Distance Between Skew Lines

The shortest distance between skew lines is equal to the length of the perpendicular between the two lines. This lesson lets you understand the meaning of skew lines and how the shortest distance between them can be calculated. We will look at both, Vector and Cartesian equations in this topic. Let’s Begin!

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Skew Lines in 3-D Geometry

In three-dimensional geometry, we are always dealing with objects in the three-dimensional Cartesian space. One of the key elements of three-dimensional geometry is the straight line, also sometimes simply referred to as a line.

There can be various ways in which two lines are related in the three-dimensional space. Our focus in the following section shall be on skew lines. The objective is to find out how to measure the distance between such skew lines.

Note that in case the two skew lines are intersecting, the shortest distance between them must necessarily be zero. The other cases are that of parallel lines and skew lines. Skew Lines are basically, lines that neither intersect each other nor are they parallel to each other in the three-dimensional space.

Browse more Topics under Three Dimensional Geometry

• Angle Between a Line and a Plane
• Angle Between Two Lines
• Coplanarity of Two Lines
• Angle Between Two Planes
• Direction Cosines and Direction Ratios of a Line
• Distance Between Parallel Lines
• Distance of a Point from a Plane
• Equation of a Plane in Normal Form
• Equation of a Plane Perpendicular to a Given Vector and Passing Through a Given Point
• The Equation of Line for Space
• Equation of Plane Passing Through Three Non Collinear Points
• Intercept Form of the Equation of a Plane
• Plane Passing Through the Intersection of Two Given Planes

The Distance Between Skew Lines

Look at the figure below. You can see two lines from the three-dimensional Cartesian plane. As is evident from the figure, the shortest distance between the lines is one which is perpendicular to both the lines as compared to any other lines that joins these two skew lines.

(Source: Mathematics Stack Exchange)

We will now move on to how the shortest distance i.e. the length of the perpendicular to two skew lines can be calculated in Vector form and in Cartesian form.

Vector Form

We shall consider two skew lines, say l₁ and l­₂ and we are to calculate the distance between them. The equations of the lines are:

$$ \vec{r}_1 = \vec{a}_1 + t.\vec{b}_1 $$

$$ \vec{r}_2 = \vec{a}_2 + t.\vec{b}_2 $$

P = \(\vec{a}_1\) is a point on line l₁ and Q = \(\vec{a}_2\) is a point on line l₁. The vectro from P to Q will be \(\vec{a}_2 – \vec{a}_1\). The unit vector normal to both the lines is given by,

$$  (\vec{b}_1 \times \vec{b}_2) / | \vec{b}_1 \times \vec{b}_2 | $$

Then, the shortest distance between the two skew lines will be the projection of PQ on the normal, which is given by

$$ d = | (\vec{a}_2 – \vec{a}_1) .  (\vec{b}_1 \times \vec{b}_2) | / | \vec{b}_1 \times \vec{b}_2 | $$

where d measures the distance or the length of the perpendicular.

Cartesian Form

Let us consider two lines whose equations are given by:

$$ ( x – x_1 ) / a_1 = ( y – y_1 ) / b_1  = ( z – z_1) / c_1 $$

$$ ( x – x_2 ) / a_2 = ( y – y_2 ) / b_2  = ( z – z_2 ) / c_2 $$

Then the shortest distance between these lines, when calculated using the Cartesian equations,  is given by

d = \( \begin{vmatrix} x_2 – x_1 & y_2 – y_1 & z_2 – z_1\\ a_1 & b_1 & c_1\\ a_2 & b_2 & c_2 \end{vmatrix} \) / \({[(b_1c_2 – b_2c_1)^2  + ( c_1a_2 – a_2c_1)^2 + (a_1b_2– b_2a_1)^2}]^{1/2} \)

Solved Example for You

Question 1: Find the shortest distance between the lines whose equations are:

$$ \vec{r}_1 = \vec{i} + \vec{j} + \lambda (2 \vec{i} –  \vec{j} + \vec{k} ) $$

$$ \vec{r}_2 = 2 \vec{i} +  \vec{j} – \vec{k}+ \mu (3\vec{i} – 5  \vec{j} + 2 \vec{k} ) $$

Answer: We shall compare the given equations with the standard form i.e. \vec{r}_1 = \vec{a}_1 + \lambda \vec{b}_1 and vec{r}_2 = \vec{a}_2 + \mu \vec{b}_2. Accordingly, we have:

$$ a_1 = \vec{i} + \vec{j} , b_1 = 2 \vec{i} –  \vec{j} + \vec{k} $$

$$ a_2 = 2 \vec{i} +  \vec{j} – \vec{k} , b_2 = 3\vec{i} – 5  \vec{j} + 2 \vec{k} $$

So, we can find the shortest distance as :

$$ d = | [(2 \vec{i} –  \vec{j} + \vec{k}) \times (3\vec{i} – 5  \vec{j} + 2 \vec{k} )] . (\vec{i} – \vec{k}) | / | (2 \vec{i} – \vec{j} + \vec{k}) \times (3 \vec{i} – 5  \vec{j} + 2 \vec{k}) | $$

= | 3 – 0 + 7 | / (59)^1/2

= |10| / (59)^1/2  is the shortest distance between the given lines.

Question 2: How does the skew lines look like?

Answer: The skew lines are those lines which do not intersect, but also never lie on a similar plane. They look like they are running in similar directions and even look totally random.

Question 3: Can skew lines be perpendicular?

Answer: A line is said to be perpendicular to the other line only if they are intersecting at a right angle, and we know that the skew lines never intersect or meet, so, the skew lines can never become a perpendicular.

Question 4: What are the dissimilarities between the parallel lines and the skew lines?

Answer: The most common difference between the parallel lines and the skew lines is that the parallel lines lie in a similar plane whereas the skew lines lie in dissimilar planes.

Question 5: How can we prove the skew lines?

Answer: Firstly, we have to show that they are not parallel to each other. For this, we will be taking the direction vectors and make sure that one is not a multiple of the other one.