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Could someone please spot the error I've made in this question:

Q: A theme park has two zip wires. Sarah models the two zip wires as straight lines $(L_1, L_2)$ using coordinates in metres. The ends of one wire are located at $(0,0,0)$ and $(0,100,-20)$. The ends of the other wire are located at $(10,0,20)$ and $(-10,100, -5)$. Use Sarah's model to find the shortest distance between the zip wires:

$$L_1: \vec{r_1}(s) = s(100\hat{j} - 20\hat{k})$$

$$L_2: \vec{r_2}(t) = (10\hat{i}+ 20\hat{k}) + t(-20\hat{i} + 100\hat{j} - 25\hat{k})$$

Let $P$ be a general point on line $L_1$ and $Q$ be a general point on line $L_2$:

Then: $$P = (100s\hat{j} - 20s\hat{k})$$ $$Q = (10-20t)\hat{i} + (100t)\hat{j} + (20 - 25t)\hat{k}$$

Vector $PQ$ is

$$\vec{PQ} = Q - P = ((10-20t)\hat{i} + (100t)\hat{j} + (20 - 25t)\hat{k}) - (100s\hat{j} - 20s\hat{k})$$

$$ \vec{PQ} = (10-20t)\hat{i} + (100t - 100s)\hat{j} + (20 - 25t + 20s)\hat{k}$$

If $\vec{PQ}$ is perpendicular to both lines then: $$\vec{PQ} \cdot \vec{r_1}(s) = \Big((10-20t)\hat{i} + (100t - 100s)\hat{j} + (20 - 25t + 20s)\hat{k}\Big) \cdot (100\hat{j} - 20\hat{k}) = 0$$

$$10000t - 10000s - 400 + 500t - 40s = 0$$ $$10500t - 10040s = 400$$

$$\vec{PQ} \cdot \vec{r_2}(t) = \Big((10-20t)\hat{i} + (100t - 100s)\hat{j} + (20 - 25t + 20s)\hat{k}\Big) \cdot (-20\hat{i} + 100\hat{j} - 25\hat{k}) = 0$$

$$-200 + 400t + 10000t - 10000s - 500 + 625t - 500s = 0$$

$$11025t - 10500s = 700$$

Is this correct? Solving simultaneously gives $t = \dfrac{404}{63}$ and $s = \dfrac{20}{3}$. Plugging this back into vector $\vec{PQ}$ and doing the length of that gives an answer of $121.15$... when I'm fairly certain the answer is $\dfrac{50}{3}$.

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  • $\begingroup$ I agreed until 'If vector PQ is perpendicular to both lines then: ((10-20t)i + (100t - 100s)j + (20 - 25t + 20s)k) . (100j - 20k) = 0' Could you explain your thinking here? Are you familiar with the cross product, for it is useful in situations like these? Also, You might want to neaten up your expressions/equations for L1 and L2. I'd write $r_1(s) = \dots$ for L1, and $r_2(t) = \dots$ for L2. $\endgroup$
    – DanLewis3264
    Apr 18, 2019 at 18:47
  • $\begingroup$ My thinking if PQ is perpendicular to both lines then the minimum distance between both lines will be the distance PQ when PQ is perpendicular to both lines. And also yes the cross product is a good idea BUT i'm doing a Year 1 paper, and cross product isn't taught until Year 2 so i'm trying to see how they'd intend for the student to do it (i.e with dot product) $\endgroup$
    – user634745
    Apr 18, 2019 at 18:51
  • $\begingroup$ Just as a note, on your $\vec{PQ} \cdot \vec{r_1}(s)$ term, you have $20 \cdot -20 = -40$. This should be $-400$. $\endgroup$
    – JacobCheverie
    Apr 18, 2019 at 19:08

1 Answer 1

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I took the liberty of editing beginning of the post so that it was possible to follow your arithmetic. The first set of errors appears in the multiplication marked in red: $20\times 20 = 400$, not $40$. That might have been just a writing error, but in solving for $t$ you should have obtained $t= \frac{44}{63}$ not $\frac{404}{63}$.

I stopped checking after those two errors.

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