Shortest Distance Between a Point and a Line (DP IB Analysis & Approaches (AA): HL): Revision Note

Shortest distance between a point and a line

What is the shortest distance from a point to a line?

• The shortest distance from any point to a line is always the perpendicular distance

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

  • Let $P$ be a point that does not lie on the line

• The shortest distance between the point and the line is sometimes referred to as the length of the perpendicular

• The point on the line that is closest to $P$ is sometimes referred to as the foot of the perpendicular

How do I find the shortest distance from a point to a line?

• For example, consider

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

  • the point $P\left(10,5,-10\right)$

• STEP 1
  Sketch a diagram showing the point $F$ on the line $l$ that is closest to the point $P$

  • The vector $\overrightarrow{FP}$ will be perpendicular to the line $l$

  • The point $F$ is sometimes called the foot of the perpendicular

• STEP 2
  Use the equation of the line to find the position vector of the point $F$  in terms of $\lambda$

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

• STEP 3
  Use this to find the displacement vector $\overrightarrow{FP}$ in terms of $\lambda$

  • $\overrightarrow{FP}=\left(\begin{array}{c}10\\ 5\\ -10\end{array}\right)-\left(\begin{array}{c}3+\lambda \\ -1-2\lambda \\ -2\end{array}\right)=\left(\begin{array}{c}7-\lambda \\ 6+2\lambda \\ -8\end{array}\right)$

• STEP 4
  Set the scalar product of the direction vector of the line $l$ and the displacement vector $\overrightarrow{FP}$ equal to zero

  • $\left(\begin{array}{c}7-\lambda \\ 6+2\lambda \\ -8\end{array}\right)·\left(\begin{array}{c}1\\ -2\\ 0\end{array}\right)=0\Rightarrow -5-5\lambda =0$

• STEP 5
  Solve the equation to find the value of $\lambda$

  • $\lambda =-1$

• STEP 6
  Substitute $\lambda$ into $\overrightarrow{FP}$ and find the magnitude $\left|\overrightarrow{FP}\right|$ 

  • $\left|\left(\begin{array}{c}7-\left(-1\right)\\ 6+2\left(-1\right)\\ -8\end{array}\right)\right|=12$

How can I use the vector product to find the shortest distance from a point to a line?

• The vector product can be used to find the shortest distance from any point to a line on a 2-dimensional plane

• The formula for the shortest distance is $\frac{\left|\overrightarrow{AP}\times b\right| }{\left|b\right|}$

  • $\mathit{r}=\mathit{a}+\lambda \mathit{b}$ is an equation of the line

  • $P$ is the point not on the line

  • $A$ is any point on the line