The following is the derivation of the distance between 2 skew lines in vector form, We shall consider two skew lines, say l1 and l2 and we are to calculate the distance between them. The equations of the lines are: $\vec{r}_1 = \vec{a}_1 + t.\vec{b}_1$ $\vec{r}_2 = \vec{a}_2 + t.\vec{b}_2$ P =$\vec{a}_1$is a point on line l1 and Q =$\vec{a}_2$ is a point on line l1. The vectro from P to Q will be $\vec{a}_2$ -$\vec{a}_1$. The unit vector normal to both the lines is given by, $(\vec{b}_1 \times \vec{b}_2) / | \vec{b}_1 \times \vec{b}_2 |$. Then, the shortest distance between the two skew lines will be the projection of PQ on the normal, which is given by $d = | (\vec{a}_2 – \vec{a}_1) . (\vec{b}_1 \times \vec{b}_2) | / | \vec{b}_1 \times \vec{b}_2|$.
Now in this derivation I dont understand whhy would the distance between the two lines have the same magnitude as the projection of PQ, it will have the same direction i understand that but why would it have the same magnitude can someone explain, shouldnt the distance between them be $\lambda$PQ where $\lambda$ is a constant.
Let $AB$ be the distance between the two lines ($A\in l_1$ and $B\in l_2$). Then line $AB$ is perpendicular to both $l_1$ and $l_2$. It follows that the projection of $P$ on $AB$ is $A$ and the projection of $Q$ on $AB$ is $B$.