Vector $\overrightarrow { \mathrm { A } }$ has a magnitude of 8.0 m and points 30° north of east; vector $\overrightarrow { \mathrm { B } }$ has a magnitude of 6.0 m and points 30° west of north; and vector $\overrightarrow { \mathrm { C } }$ has a magnitude of 5.0 m and points 30° west of south. The resultant vector $\overrightarrow { \mathrm { A } }$ + $\overrightarrow { \mathrm { B } }$ + $\overrightarrow { \mathrm { C } }$ is given by
A)2.7 m at an angle 74° north of east.
B)5.9 m at an angle 74° north of east.
C)4.8 m at an angle 74° east of north.
D)5.1 m at an angle 74° north of east.
E)2.1 m at an angle 66° east of north.

Question

Quizplus · Accepted Answer

The x-component of the resultant vector is:$$A_x + B_x + C_x = (8.0 \cos 30°) + (6.0 \cos 30°) + (5.0 \cos 30°) = 14.8 m$$The y-component of the resultant vector is:$$A_y + B_y + C_y = (8.0 \sin 30°) + (6.0 \sin 30°) + (5.0 \sin 30°) = 10.2 m$$The magnitude of the resultant vector is:$$R = \sqrt{A_x^2 + A_y^2} = \sqrt{14.8^2 + 10.2^2} = 18.0 m$$The angle of the resultant vector is:$$	heta = 	an^{-1}\left(\frac{A_y}{A_x}ight) = 	an^{-1}\left(\frac{10.2}{14.8}ight) = 34.3°$$Therefore, the resultant vector is 18.0 m at an angle of 34.3° north of east.

Vector A→\overrightarrow { \mathrm { A } }A Has a Magnitude of 8

Vector $\overrightarrow { \mathrm { A } }$ Has a Magnitude of 8