Distance from a point to a straight line in space

The distance between a point $P$ and a straight line $r$ , $d (P, r)$ 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 $\vec{P P^{'}}$ , 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 $\vec{Q P}$ and $\vec{v}$ has module equal to the vector product of both vectors: $S_{p} = | \vec{Q P} \times \vec{v} |$

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

Therefore, $d (P, r) = \frac{| \vec{Q P} \times \vec{v} |}{| \vec{v} |}$

Example

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 $\vec{Q P}$ and $\vec{v}$ .

$\vec{Q P} = (0, 1, 2)$

$\begin{aligned} | \vec{Q P} \times \vec{v} | = & | | \begin{array}{c} i & j & k \\ 0 & 1 & 2 \\ 1 & 2 & 1 \end{array} | | = | i + 2 j - k - 4 i | = | - 3 i + 2 j - k | \\ = & | (- 3, 2, - 1) | = \sqrt{9 + 4 + 1} = \sqrt{14} \end{aligned}$

and we can already apply the formula:

$d (P, r) = \frac{| \vec{Q P} \times \vec{v} |}{| \vec{v} |} = \frac{\sqrt{14}}{\sqrt{6}} = \sqrt{\frac{7}{3}}$