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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.

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  • $\begingroup$ Try drawing any line perpendicular to these two and calculating the distance between the two intersections you get. $\endgroup$ Oct 30 '15 at 20:22
  • $\begingroup$ 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. $\endgroup$
    – user137731
    Oct 30 '15 at 20:34
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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 $L_1$ & $L_2$ $$=\left|\vec{A_1A_2}\right|=\sqrt{\left(\frac{-22}{7}\right)^2+\left(\frac{25}{7}\right)^2+\left(\frac{-23}{7}\right)^2}$$ $$=\frac{\sqrt{1638}}{7}$$

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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).

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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 $(1,2,0),(0,-3,1)$.

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This is going to be a method intermediate to those of a number of the other posters, since there are several equivalent ways to describe the vector calculations involved. (In fact, the first way I worked out the distance was essentially that presented by Harish Chandra Rajpoot.)

The parametric expressions for the two lines are $ \ L_1 \ = \ \langle \ 4-2t \ , \ -2 + t \ , \ -3 + 3t \ \rangle \ \ $ and $ \ L_2 \ = \ \langle \ -2-2s \ , \ 3 + s \ , \ -2 + 3s \ \rangle \ \ . $ We would like to find a vector $ \ \overrightarrow{p} \ $ which is perpendicular to the line direction vector $ \ \overrightarrow{d} \ $ and connects points on the two lines. We are free to choose any third point on either of the lines to define the plane containing the two lines: we may, for instance, take the point on $ \ L_2 \ $ at which $ \ x = 0 \ \ . $ This is given by $ \ s = -1 \ \ , $ giving us $ \ P_3 \ (0 , 2 , -5) \ \ . $ But since the two lines are parallel, the vector from the given point on one line to the second point on that line will just be a scalar multiple of $ \ \overrightarrow{d} \ \ , $ as we will see below.

We are also free to choose any one of our three points as the vertex for two vectors from which we will form the cross-product. So we can take $$ \ \overrightarrow{P_2P_1} \ = \ \langle \ 4 - (-2) \ , \ (-2) - 3 \ , \ (-3) - (-2) \ \rangle \ = \ \langle \ 6 \ , \ -5 \ , \ -1 \ \rangle $$ and $$ \ \overrightarrow{P_2P_3} \ = \ \langle \ 0 - (-2) \ , \ 2 - 3 \ , \ (-5) - (-2) \ \rangle \ = \ \langle \ 2 \ , \ -1 \ , \ -3 \ \rangle \ = \ -\overrightarrow{d} \ . $$ So in our cross-product calculation, we will simply use $ \ \overrightarrow{d} \ $ as the second vector. A normal vector to the plane is then $ \ \overrightarrow{n} \ = \ \overrightarrow{P_2P_1} \ \times \ \overrightarrow{d} \ = \ \langle \ -14 \ , \ -16 \ , \ -4 \ \rangle \ \ . $ For simplifying a calculation by hand, we may extract the common factor of $ \ -2 \ $ from these components, as long as we continue using that version of the normal vector hereafter. So we will take $ \ \overrightarrow{n} \ = \ \langle \ 7 \ , \ 8 \ , \ 2 \ \rangle \ \ . $

The particular orthogonal vector to $ \ \overrightarrow{d} \ $ that we want is then given by the cross-product of $ \ \overrightarrow{n} \ $ with $ \ \overrightarrow{d} \ \ , $ $$ \ \overrightarrow{p} \ = \ \overrightarrow{n} \ \times \ \overrightarrow{d} \ = \ \langle \ 22 \ , \ -25 \ , \ 23 \ \rangle \ \ . $$

[And if we look at Harish Chandra Rajpoot's solution, we can see what's coming... We can check that we've done this calculation correctly so far by observing that $$ \ \overrightarrow{p} \ \cdot \ \overrightarrow{d} \ = \ 22 · 2 \ + \ (-25) · 1 \ + \ 23 · 3  \ \ = \ \ -44 - 25 + 69 \ \ = \ \ 0 \ \ . ] $$

