Chapter 8 - 8.3 - Vectors in the Plane - 8.3 Exercises - Page 586: 71

$\lt 2, 2 \sqrt 3+4 \gt$

Work Step by Step

The component form of the vector $u$ is: $u=||u|| \lt \cos \theta_{u} , \sin \theta_{u} \gt =4 \lt \cos 60^{\circ},\sin 60^{\circ} \gt \\= 4 \lt \dfrac{1}{2},\dfrac{\sqrt {3}}{2} \gt \\= \lt 2 , 2 \sqrt 3 \gt$ Now, the component form of the vector $v$ is: $v=||v|| \lt \cos \theta_{v} , \sin \theta_{v} \gt =4 \lt \cos 90^{\circ},\sin 90^{\circ} \gt \\= 4 \lt 0,1 \gt \\= \lt 0,4 \gt$ Thus, $u+v=\lt 2 , 2 \sqrt 3 \gt+\lt 0,4 \gt = \lt 2, 2 \sqrt 3+4 \gt$

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