I tried using a point on the first line called P and entered $s=0$ and $s=2$ to get q and r. I then took the cross product of qr and qp then used the distance formula but this didn"t seem to work.

You are watching: Find the distance between the skew lines with parametric equations


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The squared distance between a point on each line is given by

$$d_{ts}^2=(1+t-2s)^2+(-5+6t-14s)^2+(1+2t-5s)^2.$$

Now you need to find the shortest distance and it suffices to cancel the partial derivaties on $t$ and $s$:

$$\begin{cases}2(1+t-2s)+12(-5+6t-14s)+4(1+2t-5s)=0\\-4(1+t-2s)-28(-5+6t-14s)-10(1+2t-5s)=0.\\\end{cases}$$

By solving the linear system, you get $t$ and $s$, then the points and the distance.


You can write a point on the first line as $P = (2,1,0) + t(1,6,2)$ and a point on the second line as $Q = (1,6,-1) + s(2,14,5)$. If you take the difference $P - Q$ you get $(1,-5,1) + t(1,6,2) + s(-2,-14,-5)$.

So your problem is equivalent to finding the distance between the plane given by $(1,-5,1) + t(1,6,2) + s(-2,-14,-5)$ and the origin. The first step here is to find a normal vector $\katifund.orgbf{n}$ to the plane, you can use $\katifund.orgbf{n} = (1,6,2) \times (-2,-14,-5) = (-2,1,-2)$. Now take any vector $\katifund.orgbf{v}$ in the plane (for example, $\katifund.orgbf{v} = (1,-5,1))$ you can compute the distance as:

$$D = \frac{|\katifund.orgbf{v} \cdot \katifund.orgbf{n}|}{||\katifund.orgbf{n}||} = \frac{|-2 - 5 - 2|}{\sqrt{4 + 1 + 4}} = \frac{9}{3} = 3$$

In general, if your lines are given by $\katifund.orgbf{p} + t\katifund.orgbf{v}$ and $\katifund.orgbf{q} + s\katifund.orgbf{w}$, you can compute:

$$D = \frac{|(\katifund.orgbf{p} - \katifund.orgbf{q}) \cdot \katifund.orgbf{n}|}{||\katifund.orgbf{n}||}; \qquad \qquad \katifund.orgbf{n} = \katifund.orgbf{v} \times \katifund.orgbf{w}$$


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answered Oct 17 "16 at 20:28
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Alex ZornAlex Zorn
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