Shortest Distance Between Two Lines: The meaning of distance between two lines is how far the lines are located from each other. A line is formed when any \(2\) points are connected, and both the ends of a line are extended to infinity. The distance between two lines is the perpendicular distance between them.

Generally, we find the distance between two parallel lines. For the two non-intersecting lines which lie in the same plane, the shortest distance between them is the shortest distance between two points on the lines. We will learn more about the shortest distance between the two lines in this article.

Forms of Line

The various forms of the line are given below:

General Form of a Line

If \({P_1}\) and \({P_2}\) are two intersecting planes, then they always intersect along a line. If \({P_1} = {a_1}x + {b_1}y + {c_1}z + {d_1} = 0\) and \({P_2} = {a_2}x + {b_2}y + {c_2}z + {d_2} = 0\) are two non-parallel lines.

Then, \({a_1}x + {b_1}y + {c_1}z + {d_1} = 0 = {a_2}x + {b_2}y + {c_2}z + {d_2}\) represents the equation of the line of intersection of two given planes \({P_1}\) and \({P_2}.\)

This form of representation is also known as the non-symmetrical form or general form of a line. Hence, a straight line is represented by two equations of first degree in three variables \(x,y\) and \(z.\)

Symmetrical Form of Equation of a Line

1. Equation of line passing through a fixed point and given direction ratios

a. Vector form

Let the line pass through a fixed point \(A\) whose position vector is \(\overrightarrow a \) and a line is parallel to the vector \(\overrightarrow b .\) Let \(P\left( {\overrightarrow r } \right)\) be a variable point on the line. Then the equation of line in the vector form is
\(\vec r = \vec a + \lambda \vec b\)

b. Cartesian form or parametric form

Let the line pass through a fixed point \(A\left( {{x_1},{y_1},{z_1}} \right)\) and parallel to \(\overrightarrow b \) whose direction cosines are \(l,m,n.\) If \(P\left( {x,y,z} \right)\) be a point on the line, then the equation of the line is \(\frac{{x – {x_1}}}{l} = \frac{{y – {y_1}}}{m} = \frac{{z – {z_1}}}{n} = \lambda \)

2. Equation of line passing through two fixed points

a. Vector form

Let the line pass through two fixed points \(A\) and \(B,\) whose position vectors are \(\overrightarrow a \) and \(\overrightarrow b ,\) respectively. Let \(P\) be a variable point on the line whose position vector is \(\overrightarrow r .\) Then the equation of the line in vector form is \(\vec r = \vec a + \lambda \left( {\vec b – \vec a} \right)\)

b. Cartesian form or parametric form

Let the line pass through two fixed points \(A\left( {{x_1},{y_1},{z_1}} \right)\) and \(B\left( {{x_2},{y_2},{z_2}} \right).\)
If \(P\left( {x,y,z} \right)\) is a variable point on the line, then the equation of the line is
\(\frac{{x – {x_1}}}{{{x_2} – {x_1}}} = \frac{{y – {y_1}}}{{{y_2} – {y_1}}} = \frac{{z – {z_1}}}{{{z_2} – {z_1}}} = \lambda \)

Shortest Distance Between Two Lines

1. If two lines in space intersect at a point, then the shortest distance between them is zero.
2. If two lines in space are parallel, then the shortest distance between them will be the perpendicular distance from any point on the first line to the second line.

Note: In space, there are lines that are neither intersecting nor parallel. Such pair of lines are non-coplanar and are called skew lines.

Skew Lines

Two straight lines in the space which are neither intersecting nor parallel are said to be skew lines. Thus, the two skew lines in space are never coplanar.

Line of Shortest Distance

If \({l_1}\) and \({l_2}\) are two skew lines, then there is exactly one line which is perpendicular to both the lines \({l_1}\) and \({l_2}\) that is known as the line of shortest distance.

Shortest Distance Between Two Skew Lines

Vector Form:

Let \({l_1}\) and \({l_2}\) be two lines whose equations are:
\(\vec r = \overrightarrow {{a_1}} + \lambda \overrightarrow {{b_1}} \) and \(\vec r = \overrightarrow {{a_2}} + \mu \overrightarrow {{b_2}} \) respectively.
clearly \({l_1}\) and \({l_2}\) pass through the points \(A\) and \(B\) with position vectors \(\overrightarrow {{a_1}} \) and \(\overrightarrow {{a_2}} \) respectively and are parallel to the vectors \(\overrightarrow {{b_1}} \) and \(\overrightarrow {{b_2}} \) respectively.

