Intersection of 2 vectors in 3D, knowing the angle between them

[calcioitalia0]

** I accidentally posted this in the pre-calc math section first, but I think it's more applicable here...sorry**

Homework Statement

I need to find the intersection point of two vectors.

For vector A, I have it's start point (0,0,0) and it's magnitude in components (-.41, .28, -.08).

For vector B, I only know it's start point (-2.70, -.45, -.21)

I also know that the angle between the two vectors is 38.3 degrees.

I know this system has a unique solution, but I am having trouble arriving there mathematically.

Homework Equations

I am guessing I need to use some combination of AB=|A||B|cos(theta) and setting the equations of the lines equal... but haven't had any success yet.

The Attempt at a Solution

Like I said, I've tried writing equations for the two vectors and solving for them equal to each other, as well as a geometric approach.

 

[gabbagabbahey]

** I accidentally posted this in the pre-calc math section first, but I think it's more applicable here...sorry**

Homework Statement

I need to find the intersection point of two vectors.

For vector A, I have it's start point (0,0,0) and it's magnitude in components (-.41, .28, -.08).

For vector B, I only know it's start point (-2.70, -.45, -.21)

I also know that the angle between the two vectors is 38.3 degrees.

I know this system has a unique solution, but I am having trouble arriving there mathematically.

Homework Equations

I am guessing I need to use some combination of AB=|A||B|cos(theta) and setting the equations of the lines equal... but haven't had any success yet.

The Attempt at a Solution

Like I said, I've tried writing equations for the two vectors and solving for them equal to each other, as well as a geometric approach.

Hint: If a point [itex]\textbf{r}[/itex] lies along the first vector, then

[tex]\textbf{r}=\begin{pmatrix} 0 \\ 0 \\ 0\end{pmatrix} + \begin{pmatrix} -0.41 \\ 0.28 \\ -0.08\end{pmatrix}t[/tex]

For some value of [itex]t[/itex] between zero and one. (Do you see why?)...Suppose you connect the starting point of your 2nd vector to some point [itex]\textbf{r}(t)[/itex] on your first vector, what would be the angle between your 1st vector and this new vector (as a function of [itex]t[/itex] )? What value of [itex]t[/itex] makes that angle 38.3 degrees?

 

[calcioitalia0]

Ok, I get that the point of intersection is going to be some multiple of |A|. And you're saying I can basically call my second vector B [(-.41, .28, -.08)t - (-2.70, -.45, -.21)] since that is the difference of it's two endpoints, which is what a vector is. If I do the dot product of AB then and set it equal to |A||B|cos(38.3) then I should get my t value, which in turn will give me the coordinates of intersection?

I'll go try it out.

Thanks!

 

[calcioitalia0]

If we call the point of intersection (a,b,c) and the start point of B (x,y,z), will |B| be

sqrt(((a-x)^2,(b-y)^2,(c-z)^2))?

With the t's in the values of a,b,c that is going to get messy. Can I call |B| = 1 since I'm really just worried about where a vector going in the DIRECTION of B would intersect with A, so is the magnitude necessary?

If I end up doing it the first way I describe I will end up with a quadratic which I don't think make sense physically... but maybe I just have to throw one of the solutions out?

 

[gabbagabbahey]

If we call the point of intersection (a,b,c) and the start point of B (x,y,z), will |B| be

sqrt(((a-x)^2,(b-y)^2,(c-z)^2))?

With the t's in the values of a,b,c that is going to get messy. Can I call |B| = 1 since I'm really just worried about where a vector going in the DIRECTION of B would intersect with A, so is the magnitude necessary?

There's no guarantee that the point of intersection will be the endpoint of your second vector, but the vector from (x,y,z) to (a,b,c) will be part of your 2nd vector and will thus make the same angle with your 1st vector. The length of that vector will be [itex]\sqrt{(a-x)^2+(b-y)^2+(c-z)^2}[/itex]. Let's call that vector "C", clearly its length will be fixed, so you can't just set it equal to one.

If I end up doing it the first way I describe I will end up with a quadratic which I don't think make sense physically... but maybe I just have to throw one of the solutions out?

Not all quadratics have two unique solutions ([itex]x^2 = 0[/itex] for example), but even if this one does, I'm sure you'll be able to tell which one is correct.