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Whilst working on vectors I have come across a lot of problems like this. I am able to work it out for the shortest distance from a vector to a point, but not from a vector to a vector. Here is my usual method for a question asking the shortest distance from a vector (passing through $A$ and $B$) to a point $C$:
$$vec{OA}=a$$
$$vec{OB}=b$$
$$vec{OC}=c$$
Where $O$ is the origin. We know that the equation for a line passing through $A$ and $B$ is:
$$vec{r}=mu(b-a)+a$$
we also know that at the closest distance a line from $C$ to $vec{r}$ is perpendicular to $vec{r}$.
I would now define:
$$vec{r}=begin{pmatrix}mu(b_1-a_1)+a_1\mu(b_2-a_2)+a_2\mu(b_3-a_3)+a_3end{pmatrix}=begin{pmatrix}d_1\d_2\d_3end{pmatrix}$$
so the distance from $vec{r}$ to $C$ is:
$$l=sqrt{(d_1-c_1)^2+(d_2-c_2)^2+(d_3-c_3)^2}$$
now find the point at which $frac{dl}{dmu}=0$ solving for $mu$ and subbing into the equation.

However, I am aware that there are much easier methods for find the point $N$ and the shortest distance $|vec{NC}|$ involving the fact that $vec{r}bulletvec{NC}=0$ or potentially cross product as well. Does anyone have a tutorial for this method? Also, how would I solve this same problem but finding the minimum distance between two vectors?
Thanks

For problems like this one you don't need derivatives.

Suppose that you know the coordinates of points $A(x_A, y_A, z_A)$, $B(x_B, y_B, z_B)$ and components of vectors $vec a=(a_x,a_y,a_z)$, $vec b=(b_x,b_y,b_z)$. The shortest distance between lines is represented with segment $CD$ and that segment is prependicular both to $vec a$ and $vec b$.

Now you have:

$$AC=mu vec a$$

$$BD=lambda vec b$$

$$vec {CD} bot vec a implies vec{CD}cdot vec a=0$$

$$vec {CD} bot vec b implies vec{CD}cdot vec b=0$$

...or, in scalar form:

$$x_C-x_A=lambda a_x$$

$$y_C-y_A=lambda a_y$$

$$z_C-z_A=lambda a_z$$

$$x_D-x_B=mu b_x$$

$$y_D-y_B=mu b_y$$

$$z_D-z_B=mu b_z$$

$$(x_D-x_C)a_x+(y_D-y_C)a_y+(z_D-z_C)a_z=0$$

$$(x_D-x_C)b_x+(y_D-y_C)b_y+(z_D-z_C)b_z=0$$

You have 8 linear equations and 8 unknowns: $x_C, y_C, z_C, x_D, y_D, z_D, lambda, mu$:

• From the first three equations express $x_C, y_C, z_C$ in terms of $lambda$.


• From the next three equations express $x_D, y_D, z_D$ in terms of $mu$.


• Replace all that into the last two equations and you have a system of two equations with two unknowns $(lambda,mu)$.


• Solve, find coordinates of points $C,D$


• Calculate distance CD.

For the shortest distance between a pair of lines $L_1$ and $L_2$ in $mathbb R^3$, you can use symmetry and projections to develop a simple formula. You already know that the closest point on a line to a point $P$ not on the line lies on the perpendicular through $P$. So by symmetry, the nearest points to each other on a pair of lines must lie on a line $M$ that’s perpendicular to them both.

Working backwards for a moment, suppose that $Q_1$ and $Q_2$ are the nearest points to each other on the two lines. Since $L_1perp M$, the orthogonal projection of any point on $L_1$ onto $M$ is $Q_1$ and similarly, the orthogonal projection of any point on $L_2$ onto $M$ is $Q_2$. Hence, given any pair of points on the two lines, the orthogonal projection onto $M$ of the segment joining them is $overline{Q_1Q_2}$. This means that the shortest distance between $L_1$ and $L_2$ can be found by taking any pair of points on the respective lines and projecting them onto any line that’s perpendicular to $L_1$ and $L_2$.

Now, since $M$ is perpendicular to both $L_1$ and $L_2$, if you represent the two lines in parametric form as $P_1+svec v_1$ and $P_2+tvec v_2$, respectively, then the cross product $vec v = vec v_1timesvec v_2$ is a direction vector for $M$. For points on the lines, we can take $P_1$ and $P_2$, and so the distance between $L_1$ and $L_2$ is the length of the orthogonal projection of $P_1-P_2$ onto $vec v$: $${|(P_1-P_2)cdot(vec v_1timesvec v_2)| over |vec v_1timesvec v_2|}.$$

You can also develop a simple formula for 3-D point-line distance using geometric considerations. Again, let the line be given in parametric form as $P+tvec v$ and $Q$ be an arbitrary point. Consider the triangle formed by $P$, $Q$ and $R=Q+vec v$. Taking $overline{QR}$ as the base of the triangle, its area is $frac12bh = frac12 |vec v| h$, but the altitude $h$ is the perpendicular distance from $P$ to the line. The area of this triangle is $frac12|(P-Q)timesvec v|$ (it’s half the area of the paralellogram defined by these two vectors), from which we have $$h = {|(P-Q)timesvec v|over|vec v|}.$$