Since \(\overrightarrow {PQ} \) is perpendicular to both \({l_1}\) and \({l_2}\) which are parallel to \(\overrightarrow {{b_1}} \) and \(\overrightarrow {{b_2}} .\)
\(\therefore \overrightarrow {PQ} \) is parallel to \(\overrightarrow {{b_1}} \times \overrightarrow {{b_2}} \)
Let \(\hat n\) be a unit vector along \(\overrightarrow {PQ} ,\) then
\(\hat n = \pm \frac{{\overrightarrow {{b_1}} \times \overrightarrow {{b_2}} }}{{\left| {\overrightarrow {{b_1}} \times \overrightarrow {{b_2}} } \right|}}\)
\(PQ = \) Projection of \(\overrightarrow {AB} \) on \(\overrightarrow {PQ} \)
\( \Rightarrow \overrightarrow {PQ} = \overrightarrow {AB} \cdot \hat n\)
\( = \pm \left( {\overrightarrow {{a_1}} – \overrightarrow {{a_2}} } \right) \cdot \frac{{\overrightarrow {{b_1}} \times \overrightarrow {{b_2}} }}{{\left| {\overrightarrow {{b_1}} \times \overrightarrow {{b_2}} } \right|}} = \pm \frac{{\left( {\overrightarrow {{b_1}} \times \overrightarrow {{b_2}} } \right) \cdot \left( {\overrightarrow {{a_1}} – \overrightarrow {{a_2}} } \right)}}{{\left| {\overrightarrow {{b_1}} \times \overrightarrow {{b_2}} } \right|}}\)
Hence, distance \(PQ = \left| {\frac{{\left( {\overrightarrow {{b_1}} \times \overrightarrow {{b_2}} } \right) \cdot \left( {\overrightarrow {{a_1}} – \overrightarrow {{a_2}} } \right)}}{{\left| {\overrightarrow {{b_1}} \times \overrightarrow {{b_2}} } \right|}}} \right|\)

Condition on Lines to Intersect

Two lines are intersecting if
\(\left| {\frac{{\left( {\overrightarrow {{b_1}} \times \overrightarrow {{b_2}} } \right) \cdot \left( {\overrightarrow {{a_1}} – \overrightarrow {{a_2}} } \right)}}{{\left| {\overrightarrow {{b_1}} \times \overrightarrow {{b_2}} } \right|}}} \right| = 0\)
\( \Rightarrow \left( {\overrightarrow {{b_1}} \times \overrightarrow {{b_2}} } \right) \cdot \left( {\overrightarrow {{a_1}} – \overrightarrow {{a_2}} } \right) = 0\)
This is the required condition for two intersecting lines.

Cartesian Form

Let the skew lines be \(\frac{{x – {x_1}}}{{{a_1}}} = \frac{{y – {y_1}}}{{{b_1}}} = \frac{{z – {z_1}}}{{{c_1}}}\) and \(\frac{{x – {x_2}}}{{{a_2}}} = \frac{{y – {y_2}}}{{{b_2}}} = \frac{{z – {z_2}}}{{{c_2}}}.\)
Vector equations of these two lines are:
\(\overrightarrow {{r_1}} = \left( {{x_1}\hat \imath + {y_1}\hat \jmath + {z_1}\hat k} \right) + \lambda \left( {{a_1}\hat \imath + {b_1}\hat \jmath + {c_1}\hat k} \right)\)
\(\overrightarrow {{r_2}} = \left( {{x_2}\hat \imath + {y_2}\hat \jmath + {z_2}\hat k} \right) + \lambda \left( {{a_2}\hat \imath + {b_2}\hat \jmath + {c_2}\hat k} \right)\)
Shortest distance \( = \left| {\frac{{\left( {\overrightarrow {{b_1}} \times \overrightarrow {{b_2}} } \right) \cdot \left( {\overrightarrow {{a_1}} – \overrightarrow {{a_2}} } \right)}}{{\left| {\overrightarrow {{b_1}} \times \overrightarrow {{b_2}} } \right|}}} \right|\)
\( \Rightarrow d = \frac{{\left| {\begin{array}{*{20}{c}} {{x_2} – {x_1}}&{{y_2} – {y_1}}&{{z_2} – {z_1}} \\ {{a_1}}&{{b_1}}&{{c_1}} \\ {{a_2}}&{{b_2}}&{{c_2}} \end{array}} \right|}}{{\sqrt {{{\left( {{m_1}{n_2} – {m_2}{n_1}} \right)}^2} + {{\left( {{n_1}{l_2} – {n_2}{l_1}} \right)}^2} + {{\left( {{l_1}{m_2} – {l_2}{m_1}} \right)}^2}} }}\)

