#### Answer

(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 $-\sqrt{3}$

#### Work Step by Step

(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 $-\sqrt{3}$