered}$$

On both sides of the equation, multiply by $d$

$$\begin{gathered} d路\tan 胃=d路\frac{PQ}{d} \\ d路\tan 胃=PQ \end{gathered}$$

Thus, the distance $PQ=d\tan 胃$

Therefore, the answer is $PQ=d\tan 胃$

The goal is to use the cross-product method to calculate the distance.

Now, suppose $Q$is the point on the line $\mathcal{L}$that is closest to the point $P$

From the figure, we can observe that,

$鈥�\stackrel{鈫�}{PQ}鈥�=||\stackrel{鈫�}{P_{0}P}||\sin 胃$, where $胃$is the angle between $\stackrel{鈫�}{P_{0}P}$and the distance $d$

The distance between a point and a perpendicular line is defined by the theorem on perpendicular lines and points.

$$\begin{gathered} ||d脳\stackrel{鈫�}{P_{0}P}||=鈥�d鈥�\stackrel{鈫�}{P_{0}P}鈥�\sin 胃 \\ \frac{||d脳\stackrel{鈫�}{P_{0}P}||}{鈥�d鈥�}=||\stackrel{鈫�}{P_{0}P}||\sin 胃 \\ ||\stackrel{鈫�}{P_{0}P}||\sin 胃=\frac{||d脳\stackrel{鈫�}{P_{0}P}||}{鈥�d鈥�} \end{gathered}$$

Thus,

We know that $鈥�\stackrel{鈫�}{PQ}鈥�=||\stackrel{鈫�}{P_{0}P}||\sin 胃$

$鈥�\stackrel{鈫�}{PQ}鈥�=\frac{||d脳{\stackrel{鈫�}{P}}_{0}P||}{鈥�d鈥�}$

Therefore, the required distance using the cross product is $\frac{||d脳\stackrel{鈫�}{P_{0}P}||}{鈥�d鈥�}$