# Chapter 8 - 8.3 - Vectors in the Plane - 8.3 Exercises - Page 586: 74

$90^{\circ}$

#### Work Step by Step

The magnitude of the vector $v$ is: $||v|| =\sqrt{1^2+2^2}=\sqrt 5$ Now, the magnitude of the vector $w$ is: $||w|| =\sqrt{2^2+(-1)^2}= \sqrt 5$ Thus, the distance between the terminal points is: $d=\sqrt{(2-1)^2+(-1-2)^2}=\sqrt {10}$ Apply the law of cosines to compute the angle between the vectors. $\cos \theta =\dfrac{(\sqrt 5)^2+(\sqrt 5)^2 -(\sqrt {10})^2}{2 (\sqrt 5)(\sqrt 5)}=0$ So, $\theta=90^{\circ}$

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.