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Shortest Distances - Lines Simplified Revision Notes

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6.1.4 Shortest Distances - Lines

Introduction to Shortest Distances

Shortest distance calculations are essential in 3D geometry for finding:

  • The perpendicular distance between two lines.

Perpendicular Distance Between Two Lines

Two skew lines in 3D space can have a minimum distance between them. Let:

r1=a1+λb1,r2=a2+μb2\mathbf{r}_1 = \mathbf{a}_1 + \lambda \mathbf{b}_1, \quad \mathbf{r}_2 = \mathbf{a}_2 + \mu \mathbf{b}_2

Perpendicular Distance Formula:

d=(a2a1)×(b1×b2)b1×b2d = \frac{|(\mathbf{a}_2 - \mathbf{a}_1) \times (\mathbf{b}_1 \times \mathbf{b}_2)|}{|\mathbf{b}_1 \times \mathbf{b}_2|}

Here:

  • b1\mathbf{b}_1 and b2\mathbf{b}_2 are the direction vectors of the lines.
  • a1\mathbf{a}_1 and a2\mathbf{a}_2 are points on the lines.

Perpendicular Distance from a Point to a Line

Let the line be:

r=a+λb\mathbf{r} = \mathbf{a} + \lambda \mathbf{b}

and the point be p\mathbf{p}.

Perpendicular Distance Formula:

d=(pa)×bbd = \frac{|(\mathbf{p} - \mathbf{a}) \times \mathbf{b}|}{|\mathbf{b}|}

Worked Example

infoNote

Example: Perpendicular Distance Between Two Lines

Find the distance between:

r1=(123)+λ(421),r2=(102)+μ(231)\mathbf{r}_1 = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} + \lambda \begin{pmatrix} 4 \\ -2 \\ 1 \end{pmatrix}, \quad \mathbf{r}_2 = \begin{pmatrix} -1 \\ 0 \\ 2 \end{pmatrix} + \mu \begin{pmatrix} 2 \\ 3 \\ -1 \end{pmatrix}

Step 1: Find b1×b2\mathbf{b}_1 \times \mathbf{b}_2:

b1×b2=ijk421231\mathbf{b}_1 \times \mathbf{b}_2 = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 4 & -2 & 1 \\ 2 & 3 & -1 \end{vmatrix} =i(2+3)j(42)+k(12+4)= \mathbf{i}(-2 + 3) - \mathbf{j}(4 - 2) + \mathbf{k}(12 + 4) =(1216)= \begin{pmatrix} 1 \\ -2 \\ 16 \end{pmatrix}

Step 2: Compute(a2a1)(b1×b2) (\mathbf{a}_2 - \mathbf{a}_1) (\mathbf{b}_1 \times \mathbf{b}_2):

a2a1=(102)(123)=(221)\mathbf{a}_2 - \mathbf{a}_1 = \begin{pmatrix} -1 \\ 0 \\ 2 \end{pmatrix} - \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} = \begin{pmatrix} -2 \\ -2 \\ -1 \end{pmatrix}(a2a1)×(1216)=(2)(1)+(2)(2)+(1)(16)(\mathbf{a}_2 - \mathbf{a}_1) \times \begin{pmatrix} 1 \\ -2 \\ 16 \end{pmatrix} = (-2)(1) + (-2)(-2) + (-1)(16) =2+416=:highlight[14]= -2 + 4 - 16 = :highlight[-14]

Step 3: Find b1×b2|\mathbf{b}_1 \times \mathbf{b}_2|:

b1×b2=12+(2)2+162|\mathbf{b}_1 \times \mathbf{b}_2| = \sqrt{1^2 + (-2)^2 + 16^2} =1+4+256=:highlight[261]= \sqrt{1 + 4 + 256} = :highlight[\sqrt{261}]

Step 4: Calculate the distance:

d=(a2a1)×(b1×b2)b1×b2d = \frac{|(\mathbf{a}_2 - \mathbf{a}_1) \times (\mathbf{b}_1 \times \mathbf{b}_2)|}{|\mathbf{b}_1 \times \mathbf{b}_2|} =14261=:highlight[14261]= \frac{|-14|}{\sqrt{261}} = :highlight[\frac{14}{\sqrt{261}}]

Note Summary

infoNote

Key Formulas:

  1. Distance Between Two Lines:
d=(a2a1)×(b1×b2)b1×b2d = \frac{|(\mathbf{a}_2 - \mathbf{a}_1) \times (\mathbf{b}_1 \times \mathbf{b}_2)|}{|\mathbf{b}_1 \times \mathbf{b}_2|}
  1. Distance from Point to Line:
d=(pa)×bbd = \frac{|(\mathbf{p} - \mathbf{a}) \times \mathbf{b}|}{|\mathbf{b}|}
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