Find the shortest distance between the lines: →r = (î + 2ĵ + k̂) + λ(î + ĵ + k̂) and →r = (2î − ĵ + 4k̂)

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Q. Find the shortest distance between the lines: \tor = (î + 2ĵ + k̂) + λ(î + ĵ + k̂) and \tor = (2î - ĵ + 4k̂) + μ(2î + ĵ + 2k̂)

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Concept understanding — Skew Lines

Skew Lines

In a plane, two straight lines have only two possibilities: they meet, or they are parallel. In three dimensions a third possibility appears — lines that neither meet nor run parallel. These are skew lines.

What Makes Lines Skew

Two lines in space are skew if they are not parallel and do not intersect. The deeper reason is that skew lines do not lie in the same plane — they are non-coplanar. Parallel lines and intersecting lines always share a plane; skew lines never do.

Note

A classic picture: one edge along the top of a room and a different edge along the floor, running in a different direction. Extend them forever and they still never touch, yet they are clearly not parallel.

The Three Cases in Space

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

How to Test for Skew Lines

Take two lines $\vec{r} = \vec{a_1} + \lambda\vec{b_1}$ and $\vec{r} = \vec{a_2} + \mu\vec{b_2}$ .

Not parallel: $\vec{b_1}$ and $\vec{b_2}$ are not proportional (so $\vec{b_1}\times\vec{b_2}\neq\vec{0}$ ).
Do not intersect: no values of $\lambda,\mu$ make the points coincide.

Both conditions are captured by one scalar triple product. The lines are skew exactly when

(\vec{a_2}-\vec{a_1})\cdot(\vec{b_1}\times\vec{b_2}) \neq 0.

If this value is zero, the lines are coplanar (they intersect or are parallel); if it is non-zero, they are skew.

Shortest Distance Between Skew Lines

Because skew lines miss each other, there is a well-defined shortest distance between them, measured along their common perpendicular:

$d = \frac{\left|(\vec{a_2}-\vec{a_1})\cdot(\vec{b_1}\times\vec{b_2})\right|}{\left|\vec{b_1}\times\vec{b_2}\right|}.$ …

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For two skew lines $\vec r=\vec a_1+\lambda\vec b_1$ and $\vec r=\vec a_2+\mu\vec b_2$ , the shortest distance is $\left|\dfrac{(\vec a_2-\vec a_1)\cdot(\vec b_1\times\vec b_2)}{|\vec b_1\times\vec b_2|}\right|$ .

Line 1: $\vec a_1 = \hat\imath+2\hat\jmath+\hat k$ , $\vec b_1 = \hat\imath+\hat\jmath+\hat k$ .

Line 2: $\vec a_2 = 2\hat\imath-\hat\jmath+4\hat k$ , $\vec b_2 = 2\hat\imath+\hat\jmath+2\hat k$ .

Step 1: $\vec a_2 - \vec a_1$

$\vec a_2-\vec a_1 = (2-1)\hat\imath+(-1-2)\hat\jmath+(4-1)\hat k = \hat\imath-3\hat\jmath+3\hat k$

Step 2: $\vec b_1 \times \vec b_2$

$\vec b_1\times\vec b_2 = \begin{vmatrix}\hat\imath&\hat\jmath&\hat k\\1&1&1\\2&1&2\end{vmatrix} = \hat\imath(1\cdot2-1\cdot1)-\hat\jmath(1\cdot2-1\cdot2)+\hat k(1\cdot1-1\cdot2)$

$= \hat\imath(1) - \hat\jmath(0) + \hat k(-1) = \hat\imath - \hat k$

Step 3: dot product …

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Q.Find the shortest distance between the lines: →r = (î + 2ĵ + k̂) + λ(î + ĵ + k̂) and →r = (2î − ĵ + 4k̂) + μ(2î + ĵ + 2k̂)

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Q.Find the shortest distance between the lines:
→r = (î + 2ĵ + k̂) + λ(î + ĵ + k̂) and
→r = (2î − ĵ + 4k̂) + μ(2î + ĵ + 2k̂)

Unlock everything free for 14 days