Calculate | $\to u$ $vec u$ + $\to v$ $vec v$ | help?

Question

Given that | $\to u$ $vec u$ | = 3 and | $\to v$ $vec v$ | = 4

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calculate the angle between $\to u$ $vec u$ and
$\to u$ $vec u$ + $\to v$ $vec v$

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Answer 1 · 2018-06-09T06:55:16

See below

First Part

$| u | = 3, | v | = 4, ϕ = \frac{2 π}{3}$ $abs bb u = 3, abs bb v = 4, phi = (2pi)/3$

${| u + v |}^{2} = (u + v) \cdot (u + v)$ $abs(bb u + bb v)^2 = (bb u + bb v )*(bb u + bb v )$

$= u^{2} + v^{2} + 2 u \cdot v$ $= u^2 + v^2 + 2 bb u * bb v$

$= 3^{2} + 4^{2} + 2 \cdot 3 \cdot 4 \cdot cos (120)$ $= 3^2 + 4^2 + 2 * 3 * 4 * cos(120)$

$= 13$ $=13$

$\Rightarrow | u + v | = \sqrt{13}$ $implies abs(bb u + bb v) = sqrt13$

Second Part

$u \cdot (u + v) = | u | \cdot | u + v | cos θ$ $bb u * (bb u + bbv) =abs bb u * abs (bb u + bbv) cos theta$

$cos θ = \frac{u \cdot (u + v)}{| u | \cdot | u + v |}$ $cos theta = (bb u * (bb u + bbv))/ (abs bb u * abs (bb u + bbv) )$

$= \frac{u^{2} + | u | | v | cos ϕ}{| u | \cdot | u + v |}$ $= (u^2 +abs bb u abs bb v cos phi)/ (abs bb u * abs (bb u + bbv) )$

$= \frac{3^{2} + 3 \cdot 4 \cdot cos (120)}{3 \cdot \sqrt{13}}$ $= (3^2 +3 * 4 * cos (120))/ (3 * sqrt(13) )$

$= \frac{1}{\sqrt{13}}$ $= 1/sqrt(13)$

$θ = {cos}^{- 1} (\frac{1}{\sqrt{13}}) = {73.9}^{o}$ $theta = cos^(-1) (1/sqrt(13)) = 73.9^o$