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Q6.

Expert-verifiedFound in: Page 646

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) **$u\xb7v$

**b) The angle between u and v to the nearest degree.**

$u=i+\sqrt{3}j,v=-\sqrt{3}i+j$

a. $u\xb7v=0$

b. $\theta =90\xb0$

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

The dot product of vectors

$\overrightarrow{u}\u200a.\overrightarrow{v}\u200a={a}_{1}{b}_{1}+{a}_{2}{b}_{2}$

$\overrightarrow{u}\u200a.\overrightarrow{v}\u200a=(1\times -\sqrt{3})+(1\times \sqrt{3})$

$\overrightarrow{u}\u200a.\overrightarrow{v}\u200a=\sqrt{3}+(-\sqrt{3})$

$\overrightarrow{u}\u200a.\overrightarrow{v}\u200a=0$

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

also from part (a) $\overrightarrow{u}\u200a\xb7\overrightarrow{v}\u200a=0$

Now,

$\mathrm{cos}\theta =\frac{\overrightarrow{u}\u200a\xb7\overrightarrow{v}\u200a}{|\overrightarrow{u}\u200a||\overrightarrow{v}\u200a|}$$|\overrightarrow{u}\u200a|=\sqrt{{1}^{2}+{\sqrt{3}}^{2}}$

$|\overrightarrow{u}\u200a|=\sqrt{4}$

$|\overrightarrow{u}\u200a|=2$

$|\overrightarrow{v}\u200a|=\sqrt{-{\sqrt{3}}^{2}+{1}^{2}}$

$|\overrightarrow{v}\u200a|=\sqrt{4}$

Hence,

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

$\mathrm{cos}\theta =0$

$\theta =90\xb0$

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