## Calculus 8th Edition

Published by Cengage

# Chapter 12 - Vectors and the Geometry of Space - 12.3 The Dot Product - 12.3 Exercises: 35

#### Answer

Direction cosines are: $\frac{1}{\sqrt {14}},\frac{-2}{\sqrt {14}}, \frac{-3}{\sqrt {14}}$ Direction angles are: $74 ^\circ, 122 ^\circ, 143 ^\circ$

#### Work Step by Step

Let $v= i-2j-3k=\lt 1,-2,-3 \gt$ $|v|=\sqrt {1^2+(-2)^2+(-3)^2}=\sqrt {14}$ Direction cosines are: $cos \alpha = \frac{1}{\sqrt {14}}, cos \beta =\frac{-2}{\sqrt {14}}, cos \gamma=\frac{-3}{\sqrt {14}}$ Thus, the direction angles are: $\alpha =cos^{-1} \frac{1}{\sqrt {14}}=74 ^\circ, \beta = cos^{-1} \frac{-2}{\sqrt {14}}=122 ^\circ, \gamma = cos^{-1} \frac{-3}{\sqrt {14}}=143^ \circ$

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