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Q 23.

Expert-verified

Calculus

Found in: Page 812

Book edition 1st

Author(s) Peter Kohn, Laura Taalman

Pages 1155 pages

ISBN 9781429241861

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Short Answer

In Exercises 20-23, find the dot product of the given pairs of vectors and the angle between the two vectors.
$u = <- 5, 1, 3>, v = <- 3, 2, 7>$

The dot product is 38 and the angle is $\cos^{- 1} (\frac{38}{\sqrt{2170}})$ .

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1. Given information.

The given pairs of vectors are:

$u = <- 5, 1, 3> and v = <- 3, 2, 7>$

Step 2. Find the dot product.

The dot product is: $u \cdot v = u_{1} v_{1} + u_{2} v_{2} + u_{3} v_{3}$

$u = <- 5, 1, 3> and v = <- 3, 2, 7> u \cdot v = (- 5) (- 3) + 1 (2) + 3 (7) = 15 + 2 + 21 = 38$

Therefore, the dot product of the given two vectors $u = <- 5, 1, 3> and v = <- 3, 2, 7>$ is 38.

Step 3. Find the angle between the two vectors.

The formula for the angle between the two vectors is:

$\cos θ = \frac{u \cdot v}{||u|| ||v||} u = <- 5, 1, 3> and v = <- 3, 2, 7> u \cdot v = 38 ||u|| = \sqrt{{(- 5)}^{2} + 1^{2} + 3^{2}} \sqrt{25 + 1 + 9} = \sqrt{35} ||v|| = \sqrt{{(- 3)}^{2} + 2^{2} + 7^{2}} \sqrt{9 + 4 + 49} = \sqrt{62}$

Then,

$\cos θ = \frac{u \cdot v}{||u|| ||v||} \cos θ = \frac{38}{\sqrt{35} \sqrt{62}} θ = \cos^{- 1} (\frac{38}{\sqrt{2170}})$

Therefore, the angle is $\cos^{- 1} (\frac{38}{\sqrt{2170}})$ .

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