Condition on Lines to Intersect
\(\left| {\begin{array}{*{20}{c}} {{x_2} – {x_1}}&{{y_2} – {y_1}}&{{z_2} – {z_1}} \\ {{a_1}}&{{b_1}}&{{c_1}} \\ {{a_2}}&{{b_2}}&{{c_2}} \end{array}} \right| = 0\)

Shortest Distance Between Parallel Lines

Let \({l_1}\) and \({l_2}\) be two lines whose equations are:
\(\vec r = \overrightarrow {{a_1}} + \lambda \overrightarrow {{b_1}} \) and \(\vec r = \overrightarrow {{a_2}} + \mu \overrightarrow {{b_2}} \) respectively.
Clearly \({l_1}\) and \({l_2}\) pass through the points \(A\) and \(B\) with position vectors \(\overrightarrow {{a_1}} \) and \(\overrightarrow {{a_2}} \) respectively and both are parallel to the vectors \(\overrightarrow b \) respectively, where \(BM\) is the shortest distance between \({l_1}\) and \({l_2}.\)

Let \(\theta \) be the angle between \(AB\) and \({l_1}.\)
\(\therefore \sin \theta = \frac{{BM}}{{AB}}\)
\( \Rightarrow BM = AB\sin \theta = \overrightarrow {\left| {AB} \right|} \sin \theta \)
Now, \(\left| {\overrightarrow {AB} \times \vec b} \right| = \left| {\overrightarrow {AB} } \right|\left| {\vec b} \right|\sin \left( {\pi – \theta } \right)\)
\( = \left| {\overrightarrow {AB} } \right|\left| {\vec b} \right|\sin \left( \theta \right) = \left( {\left| {\overrightarrow {AB} } \right|\sin \theta } \right)\left| {\vec b} \right|\)
\(\therefore \left| {\overrightarrow {AB} \times \vec b} \right| = BM\left| {\vec b} \right|\)
\( \Rightarrow BM = \frac{{\left| {\overrightarrow {AB} \times \vec b} \right|}}{{\left| {\vec b} \right|}}\)
\(\therefore BM = \frac{{\left| {\left( {\overrightarrow {{a_2}} – \overrightarrow {{a_1}} } \right) \times \vec b} \right|}}{{\left| {\vec b} \right|}}\)

Method to Determine Equation of Line of Shortest Distance

Let the equation of two skew lines be
\({L_1} = \frac{{x – {x_1}}}{{{l_1}}} = \frac{{y – {y_1}}}{{{m_1}}} = \frac{{z – {z_1}}}{{{n_1}}}\)
\({L_2} = \frac{{x – {x_2}}}{{{l_2}}} = \frac{{y – {y_2}}}{{{m_2}}} = \frac{{z – {z_2}}}{{{n_2}}}\)
As the line of shortest distance is perpendicular to both \({L_1}\) and \({L_2},\) let \(\left( {l,m,n} \right)\) are \(DR’s\) of the line of shortest distance, then
\(l{l_1} + m{m_1} + n{n_1} = 0 \ldots \left( i \right)\)
\(l{l_2} + m{m_2} + n{n_2} = 0 \ldots \left( ii \right)\)
From equations \(\left( i \right)\) and \(\left( {ii} \right)l,m,n\) can be obtained by the cross-product method.
As the line of shortest distance is a line which is coplanar with \({L_1}\) and \({L_2}\) separately.
Let the equation of the line of shortest be
\(L:\frac{{x – \alpha }}{l} = \frac{{y – \beta }}{m} = \frac{{z – \gamma }}{n}\)
Equation of plane containing \({L_1}\) and \(L\) is
\(\left| {\begin{array}{*{20}{c}} {x – {x_1}}&{y – {y_1}}&{z – {z_1}} \\ {{l_1}}&{{m_1}}&{{n_1}} \\ l&m&n \end{array}} \right| = 0…..\left( {iii} \right)\)
Equation of plane containing \({L_2}\) and \(L\) is
\(\left| {\begin{array}{*{20}{c}} {x – {x_2}}&{y – {y_2}}&{z – {z_2}} \\ {{l_2}}&{{m_2}}&{{n_2}} \\ l&m&n \end{array}} \right| = 0…\left( {iv} \right)\)
Equation of the line of shortest distance of \({L_1}\) and \({L_2}\) is the line of intersection of planes given by equations \(\left( {iii} \right)\) and \(\left( {iv} \right).\)

