I have 2 skew lines
$L_1 :\begin{cases} x = 1 + 2t\\ y = 2-2t\\ z = 3+4t\end{cases}$
$L_2 :\begin{cases} x = 2+2t\\ y = 1\\ z = t\end{cases}$
Lets take 2 points on line 1 $(A,B)$ by setting $t$ to $0$ and $1$
And 1 point from line 2 $(P)$ by setting $t$ to $0$
My question is this:
why is this formula
$$\frac{|| AB \times AP || }{ ||AB||}$$
(The formula for finding the distance between a point and a line that passes through 2 points)
giving the same result as this formula?
$$\frac{|a\cdot x_0 + b\cdot y_0 + c\cdot z_0 + d|}{ \sqrt{a^2 + b^2 + c^2}}$$
(The formula for the distance between a point and a plane)
for the given skew lines?
Additional information for those who don't understand:
. $a$, $b$ and $c$ are the components of a vector normal to the plane formed by the parallel vectors of $L_1$ and $L_2$ (ie. $<2,-2,4> \times <2,0,1>$ )
$d$ = – $ax_1$ – $by_1$ – $cz_1$ (just for those of you who don't know and are wondering how the last formula is made the $d$ is derived from the vector equation of the plane.
$$a(x-x_0)+b(y-y_0)+c(z-z_0)=0$$
The scalar equation of the plane is
$$d=ax_0+bx_0+cx_0$$
$x_0$ $y_0$ $z_0$ is just the point on line 2 $(P)$ (that we get when setting $t$ to $0$)
why are those 2 formulas giving the same result?
Best Answer
Two skew lines lie in a unique pair of parallel planes, whose normal vectors — as you said — is the cross-product of the direction vectors of the lines. The distance between the lines in the distance between those parallel planes. And you can find that by taking the distance from any point on one plane to the other plane. So you've taken the distance from any point on line 1 to the point on the plane passing through line 2.