• Support PF! Buy your school textbooks, materials and every day products Here!

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

  • Thread starter nuggets
  • Start date
  • #1
3
0
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
 

Answers and Replies

  • #2
HallsofIvy
Science Advisor
Homework Helper
41,794
925
Here's a big hint: if [itex]\theta[/itex], [itex]\phi[/itex], and [itex]\psi[/itex] are the angles a line or vector makes with the x, y, and z axes, respectively, then [itex]cos(\theta)\vec{i}+ cos(\phi)\vec{j}+ cos(\psi)\vec{k}[/itex] 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 [itex](1/2)\vec{i}+ (1/2)\vec{j}+ z\vec{k}[/itex] has length [itex]\sqrt{1/4+ 1/4+ z^2}= 1[/itex] so 1/2+ z2= 1, z2= 1/2, z= [itex]\sqrt{2}/2[/itex]. The unit vector in that direction is [itex](1/2)\vec{i}+ (1/2)\vec{j}+ (\sqrt{2}/2)\vec{k}[/itex] and the angle the vector makes with the z-axis is 45 degrees, not 60.

A unit vector in the same direction as [itex]\vec{i}+ \vec{j}+ \vec{k}[/itex] would be [itex](\sqrt{3}/3)\vec{i}+ (\sqrt{3}/3)\vec{i}+ (\sqrt{3}/3)\vec{k}[/itex] and so would make angle arccos(\sqrt{3}/3) about 54.7 degrees with each axis, not 60 degrees.
 
  • #3
3
0
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
rock.freak667
Homework Helper
6,230
31
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?
If I remember correctly, another formula is that

[tex]cos^2 \theta +cos^2 \phi + cos^2 \psi =1[/tex]
 
  • #5
HallsofIvy
Science Advisor
Homework Helper
41,794
925
Yes, that's why [itex]cos(\theta)\vec{i}+ cos(\phi)\vec{i}+ cos(\psi)\vec{k}[/itex] is a unit vector. And using that is exactly the same as "calculate the vector, solve for the component, and then take the arccosine".
 

Related Threads for: Calculate 3d vector from angles with respect to x and y axis

Replies
22
Views
36K
Replies
3
Views
1K
Replies
0
Views
1K
Replies
16
Views
2K
Replies
2
Views
11K
Replies
1
Views
888
Replies
2
Views
2K
Top