Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

(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.
(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.