# direction cosines/vector help

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• Apr 6th 2008, 07:40 AM
JennK
direction cosines/vector help
Hi everyone! I'm working on a vector question involving finding and proving direction cosines/sines/tangents of vector (a,b,c), u = magnitude

I know:
cos[alpha] = a/u, cos[beta] = b/u, cos[gamma] = c/u, and that cos2[alpha] + cos2[beta] + cos2[gamma] = 1 (the 2s are squares). But I don't understand the second part. I know it must be pretty simple and related to unit vectors, and I'm just missing something, so any help would be very much appreciated! Also, any help on the sine/tangent part would be useful, too.

Thanks so much for any and all help!
• Apr 6th 2008, 08:11 AM
earboth
Quote:

Originally Posted by JennK
Hi everyone! I'm working on a vector question involving finding and proving direction cosines/sines/tangents of vector (a,b,c), u = magnitude

I know:
cos[alpha] = a/u, cos[beta] = b/u, cos[gamma] = c/u, and that cos2[alpha] + cos2[beta] + cos2[gamma] = 1 (the 2s are squares). But I don't understand the second part. ...

Probably you mean by second part how to calculate the magnitude of the vector(?). If so:

$\displaystyle u = \sqrt{a^2+b^2+c^2}$
• Apr 6th 2008, 10:01 AM
JennK
Hi! Yes, I mean the magnitude. And I understand the formula you posted...but if u = square root of a2 + b2 + c2, why is the formula cos2[alpha] + cos2[beta] + cos2[gamma] = 1, and not the square root of (cos2[alpha] + cos2[beta] + cos2[gamma]) = 1?

Please help, and thank you!
• Apr 6th 2008, 09:09 PM
earboth
Quote:

Originally Posted by JennK
Hi! Yes, I mean the magnitude. And I understand the formula you posted...but if u = square root of a2 + b2 + c2, why is the formula cos2[alpha] + cos2[beta] + cos2[gamma] = 1, and not the square root of (cos2[alpha] + cos2[beta] + cos2[gamma]) = 1?

Please help, and thank you!

To answer this question first: $\displaystyle 1 = \sqrt{1}$

Quote:

I know:
cos[alpha] = a/u, cos[beta] = b/u, cos[gamma] = c/u, and that cos2[alpha] + cos2[beta] + cos2[gamma] = 1 (the 2s are squares).
If $\displaystyle \cos(\alpha)=\frac au ~\implies~ a =u \cdot \cos(\alpha)$ and

$\displaystyle b =u \cdot \cos(\beta)$

$\displaystyle c =u \cdot \cos(\gamma)$

Now consider the term:

$\displaystyle a^2+b^2+c^2 = (u \cdot \cos(\beta))^2 + (b =u \cdot \cos(\alpha))^2 + (u \cdot \cos(\gamma))^2 =$ $\displaystyle u^2 \cdot (\cos(\alpha))^2 + u^2 \cdot (\cos(\beta))^2 + u^2 \cdot (\cos(\gamma))^2 =$ $\displaystyle u^2((\cos(\alpha))^2 + (\cos(\beta))^2 + (\cos(\gamma))^2) = u^2 \cdot 1$

and therefore: $\displaystyle u = \sqrt{a^2+b^2+c^2}$