Book edition 7th Edition

Author(s) James Stewart, Lothar Redlin, Saleem Watson

Pages 948 pages

ISBN 9781337067508

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Short Answer

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

a. $u \cdot v = 0$

b. $θ = 90 °$

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

a.Step 1. Given information.

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

Step 2. Determining the dot product.

The dot product of vectors

$\vec{u} . \vec{v} = a_{1} b_{1} + a_{2} b_{2}$

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

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

$\vec{u} . \vec{v} = 0$

b.Step 1. Given information.

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

also from part (a) $\vec{u} \cdot \vec{v} = 0$

Now,

\cos θ = \frac{\vec{u} \cdot \vec{v}}{| \vec{u} | | \vec{v} |}

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

$| \vec{u} | = \sqrt{4}$

$| \vec{u} | = 2$

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

$| \vec{v} | = \sqrt{4}$

Hence,

$\cos θ = \frac{0}{4}$

$\cos θ = 0$

$θ = 90 °$

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Precalculus Mathematics for Calculus

Answers without the blur.

Short Answer

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

Step by Step Solution

a.Step 1. Given information.

Step 2. Determining the dot product.

b.Step 1. Given information.

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