Shortest Distance Between Two Lines (DP IB Applications & Interpretation (AI): HL): Revision Note

Shortest distance between two lines

What is the shortest distance between two skew lines?

• The shortest distance between two lines is always the perpendicular to both lines

  • Let $l_1$ be a line with equation $\mathit{r}={\mathit{a}}_1+\lambda {\mathit{b}}_1$  

  • Let $l_2$ be a line with equation $\mathit{r}={\mathit{a}}_2+\mu {\mathit{b}}_2$  

• The vector product of the two direction vectors is perpendicular to both lines

How do I find the shortest distance between two skew lines?

• For example, consider the lines with equations:

  • $l_1:\mathit{r}=\left(\begin{array}{c}3\\ -1\\ -2\end{array}\right)+\lambda \left(\begin{array}{c}1\\ -2\\ 0\end{array}\right)$

  • $l_2:\mathit{r}=\left(\begin{array}{c}9\\ 9\\ -9\end{array}\right)+\mu \left(\begin{array}{c}-1\\ 4\\ 1\end{array}\right)$

Using the vector product

• STEP 1
  Find the vector product of the direction vectors ${\mathit{b}}_1$ and ${\mathit{b}}_2$

  • $\mathit{d}=\left(\begin{array}{c}1\\ -2\\ 0\end{array}\right)\times \left(\begin{array}{c}-1\\ 4\\ 1\end{array}\right)=\left(\begin{array}{c}-2\\ -1\\ 2\end{array}\right)$

• STEP 2
  Find an expression for the position vector of a point $A$ on $l_1$ and a point $B$ on $l_2$  in terms of $\lambda$ and $\mu$

  • $\overrightarrow{OA}=\left(\begin{array}{c}3+\lambda \\ -1-2\lambda \\ -2\end{array}\right)$

  • $\overrightarrow{OB}=\left(\begin{array}{c}9-\mu \\ 9+4\mu \\ -9+\mu \end{array}\right)$

• STEP 3
  Find the displacement vector between the general points

  • $\overrightarrow{AB}=\left(\begin{array}{c}9-\mu \\ 9+4\mu \\ -9+\mu \end{array}\right)-\left(\begin{array}{c}3+\lambda \\ -1-2\lambda \\ -2\end{array}\right)=\left(\begin{array}{c}6-\mu -\lambda \\ 10+4\mu +2\lambda \\ -7+\mu \end{array}\right)$

• STEP 4
  Set the displacement vector parallel to the vector product and set up a system of three linear equations in terms of $k, \lambda$ and $\mu$

  • $\left(\begin{array}{c}6-\mu -\lambda \\ 10+4\mu +2\lambda \\ -7+\mu \end{array}\right)=k\left(\begin{array}{c}-2\\ -1\\ 2\end{array}\right)\Rightarrow \begin{array}{c}-\mu -\lambda +2k=-6\\ 4\mu +2\lambda +k=-10\\ \mu -2k=7\end{array}$

• STEP 5
  Solve the system to find the value of $k$

  • $k=-4$

• STEP 6
  Multiply the vector product by $k$and find the magnitude $\left|k\mathit{d}\right|$ 

  • $\left|-4\left(\begin{array}{c}-2\\ -1\\ 2\end{array}\right)\right|=12$

Using the scalar product

• STEP 1
  Sketch a diagram showing the point $F_1$ on the line $l_1$ that is closest to the point $F_2$ on the line $l_2$

  • The vector $\overrightarrow{F_1{F}_2}$ will be perpendicular to both lines

• STEP 2
  Use the equations of the lines to find the position vector of the point $F_1$  in terms of $\lambda$ and $F_2$ in terms of $\mu$

  • $\overrightarrow{O{F}_1}=\left(\begin{array}{c}3+\lambda \\ -1-2\lambda \\ -2\end{array}\right)$

  • $\overrightarrow{O{F}_2}=\left(\begin{array}{c}9-\mu \\ 9+4\mu \\ -9+\mu \end{array}\right)$

• STEP 3
  Use this to find the displacement vector $\overrightarrow{F_1{F}_2}$ in terms of $\lambda$ and $\mu$

  • $\overrightarrow{F_1{F}_2}=\left(\begin{array}{c}9-\mu \\ 9+4\mu \\ -9+\mu \end{array}\right)-\left(\begin{array}{c}3+\lambda \\ -1-2\lambda \\ -2\end{array}\right)=\left(\begin{array}{c}6-\mu -\lambda \\ 10+4\mu +2\lambda \\ -7+\mu \end{array}\right)$

• STEP 4
  Form two equations by setting the scalar product of the direction vector of each line and the displacement vector $\overrightarrow{F_1{F}_2}$ equal to zero

  • $\left(\begin{array}{c}6-\mu -\lambda \\ 10+4\mu +2\lambda \\ -7+\mu \end{array}\right)·\left(\begin{array}{c}1\\ -2\\ 0\end{array}\right)=0\Rightarrow -9\mu -5\lambda =14$

  • $\left(\begin{array}{c}6-\mu -\lambda \\ 10+4\mu +2\lambda \\ -7+\mu \end{array}\right)·\left(\begin{array}{c}-1\\ 4\\ 1\end{array}\right)=0\Rightarrow 18\mu +9\lambda =-27$

• STEP 5
  Solve the equations simultaneously to find the values of $\lambda$ and $\mu$

  • $\lambda =-1$ and $\mu =-1$

• STEP 6
  Substitute $\lambda$ and $\mu$ into $\overrightarrow{F_1{F}_2}$ and find the magnitude $\left|\overrightarrow{F_1{F}_2}\right|$ 

  • $\left|\left(\begin{array}{c}6-\left(-1\right)-\left(-1\right)\\ 10+4\left(-1\right)+2\left(-1\right)\\ -7+\left(-1\right)\end{array}\right)\right|=12$

How do I find the shortest distance between two parallel lines?

• Pick any point on one of the lines

• Find the shortest distance between that point and the other line

How do I find the shortest distance between a point on a line and an interesting line?

• You can use the steps to find the shortest distance between a point and a line

• It might be quicker to use right-angled trigonometry if you also know the point of intersection and/or the angle between the lines

  • The shortest length between the point and the line is perpendicular to the line