# Collision problem using vectors

1. Dec 28, 2011

### Lucy Yeats

1. The problem statement, all variables and given/known data

Two particles are moving on trajectories given by r=a+ut and r=b+vt respectively where a, b, u and v are constant vectors and t is the time. Show that the particles will collide if v.(bxu)=v.(axu) Obtain an expression for the time of the collision in terms of a, b, u and v. (I think I've done the parts so far.) Hence, or otherwise, show that the collision will take place at position r=b+((a.(bxu))/(v.(bxu))v

2. Relevant equations

If three vectors a, b, c are linearly dependent, a.(bxc)=o

3. The attempt at a solution

a+ut=b+vt
ut=(b-a)+vt
ut, vt, and b-a are coplanar. Since t is a scalar multiple, u, v, and b-a are coplanar.
v.((b-a)xu)=0
v.(bxu-axu)=o
v.(bxu)=v.(axu)

ut-vt=b-a
(u-v)t=b-a
t=mod(b-a)/mod(u-v)

I can't see how this expression for to is equivalent to ((a.(bxu))/(v.(bxu)), so I'm stuck.

2. Dec 28, 2011

### Lucy Yeats

Any suggestions would be great. :)

3. Dec 28, 2011

### Dick

Ok, here's a sample of how to do it. I haven't worked through the permutations to show that the answer is what same as what the book gives but here's a start.

(u-v)t=(b-a)

cross both sides with a, so,

(uxa-vxa)t=(bxa)

Now we want to turn both sides into scalars so we can divide, let's dot with v.

((uxa).v-(vxa).v))t=(bxa).v

(vxa).v=0 so,

(uxa).v)t=(bxa).v, hence,

t=((bxa).v)/((uxa).v). Now use cyclic permutations and the result from the first part to turn that into the books answer, if you can. You might. I'm not sure. I haven't chased it that far yet. Nor am I sure the book answer is even right. I haven't checked that either.

4. Dec 28, 2011

### ehild

Start to cross both sides with u instead:
-(uxv)t=(uxb)-(uxa),

then dot product with a :

-a(uxv)t=a(uxb)

as (uxa) is perpendicular to a. The triple products are scalars, so

t=-a(uxb)/a(uxv)=a(bxu)/a(uxv).

You need to change the denominator to v(bxu) by cyclical permutation of the vectors and using the condition v.(bxu)=v.(axu)

ehild

5. Dec 28, 2011

### Dick

Yeah, I figured there was some permutation of those things that would work. But I was kind of hoping Lucy Yeats would figure it out instead.

6. Dec 29, 2011

### SammyS

Staff Emeritus
Instead of dotting with v, dot with u.

I.E. look at:
$((\vec{u}\times\vec{a})\cdot\vec{u}-(\vec{x}\times\vec{a})\cdot\vec{u})t=(\vec{b} \times\vec{a})\cdot\vec{u}$​
Of course, $(\vec{u}\times\vec{a})\cdot\vec{u}=0$

I must have been asleep (as in not paying attention) when ehild made this same suggestion!

Last edited: Dec 29, 2011
7. Dec 29, 2011

### ehild

Sorry Dick. But you directed the OP to a more difficult way than necessary, and I did not say more than you.

ehild

Last edited: Dec 29, 2011
8. Dec 29, 2011

### Dick

Just trying to explain why I was being too lazy to work out the exact solution.

9. Dec 29, 2011

### SammyS

Staff Emeritus
BTW:

The condition v.(bxu)=v.(axu), can be obtained in a similar manner. Start with
$\vec{a}+\vec{u}t=\vec{b}+\vec{v}t$​
Cross that with vector $\vec{u}$ then dot that with vector $\vec{v}\,.$