# Calculate 3d vector from angles with respect to x and y axis

1. Sep 22, 2008

### nuggets

1. I am considering the problem of getting the equation for a 3d vector(magnitude irrelevant) in cartesian coordinates when given 2 angles with respect to arbitrary axes.

2. I think that 180-angle1-angle2=angle3

3. So if the angle between the x axis and the vector is 60deg, and the angle between the y axis and the vector is 60 deg, then the angle with the z axis will be 60d. So a possible solution vector would be <1,1,1>

Am I correct?

Thanks

2. Sep 23, 2008

### HallsofIvy

Staff Emeritus
Here's a big hint: if $\theta$, $\phi$, and $\psi$ are the angles a line or vector makes with the x, y, and z axes, respectively, then $cos(\theta)\vec{i}+ cos(\phi)\vec{j}+ cos(\psi)\vec{k}$ is a unit length vector in that direction. If you know two of the angles, take the cosines of those as two components and find the third component by the condition that the vector has length 1.

In particular, You are mistaken in thinking that if the angles with the x and y axes are 60 degrees, then so is the angle with the z axis. If the angles with the x and y-axes are 60 degrees, then the x and y-components will be cos(60)= 1/2. The vector $(1/2)\vec{i}+ (1/2)\vec{j}+ z\vec{k}$ has length $\sqrt{1/4+ 1/4+ z^2}= 1$ so 1/2+ z2= 1, z2= 1/2, z= $\sqrt{2}/2$. The unit vector in that direction is $(1/2)\vec{i}+ (1/2)\vec{j}+ (\sqrt{2}/2)\vec{k}$ and the angle the vector makes with the z-axis is 45 degrees, not 60.

A unit vector in the same direction as $\vec{i}+ \vec{j}+ \vec{k}$ would be $(\sqrt{3}/3)\vec{i}+ (\sqrt{3}/3)\vec{i}+ (\sqrt{3}/3)\vec{k}$ and so would make angle arccos(\sqrt{3}/3) about 54.7 degrees with each axis, not 60 degrees.

3. Sep 23, 2008

### nuggets

Hey thank you. I got it now.

Is there an easy way to get the third angle or do I have to calculate the vector, solve for the component, and then take the arccosine?

4. Sep 26, 2008

### rock.freak667

If I remember correctly, another formula is that

$$cos^2 \theta +cos^2 \phi + cos^2 \psi =1$$

5. Sep 27, 2008

### HallsofIvy

Staff Emeritus
Yes, that's why $cos(\theta)\vec{i}+ cos(\phi)\vec{i}+ cos(\psi)\vec{k}$ is a unit vector. And using that is exactly the same as "calculate the vector, solve for the component, and then take the arccosine".