# Intersection of 2 vectors in 3D knowing the angle between the two

1. Sep 3, 2010

### calcioitalia0

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

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.

2. Relevant 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.

3. 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.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Sep 3, 2010

### HallsofIvy

Staff Emeritus
"Vectors" do not have a point of intersection since they are "movable". What you want is the intersection point of two lines, one of which contains the point (0, 0, 0)has direction vector <-.41, .28, -.08>. That first line can be written in parametric equations as x= -.41t, y= .28t, z= -.08t.

Now, you are told that the second line contains the point (-2.70, -.45, -.21)
and makes an angle of 38.3 degrees with the first line. Now the problem is that in three dimensions there exist an infinite number of lines that make an angle of 38.3 degrees with that first line- Take any cone at an angle of 38.3 degree to that line at any point on the line.

I think what you can do is this: let (x0, y0, z0) be a point on the first line. A vector from that line to (-2.70, -.45, -.21) is of the form <-2.7- x0, -.45- y0, -.21- z0>.

Since the dot product of two vectors, u and v, is given by $u\cdot v= |u||v|cos(\theta)$ we have
$$(-2.7- x0)(-2.7)+ (-.45- y0)(-.45)+ (-.21- z0)(-.21)$$$$= \sqrt{(-2.7- x0)^2+ (-.45- y0)^2+ (-.21- z0)^2}\sqrt{(2.7)^2+ (.45)^2+ (.21)^2}$$

Replace x0, y0, and z0 with x0= -.41t, y0= .28t, z0= -.08t and that becomes an equation in the single variable t. Solve for t and use that t to find x, y, and z.