Vector problem - solving x y z using angles between vector and axis

[Ascendant78]

Homework Statement

Given that a vector has a magnitude of 6.0 units and makes an angle of 45° and 85° with the x and y axes, respectively, find the x and y components of this vector. Does the given information determine the z component? What can you say about the z component?

Homework Equations

equations included in #3 below

The Attempt at a Solution

What I did was this:
solve for x: 6cos45° = 4.24
solve for y: 6cos85° = 5.99

solve for z:
sqrt(x^2 + y^2 + z^2) = 6
17.98 + 35.88 + z^2 = 36
z^2 = -17.86
z = ±4.23

Although I got the right answer, I am not 100% sure why. I just tried this to see if it would work and it did. However, I am not sure exactly how the angle on each side is viewed? Is it the angle that the vector projects onto that side of the vector, or is it the angle projected onto the actual axis itself? I'm not sure if I'm even wording that in a way that makes sense, but hoping someone might know of a site or diagram I can look at that shows how the angle is evaluated for each side. I am just a bit confused as to how I get the right answer by multiplying the angle by 6, even though a magnitude of 6 will not necessarily project itself as 6 onto the x axis. I also know something is wrong because when I work backwards to solve for the magnitude of the angle, the value is wrong. I just want to make sure I know how to visualize this properly for related problems.

 

[jedishrfu]

The strategy is to understand the angle with the x-axis and y-axis means to take your vector and do an inner product with the x-unit-vector and with the y-unit-vector:

Vx = V . ux = |V| |ux| cos(45) = |V| cos(45) since |ux|=1

Vy = V . uy = |V| |uy| cos(85) = |V| cos(85) since |uy|=1

|V| = sqrt( Vx^2 + Vy^2 + Vz^2) and knowing Vx and Vy you can now solve for Vz.

Does that help?

By the way, the angle cosines are known as the direction cosines of V.

 

[Chestermiller]

Your calculated component in the y-direction is incorrect. The cosine of 85 degrees is very small, not close to 1. You might have suspected something was wrong when you found that the square of the z component was negative, giving an imaginary value for the z component.

Regarding your question, the answer is "the angle projected onto the axis itself."

Chet

 

[Ascendant78]

Apologies for the incorrect data I tossed up there earlier. I was in a rush and have no idea how I got that information. Anyway, I corrected my y value and used the method that you gave jedishrfu to solve it. However, I am still not exactly sure why what I did works? The book I am using to learn dot products doesn't give a very good explanation of them, so I am not so much concerned about how to use it as to why the calculations work the way they do? I am still a bit lost as to how I can multiply the magnitude (6) to the cos of the angle formed with the axis and get the value for that axis when the full 6 is most likely not projected onto that axis? Again, sorry if I am not wording this all well, as I am finding it difficult explaining exactly what I am trying to understand here.

 

[Chestermiller]

Apologies for the incorrect data I tossed up there earlier. I was in a rush and have no idea how I got that information. Anyway, I corrected my y value and used the method that you gave jedishrfu to solve it. However, I am still not exactly sure why what I did works? The book I am using to learn dot products doesn't give a very good explanation of them, so I am not so much concerned about how to use it as to why the calculations work the way they do? I am still a bit lost as to how I can multiply the magnitude (6) to the cos of the angle formed with the axis and get the value for that axis when the full 6 is most likely not projected onto that axis? Again, sorry if I am not wording this all well, as I am finding it difficult explaining exactly what I am trying to understand here.

The full 6 is projected onto each of the axes using the appropriate cosine of the angle that the vector makes with that axis.

Chet

 

[jedishrfu]

Dot products help you find the projection of one vector on another so A . B means the projection of A on B ie the component of A parallel to B but it can also mean the projection of B on A.

Its especially useful if B is a unit vector ie magnitude is 1 so that the A . Ux = |A| cos (ab-angle) which is the magnitude of the A component parallel to the x-axis. Visually you have a triangle in 3-space the hypotenuse is A and one side is Ux.