# Shortest distance between two parallel lines

Let $L_1$ be the line passing through the point $P_1=(4, −2, −3)$ with direction vector $\overrightarrow{d}=[−2, 1, 3]T$, and let $L_2$ be the line passing through the point $P_2=(−2, 3, −2)$ with the same direction vector. Find the shortest distance d between these two lines, and find a point $Q_1$ on $L_1$ and a point $Q_2$ on $L_2$ so that d(Q1,Q2) = d. Use the square root symbol '√' where needed to give an exact value for your answer.

I tried $p_1p_2$, and projected to direction vector $d$, calculated distance wrong. I don't know how to solve the problem.

• Try drawing any line perpendicular to these two and calculating the distance between the two intersections you get. – I want to make games Oct 30 '15 at 20:22
• Draw a picture. What you want to do is subtract the points $P_1$ from the point $P_2$ to get the vector $\vec{P_1P_2}$. Then realize that vector you want (the vector pointing from a point on $L_1$ to the corresponding point on $L_2$) and the vector $\vec{P_1P_2}$ are in the plane $\operatorname{span}(\vec{P_1P_2}, \vec d)$ and the vector you want is orthogonal to $\vec d$. So subtract the projection of $\vec{P_1P_2}$ onto $\operatorname{span}(\vec d)$ from $\vec{P_1P_2}$ to get the vector you want. – user137731 Oct 30 '15 at 20:34

Notice, a parametric point lying on the line $L_1$ is $A_1(-2t_1+4, t_1-2, 3t_1-3)$ &

parametric point lying on the line $L_2$ is $A_2(-2t_2-2, t_2+3, 3t_2-2)$

Since, $\vec{A_1A_2}=[-2(t_2-t_1)-6, (t_2-t_1)+5, 3(t_2-t_1)+1 ]$ is perpendicular to the direction vector $\vec d=[-2, 1, 3]$ hence we have $(\vec{A_1A_2})\cdot \vec d=0$ (dot product of perpendicular vectors is zero) $$\implies [-2(t_2-t_1)-6, (t_2-t_1)+5, 3(t_2-t_1)+1]\cdot [-2, 1, 3]=0$$ $$14(t_2-t_1)+20=0$$$$\implies t_2-t_1=-\frac{10}{7}$$ setting the value of $t_2-t_1$, we get $$\vec{A_1A_2}=\left[\frac{-22}{7}, \frac{25}{7}, \frac{-23}{7} \right]$$

hence, the shortest distance between the parallel lines $L_1$ & $L_2$ $$=\left|\vec{A_1A_2}\right|=\sqrt{\left(\frac{-22}{7}\right)^2+\left(\frac{25}{7}\right)^2+\left(\frac{-23}{7}\right)^2}$$ $$=\frac{\sqrt{1638}}{7}$$

Hint: Let $\vec p_1=[4,-2,-3]^T$ and $\vec p_2=[-2,3,-2]^T$

The plane orthogonal to the first line passing thorough $P_1$ has equation $\langle\vec d, (\vec x -\vec p_1)\rangle=0$

Intersect this plane with the line passing through $P_2$ tha has equation $\vec d t+\vec p_2=0$ and find the point $Q$ . The distance $P_1Q$ is the distance between the two lines ( and $P_1$ and $Q$ is the couple of point searched).

Here's my very custom take on it.

You need to find a segment going from $L_1$ to $L_2$ and perpendicular to both. Consider the plane $\Pi$ orthogonal to the vector $\overrightarrow{d}$ and, for the sake of easiness, going through $(0,0,0)$. $L_1$ and $L_2$ intersect this plane at points $Q_1$ and $Q_2$. The segment between $Q_1$ and $Q_2$ is the segment you're looking for. (It joins $L_1$ to $L_2$ and is perpendicular them since it belongs to an orthogonal plane). So just compute the distance between $Q_1$ and $Q_2$.

Notice that $Q_1$ and $Q_2$ are respectively the orthogonal projection of $P_1$ and $P_2$ on $\Pi$, as they are the orthogonal projection respectively of any point of $L_1$ or $L_2$.

To find the formula of $\Pi$, remember it is orthogonal to $d$ i.e. have a dot product null with it. Thus you have to the equation

$$\Pi \equiv (x,y,z)\cdot\overrightarrow d = -2\cdot x + y + 3\cdot z = 0$$

Setting $x=1,z=0$ and then $x=0,z=1$, you end up with the basis $(1,2,0),(0,-3,1)$.