A line has direction vec p and passes through M_1 . Prove…

A line has direction $\vec p$ and passes through $M_1$. Prove that the distance of $M_0$ from the line is $d=\frac{|\vec p\times\overrightarrow{M_1M_0}|}{|\vec p|} $.

Explanation of Method

Proof of the cross-product formula for the distance from a point to a line.

Step 1: Proof

Let $F$ be the foot of the perpendicular from $M_0$, and $\varphi$ the angle between $\vec p$ and $\vec v=\overrightarrow{M_1M_0}$. In the right triangle $M_1FM_0$, $d=|\vec v|\sin\varphi$. Since $|\vec p\times\vec v|=|\vec p||\vec v|\sin\varphi$:

\[d=|\vec v|\sin\varphi=\frac{|\vec p\times\overrightarrow{M_1M_0}|}{|\vec p|}\qquad\blacksquare\]

Equivalently, the parallelogram on $\vec p$ and $\vec v$ has area $|\vec p\times\vec v|$, base $|\vec p|$ and height $d$.

Independence of $M_1$: for another point $M_1'$ we have $\overrightarrow{M_1'M_0}=\overrightarrow{M_1'M_1}+\overrightarrow{M_1M_0}$. Since $\vec p\times\overrightarrow{M_1'M_1}=\vec0$ (the vectors are parallel), the cross product does not change.

$\vec p$ $\overrightarrow{M_1M_0}$ $d$ $\varphi$ $M_1$ $M_0$ $F$ $\ell$

The parallelogram on $\vec p$ and $\overrightarrow{M_1M_0}$ (shaded) has area $|\vec p\times\overrightarrow{M_1M_0}|$ = base $|\vec p|$ × height $d$.

Result

\[\boxed{\,d=|\vec v|\sin\varphi=\frac{|\vec p\times\overrightarrow{M_1M_0}|}{|\vec p|}\qquad\blacksquare\,}\]