Q.Find the shortest distance between the lines $\frac{x-3}{3}=\frac{y-8}{-1}=\frac{z-3}{1}$ and $\frac{x+3}{-3}=\frac{y+7}{2}=\frac{z-6}{4}$. Also find the equation of the line of the shortest distance.

Using the standard skew-line formula gives shortest distance $3\sqrt{30}$; solving for the common perpendicular's feet shows it joins $(3,8,3)$ and $(-3,-7,6)$.

Lines: $L_1:\dfrac{x-3}{3}=\dfrac{y-8}{-1}=\dfrac{z-3}{1}$ through $\vec a_1=(3,8,3)$, direction $\vec d_1=(3,-1,1)$.

$L_2:\dfrac{x+3}{-3}=\dfrac{y+7}{2}=\dfrac{z-6}{4}$ through $\vec a_2=(-3,-7,6)$, direction $\vec d_2=(-3,2,4)$.

Shortest distance formula:

$\text{SD}=\frac{\big|(\vec a_2-\vec a_1)\cdot(\vec d_1\times\vec d_2)\big|}{|\vec d_1\times\vec d_2|}$

$\vec a_2-\vec a_1=(-6,-15,3)$

$\vec d_1\times\vec d_2=\begin{vmatrix}\hat i&\hat j&\hat k\\3&-1&1\\-3&2&4\end{vmatrix}=\hat i[(-1)(4)-(1)(2)]-\hat j[(3)(4)-(1)(-3)]+\hat k[(3)(2)-(-1)(-3)]$

$=\hat i(-4-2)-\hat j(12+3)+\hat k(6-3)=(-6,-15,3)$

Since $(\vec a_2-\vec a_1)=(-6,-15,3)=\vec d_1\times\vec d_2$ exactly:

$(\vec a_2-\vec a_1)\cdot(\vec d_1\times\vec d_2)=(-6)^2+(-15)^2+3^2=36+225+9=270$

$|\vec d_1\times\vec d_2|=\sqrt{36+225+9}=\sqrt{270}=3\sqrt{30}$

$\text{SD}=\frac{270}{3\sqrt{30}}=\frac{90}{\sqrt{30}}=\frac{90\sqrt{30}}{30}=3\sqrt{30}\text{ units}$

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