Find the shortest distance between the following pair of lines: (i) $\dfrac{x-3}{2} = \dfrac{y-4}{1} = \dfrac{z+1}{-3}$ and $\dfrac{x-1}{-1} = \dfrac{y-3}{3} = \dfrac{z-1}{2}$. (ii) $\vec{r} = \hat{i}+2\hat{j}-4\hat{k}+\lambda(2\hat{i}+3\hat{j}+6\hat{k})$ and $\vec{r} = 3\hat{i}+

Question

Find the shortest distance between the following pair of lines: (i) $\dfrac{x-3}{2} = \dfrac{y-4}{1} = \dfrac{z+1}{-3}$ and $\dfrac{x-1}{-1} = \dfrac{y-3}{3} = \dfrac{z-1}{2}$. (ii) $\vec{r} = \hat{i}+2\hat{j}-4\hat{k}+\lambda(2\hat{i}+3\hat{j}+6\hat{k})$ and $\vec{r} = 3\hat{i}+3\hat{j}-5\hat{k}+\mu(2\hat{i}+3\hat{j}+6\hat{k})$.

Accepted Answer

>

> [!TLDR]
> Part (i) is a pair of skew lines — use the skew-line shortest-distance formula. Part (ii) has identical direction vectors, so the lines are parallel — use the parallel-line distance formula instead.
>
> **(i)** Lines: $\dfrac{x-3}2=\dfrac{y-4}1=\dfrac{z+1}{-3}$ through $A_1=(3,4,-1)$ with direction $\vec d_1=(2,1,-3)$, and $\dfrac{x-1}{-1}=\dfrac{y-3}3=\dfrac{z-1}2$ through $A_2=(1,3,1)$ with direction $\vec d_2=(-1,3,2)$.
>
> $$\vec{A_1A_2} = (-2,-1,2)$$
> $$\vec d_1	imes\vec d_2 = \begin{vmatrix}\hat i&\hat j&\hat k\2&1&-3\-1&3&2\end{vmatrix} = (11,-1,7),\qquad |\vec d_1	imes\vec d_2|=\sqrt{121+1+49}=\sqrt{171}=3\sqrt{19}$$
>
> $$\vec{A_1A_2}\cdot(\vec d_1	imes\vec d_2) = (-2)(11)+(-1)(-1)+(2)(7) = -22+1+14=-7$$
>
> $$	ext{SD} = \dfrac{|\vec{A_1A_2}\cdot(\vec d_1	imes\vec d_2)|}{|\vec d_1	imes\vec d_2|} = \dfrac{7}{3\sqrt{19}}$$
> …

Lines	Directions	Do they meet?	Coplanar?
Intersecting	different	yes, at one point	yes
Parallel	same (proportional)	no	yes
Skew	different	no	no

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