# Where two vector lines intersect

I'm not quite sure why I cannot seem to get this question. It is regarding the point of intersection of two lines using dot and cross products as opposed to considering the simultaneous equations from equating components.

Show that a necessary condition for the lines

$\textbf{r} = \textbf{a} + s\textbf{m}$ , $\textbf{r} = \textbf{b} + t\textbf{n}$ to intersect is

$[(\textbf{a} − \textbf{b}), \textbf{m}, \textbf{n}] = 0$.

Find the values of s and t at the point of intersection in terms of triple products of $\textbf{a, b, m}$ and $\textbf{n}$ assuming that the intersection occurs and that $[\textbf{a, m, n}] \neq 0$.

Also, what are you first thoughts when you a see aproblem like this? I initially just jumped in and was trying to take various dot and cross products of the equations of the two ines, and I feel like I was running myself in circles. I then tried to think about what I would like to happen: I think I want a $[(\textbf{a} − \textbf{b}), \textbf{m}, \textbf{n}] = 0$ to appear multiplied by either s or n to eliminate one of the variables. The problem is I cannot seem to get this to happen without getting 0=0.

One of the approaches I have started off with a few times is:

At the point of intersection $\textbf{a}-\textbf{b}+s\textbf{m}-t\textbf{n}=\textbf{0}$

Take the cross product with either m or n

$(\textbf{a}-\textbf{b}) \times \textbf{n}+s(\textbf{m} \times \textbf{n})=\textbf{0}$

Now any other dot or cross product I seem to take with these equations just yields 0, or doesn't get me closer to the answer. Using the 'hint' I have tried

$((\textbf{a}-\textbf{b}) \times \textbf{n}) . \textbf{a} +s(\textbf{m} \times \textbf{n}) . \textbf{a}=(-\textbf{b} \times \textbf{n}). \textbf{a} +s(\textbf{m} \times \textbf{n}) . \textbf{a}=\textbf{0}$

But then I am stuck again.

I would very much appreciate some insight as to your thought processes when approaching questions like these!#

EDIT: I apologise I did not make it clear in my question above, I am fine with the derivation of the condition for intersection, $[(\textbf{a} − \textbf{b}), \textbf{m}, \textbf{n}] = 0$, and with motivating why it must be so. However I cannot figure out how to obtain expressions for t and s!

• The triple product gives the volume of the parallelepiped with edges given by the three vectors. What does this imply about the parallelpiped if its volume is zero? – amd Apr 6 '17 at 17:33
• Ah I apologise I should have made it clearer that it is the second part of the question I am stuck on: finding s and t – Meep Apr 6 '17 at 18:42

If 2 lines intersect, consider the plane $P$ containing both lines $L_1$ and $L2$. I guess you already know when we write that 'the vector equation of a line is :

$\vec r = \vec a + s \vec m$, we mean that the line passes through the point $\vec a$ and is prallel to $\vec m$. Now in this case , $\vec m, \vec n$ and $\vec a-\vec b$ are all on the same plane $P$. Hence, their triple product:

[$(\vec a - \vec b) ,\vec m,\vec n] = (\vec a - \vec b) \cdot (\vec m \times \vec n) = 0$, since obviously ($\vec m \times \vec n$) is perpendicular to the first vector(it is the normal of the plane containing the 2 lines).

• Thank you for your reply. This result I was able to get quite quickly- It is the values of s and t that are driving me mad (second part of question); i apologise i think that was not very clear in my question – Meep Apr 6 '17 at 18:44
• s and t are just independent parameters, whose values give the line, they can take any value. – Lelouch Apr 7 '17 at 2:25
• Hi Lelouch, I know that s and t can take on any value along each line, however they have a specific value at the point of intersection which is certainly dependent on the vectos equation of the two lines in question. I'll re-quote the last part of my question: "Find the values of s and t at the point of intersection in terms of triple products of $\textbf{a, b, m}$ and $\textbf{n}$ assuming that the intersection occurs and that $[\textbf{a, m, n}] \neq 0$" – Meep Apr 7 '17 at 14:16

Use the common normal $$\mathbf{k} = \mathbf{m} \times \mathbf{n}$$

The difference is position of the two lines projected along the common normal is proportional to the minimum distance of the two lines

$$d =\mathbf{k} \cdot ( \mathbf{r}_1 - \mathbf{r}_2 ) = \mathbf{k} \cdot ( \mathbf{a} - \mathbf{b} )$$

since $\mathbf{k} \cdot \mathbf{m} = \mathbf{k} \cdot \mathbf{n}=0$

For the lines to intersect you need $d=0$

$$\boxed{ \left( \mathbf{m} \times \mathbf{n} \right) \cdot \left( \mathbf{a} - \mathbf{b} \right)=0 }$$