Solved Examples on Shortest Distance Between Two Lines

Q.1. Find the coordinates of the foot of the perpendicular drawn from point \(P\left( {1,0,3} \right)\) to the join of points \(Q\left( {4,7,1} \right)\) and \(R\left( {3,5,3} \right).\)
Ans: Let \(D\) be the foot of the perpendicular and let it divide \(QR\) in the ratio \(\lambda :1.\)

Then, the coordinates of \(D\) are \(\frac{{3\lambda + 4}}{{\lambda + 1}},\frac{{5\lambda + 7}}{{\lambda + 1}}\) and \(\frac{{3\lambda + 1}}{{\lambda + 1}}.\)
Now, \(\overrightarrow {PD} \bot \overrightarrow {QR} \Rightarrow \overrightarrow {PD} \cdot \overrightarrow {QR} = 0\)
\( \Rightarrow \left( {2\lambda + 3} \right) + 2\left( {5\lambda + 7} \right) + 4 = 0\)
\( \Rightarrow \lambda = – \frac{7}{4}\)
\(\therefore \) Coordinates of \(D\) are \(\frac{5}{3},\frac{7}{3}\) and \(\frac{{17}}{3}.\)

Q.2. Find the foot of the perpendicular from \(P\left( {1, – 3,1} \right)\) to the line \(\frac{{x + 1}}{1} = \frac{{y – 3}}{3} = \frac{{z + 2}}{{ – 1}}.\)
Ans:
Given: Line is \(\frac{{x + 1}}{1} = \frac{{y – 3}}{3} = \frac{{z + 2}}{{ – 1}}\) and \(P\left( {1, – 3,1} \right)….\left( i \right)\)

Coordinates of any point on the line \(\left( i \right)\) may be taken as \(\left( {r – 1,3r + 3, – r – 2} \right).\)
Let the foot of the perpendicular is \(Q = \left( {r – 1,3r + 3, – r – 2} \right)\)
\(Dr’s\) of \(\overrightarrow {PQ} \) are \(\left( {r – 2,3r + 6, – r – 3} \right)\)
\(Dr’s\) of \(\overrightarrow {AB} \) are \(\left( {1,3, – 1} \right)\)
Since \(\overrightarrow {PD} \bot \overrightarrow {AB} \)
\(1\left( {r – 2} \right) – 3\left( {3r + 6} \right) – 1\left( { – r – 3} \right) = 0\)
\( \Rightarrow \left( {r – 2} \right) – \left( {9r + 18} \right) + \left( {r + 3} \right) = 0\)
\( \Rightarrow – 7r – 17 = 0\)
\(\therefore r = – \frac{{17}}{7}\)
Hence, \(Q = \left( { – \frac{{31}}{7}, – \frac{9}{7},\frac{3}{7}} \right).\)

Q.3. Show that the two lines \(\frac{{x – 1}}{2} = \frac{{y – 2}}{3} = \frac{{z – 3}}{4}\) and \(\frac{{x – 4}}{5} = \frac{{y – 1}}{2} = z,\) intersect.
Ans:
Here, \(\frac{{x – 1}}{2} = \frac{{y – 2}}{3} = \frac{{z – 3}}{4}…\left( i \right)\)
and \(\frac{{x – 4}}{5} = \frac{{y – 1}}{2} = \frac{{z – 0}}{4}…\left( {ii} \right)\)
Let \(P\) be any point on the line \(\left( i \right)\) and \(P\left( {2r + 1,3r + 2,4r + 3} \right)\)
Let \(Q\) be any point on the line \(\left( ii \right)\) and \(2\left( {5\lambda + 4,2\lambda + 1,\lambda } \right)\)
The intersect if and only if \(2r + 1 = 5\lambda + 4,3r + 2 = 2\lambda + 1,4r + 3 = \lambda \))
On solving \(r = – 1\) and \(\lambda = – 1.\)
Clearly, for these values of \(r\) and \(\lambda ,P\left( { – 1, – 1, – 1} \right).\)
Hence, lines \(\left( i \right)\) and \(\left( ii \right)\) intersect at \(\left( { – 1, – 1, – 1} \right).\)

