Answered: Find the angle, in degrees, between v =…

Trigonometry (MindTap Course List)

10th Edition

ISBN:9781337278461

Author:Ron Larson

Publisher:Ron Larson

Chapter3: Additional Topics In Trigonometry

Section3.4: Vectors And Dot Products

Problem 3ECP: Find the angle between u=2,1 and v=1,3.

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Question

Find the angle, in degrees, between v = 3cos 5π/3i + 3sin 5π/3j and w =2cos πi + 2sin πj.

Expert Solution

Step 1

Given: $v = 3 \cos (\frac{5 π}{3}) i + 3 \sin (\frac{5 π}{3}) j, w = 2 \cos (π) i + 2 \sin (π) j$

We know that angle $(θ)$ between two vectors $(\vec{u}, \vec{v})$ is given by

$θ = \cos^{- 1} (\frac{\vec{u} . \vec{v}}{|\vec{u}| |\vec{v}|}) . . (1)$

Step 2

Similarly, by using equation $(1)$ angle $(θ)$ between given two vectors will be

$\begin{array}{rcl} θ & = & \cos^{- 1} (\frac{v . w}{|v| |w|}) \\ = & \cos^{- 1} (\frac{[3 (\cos (\frac{5 π}{3}) i + \sin (\frac{5 π}{3}) j)] . [2 (\cos (π) i + \sin (π) j)]}{|3 (\cos (\frac{5 π}{3}) i + \sin (\frac{5 π}{3}) j)| |2 (\cos (π) i + \sin (π) j)|}) \\ = & \cos^{- 1} (\frac{6 [\cos (\frac{5 π}{3}) \cos (π) + \sin (\frac{5 π}{3}) \sin (π)]}{(3 \sqrt{\cos^{2} (\frac{5 π}{3}) + \sin^{2} (\frac{5 π}{3})}) (2 \sqrt{\cos^{2} (π) + \sin^{2} (π)})}) (∵ \cos (\frac{5 π}{3}) = \frac{1}{2}, \cos (π) = - 1, \sin (\frac{5 π}{3}) = - \frac{\sqrt{3}}{2}, \sin (π) = 0, \sin^{2} x + \cos^{2} x = 1) \\ = & \cos^{- 1} (\frac{6 [(\frac{1}{2}) (- 1) + (- \frac{\sqrt{3}}{2}) (0)]}{6 [\sqrt{1}] [\sqrt{1}]}) \\ = & \cos^{- 1} (\frac{6 (- \frac{1}{2} + 0)}{6 (1) (1)}) \\ = & \cos^{- 1} (\frac{- 3}{6}) \\ = & \cos^{- 1} (- \frac{1}{2}) (∵ \cos^{- 1} (- \frac{1}{2}) = \frac{2 π}{3}) \\ = & \frac{2 π}{3} \end{array}$

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