Distance from a point to a straight line in space

The distance between a point $P$ and a straight line $r$, $\text{d}(P,r)$ is the minimal distance between $P$ and any point of the straight line $r$.

• If $P$ is a point of the straight line $r$, the distance is zero.
• If $P$ is not on the straight line $r$, the distance from $P$ to $r$ will be the module of the vector $\overrightarrow{PP'}$, where $P'$ is the orthogonal projection of $P$ on the straight line $r$.

Nevertheless, there exists a simpler way to calculate the distance between a point $P$ and a straight line $r$ if the point does not belong to the straight line. Let's consider a point $Q$ on the straight line $r$ and the governing vector of the straight line, $\vec{v}$. The area of the parallelogram determined by the vector $\overrightarrow{QP}$ and $\vec{v}$ has module equal to the vector product of both vectors: $$S_p=|\overrightarrow{QP}\times\vec{v}|$$

But the area of a parallelogram is also given by the product of the base times the height. Then: $$|S_p=|\vec{v}|\cdot\text{d}(P,r)$$

Therefore, $$\text{d}(P,r)=\dfrac{|\overrightarrow{QP}\times\vec{v}|}{|\vec{v}|}$$

Calculate the distance between the point $P = (2, 4, 1)$ and the straight line $r: (x, y, z) = (2, 3, -1) + k\cdot(1, 2, 1)$.

We take a point of the straight line, for example $Q = (2, 3, -1)$. Now we will have to calculate the vector product of the vectors $\overrightarrow{QP}$ and $\vec{v}$.

$$\overrightarrow{QP}=(0,1,2)$$

$$\begin{array}{rl} |\overrightarrow{QP}\times\vec{v}|=& \left| \begin{vmatrix} i & j & k \\ 0& 1& 2 \\ 1& 2& 1 \end{vmatrix} \right| = |i+2j-k-4i|=|-3i+2j-k| \\ =& |(-3,2,-1)|=\sqrt{9+4+1}=\sqrt{14} \end{array}$$

and we can already apply the formula:

$$\text{d}(P,r)=\dfrac{|\overrightarrow{QP}\times\vec{v}|}{|\vec{v}|}= \dfrac{\sqrt{14}}{\sqrt{6}}=\sqrt{\dfrac{7}{3}}$$