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Show that two nonzero vectors $\vec{v_1}$,$\vec{v_2}$ ∈ $\mathbb{R_3}$ are orthogonal if and only if their direction angles satisfy $$\cos α_1 \cos α_2 + \cos β_1 \cosβ_2 + \cosγ_1 \cosγ_2 =0$$

Note: I tried to turn all of the $\cos[\textrm{angle}]_2$ to sin[angle] and then convert all of the cos[angle] terms to $v_x$/||$\vec{v}$|| and sin[angle] terms to $v_y$/||$\vec{v}$||, but that was where I got stuck.

Any ideas?

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  • $\begingroup$ This looks like a dot product. $\endgroup$ – Jean-François Gagnon Nov 14 '15 at 6:25
  • $\begingroup$ How? dot product is in the form magnitudemagnitudecos([angle]) $\endgroup$ – Rloc Nov 14 '15 at 6:27
  • $\begingroup$ The component wise definition. $\endgroup$ – Jean-François Gagnon Nov 14 '15 at 6:28
  • $\begingroup$ Okay, that would be if $\vec{v_1}$ consisted of <cos($\alpha_1$), cos($\beta_1$), cos(γ1)>, etc., which it does not, since the angles given are the angle between $\vec{v_1}$ and $\vec{v_2}$. $\endgroup$ – Rloc Nov 14 '15 at 6:30
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Remember that if $\vec v =(a,b,c)$, its direction cosines will be

$$\begin{align} \cos \alpha &= \frac a{||v||}\Rightarrow ||v||\cos\alpha=a\\ \cos\beta&=\frac b{||v||}\\ \cos\gamma&=\frac c{||v||} \end{align} $$

Then, $v_1 \bot v_2 \Leftrightarrow \langle v_1,v_2\rangle=0=||v||^2\cos\alpha_1\cos\alpha_2+||v||^2\cos\beta_1\cos\beta_2+||v||^2\cos\gamma_1\cos\gamma_2$

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