The distance from a point to a line is defined as the length of the shortest possible segment connecting the point to the line. This shortest segment is always perpendicular to the line.

Let [latex]L[/latex] be a line and [latex]M[/latex] be any point not on the line. The distance [latex]d[/latex] from [latex]M[/latex] to [latex]L[/latex] is the length of line segment [latex]\overline{MP}[/latex], where [latex]P[/latex] is the point on [latex]L[/latex] such that [latex]\overline{MP}[/latex] is perpendicular to [latex]L[/latex].

In three dimensions, there’s no obvious way to identify which point on the line creates the perpendicular segment. We solve this by selecting an arbitrary point on the line and using vector properties.

Let [latex]P[/latex] be any point on line [latex]L[/latex] and let [latex]\textbf{v}[/latex] be a direction vector for [latex]L[/latex]. For any point [latex]M[/latex] not on the line, consider the vectors [latex]\overrightarrow{PM}[/latex] and [latex]\textbf{v}[/latex].

These vectors form two sides of a parallelogram. From the properties of cross products and geometry:

• The area of this parallelogram equals [latex]||\overrightarrow{PM} \times \textbf{v}||[/latex]
• The area also equals base × height = [latex]||\textbf{v}|| \cdot d[/latex]

Setting these equal gives us:

[latex]||\overrightarrow{PM} \times \textbf{v}|| = ||\textbf{v}|| \cdot d[/latex]

We can use this formula to find a general formula for the distance between a line in space and any point not on the line.

Find the distance between t point [latex]M=(1, 1, 3)[/latex] and line [latex]\frac{x-3}4=\frac{y+1}2=z-3[/latex].

Show Solution

From the symmetric equations of the line, we know that vector [latex]{\bf{v}}\langle4,2,1\rangle[/latex] is a direction vector for the line. Setting the symmetric equations of the line equal to zero, we see that point [latex]P(3, -1 , 3)[/latex] lies on the line. Then,

[latex]\overrightarrow{PM}=\langle1-3,1-(-1),3-3\rangle=\langle-2,2,0\rangle[/latex].

To calculate the distance, we need to find [latex]\overrightarrow{PM}\times{\bf{v}}[/latex]:

[latex]\begin{aligned} \overrightarrow{PM}\times{\bf{v}}&=\begin{vmatrix}{\bf{i}}&{\bf{j}}&{\bf{k}} \\-2&2&0 \\ 4&2&1\end{vmatrix} \\ &=(2-1){\bf{i}}-(-2-0){\bf{j}}+(-4-8){\bf{k}} \\ &=2{\bf{i}}+2{\bf{j}}-12{\bf{k}}. \end{aligned}[/latex]

Therefore, the distance between the point and the line is (Figure 4)

[latex]\begin{aligned} d&=\frac{||\overrightarrow{PM}\times{\bf{v}}||}{||{\bf{v}}||} \\ &=\frac{\sqrt{2^2+2^2+12^2}}{\sqrt{4^2+2^2+1^2}} \\ &=\frac{s\sqrt{38}}{\sqrt{21}}. \end{aligned}[/latex]