# Angle Between Two Lines?

1. Mar 10, 2013

### AXidenT

1. The problem statement, all variables and given/known data
Find the angle between
a) The line L1 given by the equations y = 2z, x = 0 and
b) The line L2 given by the equations x = 3z, y = 0.

2. Relevant equations
v.u=|v|*|u|*cos(θ)

3. The attempt at a solution

I know that I need to basically have a vector for each line to substitute into the equation above, but I am confused as to which set of the following vectors is correct:

1) L1 can be represented as 0x+1y-2z=0 so the vector/norm of it would be: (0,1,-2) (the tutor in class used norms when someone asked her about the question but I had trouble following what she was doing. :S) L2 would similarly be (1,0,-3). This produces an angle of 31.95°

2) Let z=1 in both cases, such that: L1 = (0,2,1) and L2 = (3,0,1). This when entered into the formula generates an angle of 81.87°.

Thank you!

2. Mar 10, 2013

### HallsofIvy

"Norm" is the wrong word here. For vectors, that refers to the length of the vector which is not relevant.

No, the line, L1, cannot be represented by "0x+1y-2z=0"- that's the equation of a plane. It is two dimensional because you could choose values for, say, x and y, and solve for the value of z. The vector <0, 1, -2> is perpendicular to that plane.

As originally given, y= 2z, x= 0 (equivalent to x= 0, y= 2t, z= t) is one-dimensional, a line, because, given a single value of z, we can solve for both x and y. We can write it as a vector equation $\vec{r}(t)= 0\vec{i}+ 0\vec{j}+ 0\vec{k}+ t(0\vec{i}+ 2\vec{j}+ \vec{k})$ showing a little more clearly that the line passes through the point (0, 0, 0) in the direction of the vector $0\vec{i}+ 2\vec{j}+ \vec{k}$ so that vector, or, more simply, <0, 2, 1>, not <0, 2, -1>, points in the direction of the line.

Similarly, we can write "x= 3z, y= 0" as x= 3t, y= 0, z= t or, in vector form, $\vec{r}(t)= 0\vec{i}+ 0\vec{j}+ 0\vec{k}+ t(3\vec{i}+ 0\vec{j}+ \vec{k})$. Again, that is a line through (0, 0, 0) but now in the direction of $3\vec{i}+ 0\vec{j}+ \vec{k}$ or just <3, 0, 1>.

3. Mar 10, 2013

### AXidenT

Thanks for the reply, that actually helped quite a bit!

Just to confirm - there's no way the lines can be represented as a plane, because that would negate the point of the lines intersecting (if that makes sense).

Also I tried graphing the equations in MatLab to see which angle seemed more correct and I got something that looked similar to this. After what you said, it seems the method which netted me 81° seems more correct, it appeared in the graphs i made that the 31° answer seems more correct. Have I missed something else, or do I need to get better at graphing 3D lines?

Thank you!!

EDIT: This graph shows another view (ignore the axis names, I just switched y and z around so I could see a different view) - so this angle and the other angle should somehow make up 81°? I feel like that's the more correct method... but I'm not entirely convinced since it's matching the evidence I'm seeing in these graphs...

Last edited: Mar 10, 2013