The perpendicular distance between the two parallel lines is now given by the scalar projection of $ \ \overrightarrow{P_2P_1} \ \ $ onto $ \ \overrightarrow{p} \ \ , $

$$ D \ \ = \ \ \left\vert \ \frac{\overrightarrow{P_2P_1} \ \cdot \ \overrightarrow{p}}{\Vert \ \overrightarrow{p} \ \Vert} \ \right\vert \ \ = \ \ \left\vert \ \frac{6 · 22 \ + \ (-5) · (-25) \ + \ (-1) · 23 }{\sqrt{22^2 \ + \ (-25)^2 \ + \ 23^2}} \ \right\vert \ \ = \ \ \left\vert \ \frac{132 \ + \ 125 \ - \ 23 }{\sqrt{1638}} \ \right\vert $$ $$ = \ \ \frac{234}{\sqrt{1638}} \ \ = \ \ \frac{2·3^2·13}{\sqrt{2·3^2·7·13}} \ \ = \ \ \frac{ 3 · \sqrt{2·7·13}}{ 7 } \ \ = \ \ \frac{ 3 · \sqrt{182}}{ 7 } \ \ \approx \ \ 5.78 \ \ . $$

As for the second part of the problem, there is no unique solution. We can take any point on $ \ L_1 \ $ to be $ \ Q_1 \ $ and add to it a vector of length $ \ D \ $ in the direction of $ \ -\overrightarrow{p} \ \ . $ (The sign of $ \ \overrightarrow{p} \ $ is ambiguous from the cross-product calculation; in order to "run" from $ \ L_1 \ $ to $ \ L_2 \ \ , $ we reverse its direction here.) So, for instance, the closest point on $ \ L_2 \ $ to point $ \ P_1 \ $ is

$$ (4 , -2 , -3) \ - \ D \ \hat{p} \ \ = \ \ (4 , -2 , -3) \ + \ \left( \frac{234}{\sqrt{1638}} \right) \ \frac{\langle \ -22 \ , \ 25 \ , \ -23 \ \rangle}{\sqrt{1638}} $$ $$ = \ \ (4 , -2 , -3) \ + \ \frac{\langle \ -22 \ , \ 25 \ , \ -23 \ \rangle}{7} \ \ = \ \ \left(\frac{6}{7} \ , \ \frac{11}{7} \ , \ -\frac{44}{7} \right) \ \ . $$

[We can verify that this point corresponds to $ \ s = -\frac{10}{7} \ $ on $ \ L_2 \ $ and that the distance between $ \ P_1 \ $ and this point is in fact $ \ D \ \ . $ ]

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Another way that is more analytically-based and uses vectors less is to set up a "distance-squared" function for, say, points on $ \ L_2 \ $ measured from point $ \ P_1 \ (t = 0) \ $ on $ \ L_1 \ \ : $

$$ D^2 \ = \ (4 - [-2 - 2s])^2 \ + \ ([-2] - [3 + s])^2 \ + \ ([-3] - [-2 + 3s])^2 \ = \ \Delta(s) \ \ . $$

The extremal value (minimum) is found from

$$ \frac{d \Delta}{ds} \ = \ 2 · 2 · (6 + 2s) \ + \ 2 · (-1) · ( -5 - s ) \ + \ 2 · (-3) · ( -1 - 3s ) \ = \ 0 $$

[note that this is twice the dot product of the direction vector $ \ \overrightarrow{d} \ $ and the vector from $ \ P_1 \ $ to the closest point to it on $ \ L_2 \ $ : we thereby show, rather than assume, that the closest distance between points on the two parallel lines is the perpendicular distance]

$$ \Rightarrow \ \ (12 + 4s) \ - \ (-5 - s) \ - \ (-3 - 9s) \ \ = \ \ 20 + 14s \ \ = \ \ 0 \ \ \Rightarrow \ \ s \ = \ -\frac{10}{7} \ \ , $$

with the rest of the calculation proceeding as already presented.

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