Q.4. Find the shortest distance between the lines \(\vec r = \left( {\hat \imath + 2\hat \jmath + \hat k} \right) + \lambda \left( {2\hat \imath + \hat \jmath + 2\hat k} \right)\) and \(\vec r = \left( {2\hat \imath – \hat \jmath – \hat k} \right) + \mu \left( {2\hat \imath + \hat \jmath + 2\hat k} \right).\)
Ans: Here, lines are passing through the points \(\overrightarrow {{a_1}} = \left( {\hat \imath + 2\hat \jmath + \hat k} \right)\) and \(\overrightarrow {{a_2}} = \left( {2\hat \imath – \hat \jmath – \hat k} \right).\)
Hence, the distance between the lines using the formula
\(\frac{{\left| {\left( {\overrightarrow {{a_2}} – \overrightarrow {{a_1}} } \right) \times \vec b} \right|}}{{\left| {\vec b} \right|}} = \frac{{\left| {\begin{array}{*{20}{c}} {\hat \imath }&{\hat j}&{\hat k} \\ 2&1&2 \\ 1&{ – 3}&{ – 2} \end{array}} \right|}}{3}\)
\( = \frac{{\left| {4\hat \imath – 6\hat \jmath – 7\hat k} \right|}}{3}\)
\( = \frac{{\sqrt {16 + 36 + 49} }}{3}\)
\( = \frac{{\sqrt {101} }}{3}\)

Q.5. Find the shortest distance between the line passing through the point \(\left( {2, – 1,1} \right)\) and parallel to the vector \(\left( { – 1,1,2} \right)\) and the straight line passing through \(\left( {0,3,1} \right)\) and parallel to the vector \(\left( {2,4, – 1} \right).\)
Ans:
Let \(\vec r = \vec a + t\vec b\)
where
\(\vec a = \left( {2\hat \imath – \hat \jmath + \hat k} \right)\)
\(\vec b = \left( { – \hat \imath + \hat \jmath + 2\hat k} \right)\)
Given straight line is \(\vec r = \vec c + s\vec d\)
where
\(\vec c = \left( {3\hat \jmath + \hat k} \right)\)
\(\vec d = \left( {2\hat i + 4\hat j – \hat k} \right)\)
Clearly, \(\vec a – \vec c = \left( {2\hat \imath – 4\hat j} \right)\)
\(\vec b \times \vec d = \left| {\begin{array}{*{20}{c}} {\hat \imath }&{\hat j}&{\hat k} \\ { – 1}&1&2 \\ 2&4&{ – 1} \end{array}} \right|\)
\(\therefore \vec b \times \vec d = – 9\hat \imath + 3\hat \jmath – 6\hat k\)
\(\left[ {\begin{array}{*{20}{c}} {\vec a – \vec c}&{\vec b}&{\vec d} \end{array}} \right] = \left| {\begin{array}{*{20}{c}} 2&{ – 4}&0 \\ { – 1}&1&2 \\ 2&4&{ – 1} \end{array}} \right|\)
\(\therefore \left[ {\begin{array}{*{20}{l}} {\vec a – \vec c}&{\vec b}&{\vec d} \end{array}} \right] = – 30\)
Shortest distance between the lines \( = \frac{{\left[ {\begin{array}{*{20}{l}} {\vec a – \vec c}&{\vec b}&{\vec d} \end{array}} \right]}}{{\left| {\vec b \times \vec d} \right|}}\)
\( = \frac{{\left| { – 30} \right|}}{{\sqrt {81 + 9 + 36} }}\)
\( = \frac{{30}}{{\sqrt {126} }}\)
\( = \sqrt {\frac{{50}}{7}} {\text{units}}.\)

Summary

The meaning of the distance between two lines is how far the lines are located from each other. There are two types of lines symmetric and non-symmetric. The shortest distance between two points in three-dimensional coordinate geometry can be calculated by finding the length of the line segment joining them.
Let \(A\left( {{x_1},{y_1},{z_1}} \right)\) and \(B\left( {{x_2},{y_2},{z_2}} \right)\) be two points, then the shortest distance between the points are given by \(AB = \sqrt {{{\left( {{x_2} – {x_1}} \right)}^2} + {{\left( {{y_2} – {y_1}} \right)}^2} + {{\left( {{z_2} – {z_1}} \right)}^2}} .\) If two lines in space intersect, then the shortest distance between them is zero. If the lines are parallel, the shortest distance between them will be the length of the perpendicular line between them. If \({l_1}\) and \({l_2}\) are two skew lines, there is only one perpendicular line. The length of this line is the line of shortest distance.

