Practice: Find The Distance Between The Skew Lines

I solved this problem using $xyz$ - coordinates.

Let $C$ be the origin of our system. Let $S$ be the point $(0,0,2)$ , such that $CS$ lies on the $z$ - axis. Next, let $D$ be the point $(2\sqrt{6},0,0)$ , then $\triangle{ABC}$ lies in the $xy$ - plane and is symmetric about the $x$ - axis. This fixes $E$ to be the point $(\sqrt{6},\sqrt{2},0)$ . Therefore, the parametric equation of line $SE$ is $(\sqrt{6},\sqrt{2},-2)\cdot t+(0,0,2) = (t\sqrt{6},t\sqrt{2},2-2t)$ . The shortest distance from an arbitrary point $(x,y,z)$ to the $x$ - axis is $\sqrt{y^2+z^2}$ . Using this fact, the distance from line $SE$ to $CD$ , the $x$ - axis, simply becomes $\sqrt{(t\sqrt{2})^2 + (2-2t)^2}$ . Simplifying the expression we obtain $\sqrt{6t^2-8t+4}$ . Looking only at the quadratic under the square root, we can easily find the minimum value of $\frac{4}{3}$ by completing the square.

Thus, the distance between $SE$ and $CD$ is $\sqrt{\frac{4}{3}}$ , $3+4=\boxed{7}$

$$ First, the two skew lines are expressed using two vector equations: $$ $\begin{aligned} \mathbf{L_1}&=\mathbf{a}+\mathbf{d_1}t=\langle x_{0a}, y_{0a}, z_{0a}\rangle + \langle x_{1a}-x_{0a}, y_{1a}-y_{0a}, z_{1a}-z_{0a}\rangle \,t\\ \mathbf{L_2}&=\mathbf{b}+\mathbf{d_2}t=\langle x_{0b}, y_{0b}, z_{0b}\rangle + \langle x_{1b}-x_{0b}, y_{1b}-y_{0b}, z_{1b}-z_{0b}\rangle \,t \end{aligned}$ $$ The distance between two skew lines can be obtained by using $$ $|\mathbf{d}|=\left|\frac{\mathbf{d_1}\times \mathbf{d_2}}{|\mathbf{d_1}\times \mathbf{d_2}|}\cdot(\mathbf{b}-\mathbf{a})\right|.$ $$ Set $\;D$ as the origin and it's easy to obtain that $\;D(0,0,0)$ , $\;C(0,2\sqrt{6},0)$ , $\;E(\sqrt{2},\sqrt{6},0)$ , and $\;S(0,2\sqrt{6},2)$ . Therefore $$ $\begin{aligned} \overrightarrow{DC}&=\langle 0,2\sqrt{6},0\rangle \,t\\ \overrightarrow{ES}&=\langle \sqrt{2},\sqrt{6},0\rangle+\langle -\sqrt{2},\sqrt{6},2\rangle \,t \end{aligned}$ and $\begin{aligned} |\mathbf{d}|&=\left|\frac{\langle 4\sqrt{6},0,4\sqrt{3}\rangle}{\sqrt{(4\sqrt{6})^2+0^2+(4\sqrt{3})^2}}\cdot\langle \sqrt{2},\sqrt{6},0\rangle\right|\\ &=\left|\frac{\langle 8\sqrt{3},0,0\rangle}{12}\right|\\ &=\frac{1}{12}\sqrt{(8\sqrt{3})^2+0^2+0^2}\\ &=\frac{2}{3}\sqrt{3}\\ &=\sqrt{\frac{4}{3}}. \end{aligned}$ $$ Thus, $\,a+b=4+3=\boxed{7}$ . $$ $\text{\# }\mathbb{Q}.\mathbb{E}.\mathbb{D}.\text{\#}$

Demonstration
Through C draw a line $d \in (ABC); d \perp DC$ .
Through E draw a line parallel to DC which intersects d at F.
DC and EF can be proven easily $\perp (SCF)$
Draw $CH \perp SF (CH \cap SF = \{ H \}$
So we get $CH \perp (SEF) \supset SE \Rightarrow CH \perp SE$
and $CH \in (SFC) \perp DC$
Then CH is the distance we need to find
$CF = AB/4 = \sqrt{2}$
Using the equation:
$\frac{1}{CH^2}=\frac{1}{SC^2}+\frac{1}{CF^2}$
We work out $CH = \sqrt{\frac{4}{3}}$
So $a + b = \fbox{7}$

In stereometry one idea is to find the right projection(s) so you can transform your problem to "planimetry".

The distance between skew lines can be obtained if we find a plane parallel to one of the lines containing the other line.

In this case we can find it by drawing a parallel line to $CD$ through point $E$ . This line cuts $AB$ in the midpoint $F$ of $BD$ .

Now the distance between plane $SEF$ and $CD$ is evident if we "see" the figure in the direction of $DC$ . I mean a projection in such a way that $D$ and $C$ (and $F$ and $E$ ) coincide.

If you can imagine this, you have a projected triangle $S'D'F'$ (where $D'$ is the projection of $D$ and $C$ and $F'$ is the projection of $F$ and $E$ ). The triangle is rectangle with $C'S'=2$ and $D'F'=\sqrt{2}$

The distance we are looking for is the altitude from $D'$ to hypotenuse $S'F'$ .

Now $S'F' = \sqrt{6}$

by Pythagoras and using the area of the triangle we know that $2 \sqrt{2} = d \sqrt{6}$ .

Therefore $d = \sqrt{8/6} = \sqrt{4/3}$ and $a+b=7$ .