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Find the angle between the given vectors to the nearest tenth of a degree. u = <-5, -4>, v = <-4, -3>
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\(\bf \textit{angle between two vectors }\\ \quad \\ cos(\theta)=\cfrac{u \cdot v}{||u||\ ||v||} \implies \cfrac{\text{dot product}}{\text{product of magnitudes}}\\ \quad \\ \theta = cos^{-1}\left(\cfrac{u \cdot v}{||u||\ ||v||}\right)\)
I did that and got 0.99
is that right?
i'll show you my work if you want
hmm what did you get for the dot product?
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(-5)(-4) + (-4)(-3) = 20+ 12 = 32
ok.... lemme check
ohhh i think i know what i did
one sec im going to redo it
is it 1.8?
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yeap 1.789910 degrees, thus \(\large 1.8^o\)
thanks so much!!! :)
yw
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