**Shortest Distance between Two Skew Lines**

**Definition:** l₁ and l₂ are two skew lines. If P
is a point on l₁ and Q is a point on l₂ such that PQ perpendicular to l₁ and PQ
is called shortest distance and PQ is called shortest distance line between the
lines l₁ and l₂.

**Theorem**: The shortest distance between the skew
lines r = a + tb and r = c + sd is \(\frac{\left| \left[ a-c\ \ \ b\ \ d \right]
\right|}{\left| b\times d \right|}\).

**Proof:** Let P be the point on the line r = a
+ tb … (1)

And Q be a point on the line r = c + sd … (2)

Such the PQ is the shortest distance between (1) and (2).

Thus, PQ is perpendicular to both the lines (1) and (2)

Let A, C be the point such that \(\overrightarrow{OA}\text{ }=\text{ }a\), \(\overrightarrow{OC}\text{ }=\text{ }c\).

Now the point A lies in line (1) and the point C lies in line (2)

Now b is a vector parallel to line (1) and d is a vector parallel to line (2)

Thus, b x d is perpendicular to both the lines (1) and (2)

Therefore, PQ and b x d are parallel

Hence the unit vector in the direction of PQ is ± (b x d)/ |b x d|

PQ = C component of AC on PQ is \(\pm (a-c).\frac{b\,\times \ d}{|b\ \times \ d|}=\pm \frac{\left[ a-c\ \ \ b\ d \right]}{|b\ \times \ d|}\).

Since PQ is always positive, therefore \(PQ=\pm \frac{\left| \left[ a-c\ \ \ b\ \ d \right] \right|}{\left| b\times d \right|}\).

The shortest distance between the skew lines \(\frac{x-{{x}_{1}}}{{{a}_{1}}}=\frac{y-{{y}_{1}}}{{{b}_{1}}}=\frac{z-{{z}_{1}}}{{{c}_{1}}}\) and \(\frac{x-{{x}_{2}}}{{{a}_{2}}}=\frac{y-{{y}_{2}}}{{{b}_{2}}}=\frac{z-{{z}_{2}}}{{{c}_{2}}}\) is \(\frac{\left| \begin{matrix} {{x}_{2}}-{{x}_{1}} & {{y}_{2}}-{{y}_{1}} & {{z}_{2}}-{{z}_{1}} \\ {{a}_{1}} & {{b}_{1}}

& {{c}_{1}} \\ {{a}_{2}} & {{b}_{2}} & {{c}_{2}} \\\end{matrix}

\right|}{\sqrt{{{({{b}_{1}}{{c}_{2}}-{{b}_{2}}{{c}_{1}})}^{2}}+{{({{c}_{1}}{{a}_{2}}-{{c}_{2}}{{a}_{1}})}^{2}}+{{({{a}_{1}}{{b}_{2}}-{{a}_{2}}{{b}_{1}})}^{2}}}}\)