#### Answer

(a) $B = -3\hat{i}+5\hat{j}$
(b) The magnitude of the vector is $5.8$ and the direction is an angle of $59.0^{\circ}$ above the negative x-axis.

#### Work Step by Step

(a) $A+B+C = 1\hat{j}$
$3\hat{i}+B-4\hat{j} = 1\hat{j}$
$B = -3\hat{i}+5\hat{j}$
(b) We can find the magnitude of the vector B;
$B = \sqrt{B_x^2+B_y^2}$
$B = \sqrt{(-3)^2+(5)^2}$
$B = 5.8$
We can find the angle above the negative x-axis;
$tan(\theta) = \frac{5}{3}$
$\theta = tan^{-1}(\frac{5}{3}) = 59.0^{\circ}$
The magnitude of the vector is $5.8$ and the direction is an angle of $59.0^{\circ}$ above the negative x-axis.