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I just can't see what's wrong, this should be relatively simple...

I have a vector which has length 1. Then, I have two angles. The first ($\gamma$) is the angle between the projection of the vector in the $xz$ plane and the $z$ axis. The second ($\beta$) is the angle between the projection of the vector in the $yz$ plane and the $z$ axis.

All I need now are the (cartesian) vector components $(x,y,z)$.

How do I calculate them?

I got: $x = \sin(\gamma)\\ y = \sin(\beta)\\ z = \cos(\gamma) \cdot \cos(\beta)$

But this can't be correct, because the length of this vector is not always 1. Can you help me and find the error, please?

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No, its no homework (not really). Context is: I get two angles from 4 solar panels (which means I compute those angles from the currents of the panels) –  Daniel Oertwig Nov 21 '12 at 21:12
    
Suggested a homework tag -- looks like a homework problem. –  gt6989b Nov 21 '12 at 21:12
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but nonetheless the homework tag won't hurt ^^ –  Daniel Oertwig Nov 21 '12 at 21:17
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I suppose if you are planning to install the panels on your roof, then it is literally a home-work problem. –  Gerry Myerson Nov 21 '12 at 23:16
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1 Answer

up vote 4 down vote accepted

I think you got $x=\sin\gamma$ by saying that you have a right triangle in the $xz$-plane with side $x$ opposite angle $\gamma$, but then you are also assuming the length of the hypotenuse is $1$. It isn't. The hypotenuse is a projection of a vector of length $1$, but (unless the $y$-coordinate turns out to be zero) it is itself a vector of length strictly less than $1$.

Perhaps a more fruitful approach begins with $\tan\gamma=x/z$.

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Yes! I think this would lead much more quickly to the solution than the dot product approach I used in my answer. +1. –  coffeemath Nov 22 '12 at 0:04
    
$x = z\tan\gamma$ and $y=z\tan\beta$ and $x^2+y^2+z^2=1$ so $z=\pm1/\sqrt{1+\tan^2\gamma+\tan^2\beta}$, huzzah! –  Rahul Nov 22 '12 at 0:10
    
This is indeed much quicker, but I liked coffeemath's answer, too. And there is one more good about it: actually, I do have the tangens values of both angles. –  Daniel Oertwig Nov 22 '12 at 8:41
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