Book edition 1st

Author(s) Peter Kohn, Laura Taalman

Pages 1155 pages

ISBN 9781429241861

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Short Answer

In Exercises 37–42, find $||v||$ and find the unit vector in the direction of v.
$v = <\sqrt{\frac{1}{3}}, - \sqrt{\frac{2}{3}}>$

$||v|| = 1$ and the unit vector in the direction of v is $<\sqrt{\frac{1}{3}}, - \sqrt{\frac{2}{3}}>$ .

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1. Given information.

The given vector is:

$v = <\sqrt{\frac{1}{3}}, - \sqrt{\frac{2}{3}}>$

Step 2. Find the norm of the vector.

$v = <\sqrt{\frac{1}{3}}, - \sqrt{\frac{2}{3}}> ||v|| = \sqrt{{(\sqrt{\frac{1}{3}})}^{2} + {(- \sqrt{\frac{2}{3}})}^{2}} = \sqrt{\frac{1}{3} + \frac{2}{3}} = \sqrt{\frac{3}{3}} = 1$

Therefore, the norm of the given vector is 1.

Step 3. Find the unit vector in the direction of v.

The unit vector in the direction of v is the vector:

$\frac{1}{||v||} v = \frac{1}{1} <\sqrt{\frac{1}{3}}, - \sqrt{\frac{2}{3}}> = <\sqrt{\frac{1}{3}}, - \sqrt{\frac{2}{3}}>$

Therefore, the unit vector in the direction of $v = <\sqrt{\frac{1}{3}}, - \sqrt{\frac{2}{3}}>$ is $<\sqrt{\frac{1}{3}}, - \sqrt{\frac{2}{3}}>$ .

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Calculus

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Short Answer

In Exercises 37–42, find $||v||$ and find the unit vector in the direction of v.
$v = <\sqrt{\frac{1}{3}}, - \sqrt{\frac{2}{3}}>$

Step by Step Solution

Step 1. Given information.

Step 2. Find the norm of the vector.

Step 3. Find the unit vector in the direction of v.

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Short Answer

In Exercises 37–42, find v and find the unit vector in the direction of v.v=13,-23

Step by Step Solution

Step 1. Given information.

Step 2. Find the norm of the vector.

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In Exercises 37–42, find $||v||$ and find the unit vector in the direction of v.
$v = <\sqrt{\frac{1}{3}}, - \sqrt{\frac{2}{3}}>$