# Calculating a theta using dot product in 3D coordinate

1. Dec 19, 2013

### Tah

I'm so confused about finding an angle, theta in this illustration.

With having three coordinate information, how can I calculate the theta using dot product?

I would easily find the angle by using trigonometric formula if I ignore the z-axis.

But I want to solve this problem with 3-dimensional coordinate system by using dot product formula.

2. Dec 19, 2013

### ShayanJ

There are two definitions for dot product of two vectors.we have $\vec{u}\cdot\vec{v}=u_xv_x+u_yv_y+u_zv_z=uv\cos{\theta}$.

3. Dec 19, 2013

### Tah

Thanks

Yes, I've already known that formula but could not understand exactly.

Could you help me more detail about such variables?

I was wondering about what is u_x, v_x, u_y, v_y.... in this case.

I think 'u' is (x1-x2, y1-y2, z1-z2) and 'v' is (x2-x3, y2-y3, z2-z3). Is this correct?

Thank you for your support :)

4. Dec 19, 2013

### ShayanJ

$u_x,v_y,...$ x,y and z components of the vectors u and v.They can be anything! $(x_1-x_2, y_1-y_2, z_1-z_2) and (x_2-x_3, y_2-y_3, z_2-z_3)$ are just particular examples when u and v are displacement vectors.You just have to find the components of the vectors and then their magnitudes.

5. Dec 19, 2013

### K^2

In this example, yes. And when Shyan writes $uv\ cos\theta$, here u and v are norms of the two vectors. Just in case it wasn't clear.

6. Dec 19, 2013

### Tah

Thanks

I have one more question

In this case, should I just do 180' minus the theta calculated by using dot product formula to find the theta prime(')?

Thanks!!

7. Dec 19, 2013

### K^2

Yes.

There exists a plane containing both vectors, and in that plane, all the planar geometry you know applies. So θ and θ' are complimentary angles, and therefore, their sum is 180°.