FAQs on Shortest Distance Between Two Lines

Q.1. What is the formula to find the shortest distance between two lines?
Ans: Shortest Distance between Two Lines:
Vector Form: Distance \( = \left| {\frac{{\left( {\overrightarrow {{b_1}} \times \overrightarrow {{b_2}} } \right) \cdot \left( {\overrightarrow {{a_1}} – \overrightarrow {{a_2}} } \right)}}{{\left| {\overrightarrow {{b_1}} \times \overrightarrow {{b_2}} } \right|}}} \right|\)
Cartesian Form: Shortest distance \( = \left| {\frac{{\left( {\overrightarrow {{b_1}} \times \overrightarrow {{b_2}} } \right) \cdot \left( {\overrightarrow {{a_1}} – \overrightarrow {{a_2}} } \right)}}{{\left| {\overrightarrow {{b_1}} \times \overrightarrow {{b_2}} } \right|}}} \right|\)
\(\therefore d = \frac{{\left| {\begin{array}{*{20}{c}} {{x_2} – {x_1}}&{{y_2} – {y_1}}&{{z_2} – {z_1}} \\ {{a_1}}&{{b_1}}&{{c_1}} \\ {{a_2}}&{{b_2}}&{{c_2}} \end{array}} \right|}}{{\sqrt {{{\left( {{m_1}{n_2} – {m_2}{n_1}} \right)}^2} + {{\left( {{n_1}{l_2} – {n_2}{l_1}} \right)}^2} + {{\left( {{l_1}{m_2} – {l_2}{m_1}} \right)}^2}} }}\)

Q.2. What is the shortest distance formula for two parallel lines?
Ans: Let \({l_1}\) and \({l_2}\) be two lines whose equations are:
\(\vec r = \overrightarrow {{a_1}} + \lambda \overrightarrow {{b_1}} \) and \(\vec r = \overrightarrow {{a_2}} + \mu \overrightarrow {{b_2}} \) respectively, and both are parallel to vector \(b\)
Then shortest distance between parallel lines is given by
\(\frac{{\left| {\overrightarrow {AB} \times \vec b} \right|}}{{\left| {\vec b} \right|}} = \frac{{\left| {\left( {\overrightarrow {{a_2}} – \overrightarrow {{a_1}} } \right) \times \vec b} \right|}}{{\left| {\vec b} \right|}}\)

Q.3. What is the distance between \(2\) points?
Ans: The shortest distance between two points in three-dimensional coordinate geometry can be calculated by finding the length of the line segment joining the given coordinates.
If \(A\left( {{x_1},{y_1},{z_1}} \right)\) and \(B\left( {{x_2},{y_2},{z_2}} \right)\) are two points, then the shortest distance between the points are given by
\(AB = \sqrt {{{\left( {{x_2} – {x_1}} \right)}^2} + {{\left( {{y_2} – {y_1}} \right)}^2} + {{\left( {{z_2} – {z_1}} \right)}^2}} \)

Q.4. How do you find the shortest distance in reasoning?
Ans: The meaning of the shortest distance between two lines is the joining of a point in the first line with one point on the second line so that the length of the line segment between the points is the smallest.
1. If two lines in space intersect at a point, then the shortest distance between them is zero.
2. If two lines in space are parallel, then the shortest distance between them is the perpendicular distance drawn from any point on the first line to the second line.

Q.5. How do you find the shortest distance between two lines in vector form?
Ans: Let \({l_1}\) and \({l_2}\) be two lines whose equations are:
\(\vec r = \overrightarrow {{a_1}} + \lambda \overrightarrow {{b_1}} \) and \(\vec r = \overrightarrow {{a_2}} + \mu \overrightarrow {{b_2}} ,\) respectively.
Then the shortest distance between them \( = \left| {\frac{{\left( {\overrightarrow {{b_1}} \times \overrightarrow {{b_2}} } \right) \cdot \left( {\overrightarrow {{a_1}} – \overrightarrow {{a_2}} } \right)}}{{\left| {\overrightarrow {{b_1}} \times \overrightarrow {{b_2}} } \right|}}} \right|\)