Answer
The magnitude of the net displacement is 7.5 meters.
Work Step by Step
Let $r$ be the displacement of the mouse. Then;
$r_x = 5.0~m + 3.0~m~cos(45^{\circ})$
$r_x = 7.1~m$
$r_y = -3.0~m~sin(45^{\circ})$
$r_y = -2.1~m$
$r_z = -1.0~m$
We can find the magnitude of the displacement as:
$r = \sqrt{(7.1~m)^2+(-2.1~m)^2+(-1.0~m)^2}$
$r = 7.5~m$
The magnitude of the net displacement is 7.5 meters.
