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I just started general physics, so this will probably be pretty basic for most of you. The instructor covered two dimensional vectors in lecture, and then gave us a bunch of homework involving three dimensional vectors. The book skips it completely, so I would like to know if I'm on the right track.
Vector A = 2i - 4j + k
if I ignore the z component, I can find the angle with respect to x by making a triangle in the fourth quadrant.
arctan (-4/2) = -63.4, so my angle is about 296.6.
I'm not sure how to get my other angle though. Do I make a new plane between the z axis and the hypotenuse of that first triangle? I could make a new vector of sqrt 20 in the (i, -j) direction and 1 in the z direction. arctan (1/sqrt20) is about 12.6 degrees from that plane towards the z axis.
Am I way out in left field making this super complicated or am I at least batting in the right direction?
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United States24799 Posts
I'm not 100% sure what you're asking but I think you are correct.
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What exactly is the question that was asked?
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There doesn't seem to be a question regarding the vector... it just gives it to you...
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The easiest way to find the angle between the z axis and the vector is to use a dot product.
A · B = |A| * |B| * cos(a)
A · B/(|A| * |B|) = cos(a)
So define a vector along the z axis like 0i + 0j + k, take a dot product, divide by the magnitudes, and take an arccos.
Unless you haven't covered dot products yet...
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United States24799 Posts
On October 05 2009 00:20 Thunder_Sturm wrote:
The easiest way to find the angle between the z axis and the vector is to use a dot product.
A · B = |A| * |B| * cos(a)
A · B/(|A| * |B|) = cos(a)
So define a vector along the z axis like 0i + 0j + k, take a dot product, divide by the magnitudes, and take an arccos.
Unless you haven't covered dot products yet...
They probably haven't yet. BTW HIIIII!!!!!!!!!!!!!!!!!!!!
Whenever I saw you on East Thunder_Sturm I always thought you had the coolest nick.
edit: we played a few games together I believe
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Haha, I always thought my nick was kinda lame. And yeah we've played on east a few times. Same as SCC-Micronesia right? Hey 
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plot it, x = 2, y = -4, z = 1, draw a line to it from the origin, find angle in respect to whatever axes by using arctan or whatever inverse trig function you wanna use
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Sorry, I guess I didn't really make the question clear. I'm supposed to "find the direction of A"
I'm assuming that he expects us to find both angles. Would they be the angle of the vector in the XY plane and the angle of the vector in the XZ plane (both with respect to X), or would the second angle be calculated like I did?
Maybe he just wants us to recognize that the "direction" is 2i, -4j, and 1k.
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United States24799 Posts
Usually you can split a vector into a magnitude (length of the resultant in 3d) and direction (a unit vector which is just the original vector normalized by dividing each component by the magnitude of that component).
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If you find the parallel unitary vector, the constants next to the i, j, k are the cosines of the angles to each of the axis.
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United States24799 Posts
On October 05 2009 00:57 Cloud wrote:
If you find the parallel unitary vector, the constants next to the i, j, k are the cosines of the angles to each of the axis.
Oh I didn't realize that cloud wow ty I'll take a look at that.
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On October 04 2009 23:51 Neverborn wrote:
I just started general physics, so this will probably be pretty basic for most of you. The instructor covered two dimensional vectors in lecture, and then gave us a bunch of homework involving three dimensional vectors. The book skips it completely, so I would like to know if I'm on the right track.
Vector A = 2i - 4j + k
if I ignore the z component, I can find the angle with respect to x by making a triangle in the fourth quadrant.
arctan (-4/2) = -63.4, so my angle is about 296.6.
I'm not sure how to get my other angle though. Do I make a new plane between the z axis and the hypotenuse of that first triangle? I could make a new vector of sqrt 20 in the (i, -j) direction and 1 in the z direction. arctan (1/sqrt20) is about 12.6 degrees from that plane towards the z axis.
Am I way out in left field making this super complicated or am I at least batting in the right direction?
Um, the direction of the vector in rectangle form is simply defined as the unit vector in the same direction. (in your case divide by square root of 21.)
If you need the angular direction, then there are 2 angles you need.
#1 is the angle of the component that lies on the x-y plane, which you found via arctan.
#2 is the angle that's between the positive z-axis and the vector, in which case you need to take the complement of the angle which is in between the hypotenuse of the triangle on the x-y plane, and the vector itself.
it's going to be arcsin ( sqrt20/sqrt21)
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United States24799 Posts
Oh... ok now I see what you mean. Just reading what you originally said, it didn't click for me.
I actually haven't done any real math since I graduated from college, so I'm a bit rusty.
BTW if you hadn't said "kidding lol" in the spoiler that would easily have been troll of the week. Oh god I would have raged.
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Bill307
Canada9103 Posts
On October 05 2009 01:06 Kiarip wrote:
If you need the angular direction, then there are 2 angles you need.
#1 is the angle of the component that lies on the x-y plane, which you found via arctan.
#2 is the angle that's between the positive z-axis and the vector, in which case you need to take the complement of the angle which is in between the hypotenuse of the triangle on the x-y plane, and the vector itself.
it's going to be arcsin ( sqrt20/sqrt21)
It's worth noting that these angles are useful for "spherical coordinates".
The Wikipedia article on spherical coordiantes is terrible for learning purposes, so briefly, spherical coordinates are another way to write a point or vector in space.
Normal, Cartesian coordinates show where the point is along the x-, y-, and z-axes. Spherical coordinates tell you the direction from the origin to the point, given by those 2 angles, and the distance from the origin to the point.
E.g. in Cartesian coordinates, your vector "2i - 4j + k" is found by:
1. moving 2 units along the x-axis,
2. moving -4 units along the y-axis, and
3. moving 1 unit along the z-axis.
In spherical coordinates, your vector "2i - 4j + k" is found by:
1. looking along the x-axis from the origin,
2. rotating towards the y-axis by about 296.6 degrees (keep this direction in your head...),
3. rotating down from the z-axis towards this direction by arcsin (sqrt20/sqrt21), and using this final direction,
4. moving in this direction by the length of the vector, sqrt(21).
Following either of these steps will give you the vector "2i - 4j + k".
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EPT![[image loading]](http://mtaulty.com/blog/images/SIlverlight3Projections_9DE2/487px3D_coordinate_system_svg.png)
