## College Physics (4th Edition)

(a) The magnitude of $A-B$ is $2.0$ The angle above the positive x-axis is $60^{\circ}$ (b) The x-component of $B-A$ is $-1.0$ The y-component of $B-A$ is $-\sqrt{3}$
(a) The vector $A$ has a magnitude $\sqrt{3}$ along the positive y-axis. The vector $A-B$ has an x-component that is the same magnitude as $B$ but in the opposite direction. Therefore $A-B$ has an x-component of 1.0 unit along the positive x-axis. We can find the magnitude of $A-B$: $\vert A-B \vert = \sqrt{(\sqrt{3})^2+(1.0)^2} = 2.0$ We can find the angle $\theta$ above the positive x-axis: $tan~\theta = \frac{\sqrt{3}}{1.0}$ $\theta = tan^{-1}(\frac{\sqrt{3}}{1.0})$ $\theta = 60^{\circ}$ (b) The x-component of $B-A$ is the vector $B$, which is -1.0. The y-component of $B-A$ has the same magnitude as $A$ but in the opposite direction. The y-component of $B-A$ is $-\sqrt{3}$