Questions on two lines

Question

Let $L_1$ be the line having parametric equations: $x= 3-s$, $y=-1+2s$, $z=1+s$ and let $L_2$ be the line given by $x=1+t$, $y=2+t$, $z=2t$.

a) Do the lines intersect? If so, find the point of intersection.

b) Find the point $P$ on $L_1$ that is closest to $L_2$ and find the point $Q$ on $L_2$ that is closest to $L_1$. Hint: Use the fact that the vector $PQ$ will be orthogonal to the direction vectors of both lines.

I set the equations equal to each other and tried to solve for the parameters, but it came out that they do not intersect (or so I think).

How would I go about finding the points on both lines?

Answer 1

For point a), you need to match all the coordinates, hence you need to solve the following system:

$$\begin{cases} 3-s = 1+t\\ -1+2s = 2+t\\ 1+s = 2t \end{cases} \Rightarrow \begin{cases} t = 2-s \\ t = 2s - 3\\ 1+s = 2t \end{cases} \Rightarrow \begin{cases} s = \frac{5}{3} \\ t = \frac{1}{3}\\ 1+s = 2t \end{cases}.$$

Anyway, the last equality is incompatible with $s = \frac{5}{3}$ and $t = \frac{1}{3}$. Indeed, you get a bad $$\frac{8}{3}=\frac{2}{3}.$$

So, there is no intersection between the two lines.

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For point b), you need to minimize the distance $d$ between the two lines. Let's consider the square of distance $D = d^2$:

$$D = ((3-s)-(1+t))^2 + ((-1+2s)-(2+t))^2 + ((1+s)-(2t))^2 = \\ = 6s^2 + 6t^2-6st -14s-2t + 14. $$

The gradient of $d^2$ is:

$$\nabla D = \left[\begin{array}{c} 12s-6t-14\\ 12t-6s-2 \end{array}\right].$$

Posing $\nabla D = 0$, you get $s =\frac{5}{3}$ and $t=1$.

Using $s = \frac{5}{3}$, we find that the closest point to $L_2$ is:

$$P = \left[\begin{array}{c} \frac{4}{3}\\ \frac{7}{3}\\ \frac{8}{3} \end{array}\right].$$

Finally, using $t = 1$, we find that the closest point to $L_1$ is:

$$Q = \left[\begin{array}{c} 2\\ 3\\ 2 \end{array}\right].$$