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Q6.

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Found in: Page 646

### Precalculus Mathematics for Calculus

Book edition 7th Edition
Author(s) James Stewart, Lothar Redlin, Saleem Watson
Pages 948 pages
ISBN 9781337067508

# Determine Dot products and angle between vectors.a) $u·v$b) The angle between u and v to the nearest degree.$u=i+\sqrt{3}j,v=-\sqrt{3}i+j$

a. $u·v=0$

b. $\theta =90°$

See the step by step solution

## a.Step 1. Given information.

$\begin{array}{l}\stackrel{\to }{u}=i+\sqrt{3}j\\ \stackrel{\to }{v}=-\sqrt{3}i+j\end{array}$

## Step 2. Determining the dot product.

The dot product of vectors

$\stackrel{\to }{u} .\stackrel{\to }{v} ={a}_{1}{b}_{1}+{a}_{2}{b}_{2}$

$\stackrel{\to }{u} .\stackrel{\to }{v} =\left(1×-\sqrt{3}\right)+\left(1×\sqrt{3}\right)$

$\stackrel{\to }{u} .\stackrel{\to }{v} =\sqrt{3}+\left(-\sqrt{3}\right)$

$\stackrel{\to }{u} .\stackrel{\to }{v} =0$

## b.Step 1. Given information.

$\begin{array}{l}\stackrel{\to }{u}=i+\sqrt{3}j\\ \stackrel{\to }{v}=-\sqrt{3}i+j\end{array}$

also from part (a) $\stackrel{\to }{u} ·\stackrel{\to }{v} =0$

Now,

$\mathrm{cos}\theta =\frac{\stackrel{\to }{u} ·\stackrel{\to }{v} }{|\stackrel{\to }{u} ||\stackrel{\to }{v} |}$

$|\stackrel{\to }{u} |=\sqrt{{1}^{2}+{\sqrt{3}}^{2}}$

$|\stackrel{\to }{u} |=\sqrt{4}$

$|\stackrel{\to }{u} |=2$

$|\stackrel{\to }{v} |=\sqrt{-{\sqrt{3}}^{2}+{1}^{2}}$

$|\stackrel{\to }{v} |=\sqrt{4}$

Hence,

$\mathrm{cos}\theta =\frac{0}{4}$

$\mathrm{cos}\theta =0$

$\theta =90°$