# Need to find the angle between two unit vectors vec(m) and vec(n) if the vectors vec(p)= vec(m)+2 vec(n) and vec(q)=5 vec(m)−4 vec(n) are perpendicular to each other.

I need to find the angle between two unit vectors $\stackrel{\to }{m}$ and $\stackrel{\to }{n}$ if the vectors $\stackrel{\to }{p}=\stackrel{\to }{m}+2\stackrel{\to }{n}$ and $\stackrel{\to }{q}=5\stackrel{\to }{m}-4\stackrel{\to }{n}$ are perpendicular to each other.
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gutsy8v
Calculating the dot product we get
$\stackrel{\to }{p}\cdot \stackrel{\to }{q}=5{\stackrel{\to }{m}}^{2}+10\stackrel{\to }{m}\cdot \stackrel{\to }{n}-4\stackrel{\to }{m}\cdot \stackrel{\to }{n}-8{\stackrel{\to }{n}}^{2}=0$
We get
$\stackrel{\to }{m}\cdot \stackrel{\to }{n}=\frac{1}{2}$
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Lillie Pittman
Since p,q are perpendicular and m,n have unit length,
$0=⟨p,q⟩=⟨m+2n,5m-4n⟩=5|m{|}^{2}-4⟨m,n⟩+10⟨n,m⟩-8|n{|}^{2}=-3+6⟨m,n⟩.$
This, combined with the fact that
$\mathrm{cos}\alpha =\frac{⟨m,n⟩}{|m||n|}=⟨m,n⟩$
should allow you to find the required angle $\alpha$.