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niyaz
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In the session 10 problems, 1b, we use the dot product to get the angle between Av and v. In the solutions, I don't understand how they get that (Av dot v) / Avv =1/sqrt(2).
I understand that the length is sqrt(x^2+y^2) and the dot product is (x^2+y^2)/sqrt(2).
 3 months ago
 3 months ago
niyaz Group Title
In the session 10 problems, 1b, we use the dot product to get the angle between Av and v. In the solutions, I don't understand how they get that (Av dot v) / Avv =1/sqrt(2). I understand that the length is sqrt(x^2+y^2) and the dot product is (x^2+y^2)/sqrt(2).
 3 months ago
 3 months ago

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hanson.char Group TitleBest ResponseYou've already chosen the best response.1
\[\begin{aligned} Av\cdot v &= Avv\cos\theta \\ \therefore \cos\theta &= \frac{\color{blue}{Av\cdot v}}{\color{red}{Av}\color{teal}{v}} \\ &= \color{blue}{\frac{x^2+y^2}{\sqrt{2}}} \cdot \frac{1} {\color{red}{\sqrt{x^2+y^2}}\color{teal}{\sqrt{x^2+y^2}}} \\ &= \frac{x^2+y^2}{\sqrt{2}} \cdot \frac{1} {x^2+y^2} \\ &= {1 \over \sqrt{2}} \end{aligned}\]
 3 months ago

niyaz Group TitleBest ResponseYou've already chosen the best response.0
Oh! thanks. dw:1403418695064:dw
 3 months ago
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