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Please help me solve this. I have been thinking of all sorts of ways to solve this but can't figure out how :(. Ok here's the problem: I am given a three dimensional velocity vector (i know the magnitude of this vector and I know what angle this vector makes wrt one of the axis, say the x-axis.) What I want to determine is what angle does this three-dimensional vector make with the x-y, y-z and x-z plane. Another way to look at this is if we project this three-dimensional vector in the x-y plane what is the angle between this vector and the x- axis (or the y- axis)? I do not know the velocity components of this vector in the x-, y- or z- axis. As a matter of fact, these x-, y- and z- velocity components are what I aim to calculate from determining the angle the 3-D vector makes with each plane.

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Maybe you want direction cosines: en.wikipedia.org/wiki/Direction_cosine –  Ross Millikan Jul 28 '11 at 23:19
    
Wow this post has been a big help. If I understand direction cosines right, it is similar to imagining a cone whose dimensions are given by 3-D vector and angle wrt x-axis say (lets call this vector R and angle phi). Now, if we imagine the cone completing a revolution around the x-y plane, and we ask ourselves at what angle wrt x-axis does the cone cut through the x-y plane -- it is the same angle as phi. –  Rhea Jul 29 '11 at 7:10
    
..also, the length of the cone does not change as it cuts the x-y plane-- so this reduces the problem to a 2D problem in the x-y plane with both magnitude and angle wrt to x-axis specified. Then using simple trig (R cos phi) we can find x-component of this 3-D vector R. –  Rhea Jul 29 '11 at 7:10
    
Am I on the right track on this or not? –  Rhea Jul 29 '11 at 23:02
    
you are right that given a vector length and angle with the x axis, you have a 2D problem. There will be two vectors that satisfy that: $(v \cos \theta, v \sin \theta)$ and $(v \cos \theta, -v \sin \theta)$ –  Ross Millikan Jul 30 '11 at 17:06

1 Answer 1

You need three scalar quantities to specify a vector in a three-dimensional space. If you know only the magnitude of the vector and the angle it forms with one axis, those two scalar quantities are not enough to fully determine the vector -- it could be any vector with the given magnitude on a cone around that axis. You need one more quantity to fix the vector; then you can calculate the angles and components you're looking for.

P.S.: Note that the title and body of the question are inconsistent. Specifying a direction in three-dimensional space requires two scalar quantities; since you only have one angle, you don't know the direction of the vector.

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I am sorry about that. I can see how the vector is not fixed in space and can be anywhere on a cone with the given angle wrt to x-axis. I will try to figure out if there is more information to the vector that I can find in order to fix its location in 3-D space. –  Rhea Jul 28 '11 at 22:51

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