Vector $\vec { A }$ has a magnitude of 4.0 m and points 30° south of east. Vector $\overrightarrow { \mathbf { B } }$ has a magnitude of 2.0 m and points 30° north of west. The resultant vector $\vec { A }$ + $\overrightarrow { \mathbf { B } }$ is given by
A) 10.0 m at an angle 30° south of east.
B) 10.0 m at an angle 60° east of south.
C) 2.0 m at an angle 60° south of east.
D) 2.0 m at an angle 30° south of east.
E) 1.0 m at an angle 30° east of south.

Question

Quizplus · Accepted Answer

To add vectors, we place them head to tail, meaning we place the tail of vector $\overrightarrow { \mathbf { B } }$ at the head of vector $\vec { A }$. Then, we draw the resultant vector from the tail of vector $\vec { A }$ to the head of vector $\overrightarrow { \mathbf { B } }$.

To find the magnitude and direction of this resultant vector, we can use trigonometry. We can split the resulting triangle into two right triangles, where the angle in between vector $\vec { A }$ and vector $\overrightarrow { \mathbf { B } }$ is 120° (the supplement of the given angle of 60°).

For the component of the resultant vector in the east-west direction, we have:

$$\text{Component in } x = 4.0\cos(30°) - 2.0\cos(30°) = 1.732$$

For the component in the north-south direction, we have:

$$\text{Component in } y = -4.0\sin(30°) + 2.0\sin(30°) = -0.732$$

Using the Pythagorean theorem, we can find the magnitude of the resultant vector:

$$|\vec { A } + \overrightarrow { \mathbf { B } }| = \sqrt{\text{Component in } x ^2 + \text{Component in } y ^2} = \sqrt{(1.732)^2 + (-0.732)^2} = 1.886$$

Finally, we can find the direction of the resultant vector using the arctangent function:

$$\theta = \tan^{-1} \left( \frac{\text{Component in } y}{\text{Component in } x} \right) = \tan^{-1} \left( \frac{-0.732}{1.732} \right) \approx -22.62°$$

Since vector $\vec { A }$ points 30° south of east, the resultant vector points a bit to the south of that, so our final answer is:

$$|\vec { A } + \overrightarrow { \mathbf { B } }| = 1.886 \text{ m at an angle } 30° -22.62°\text{ south of east} \approx \boxed{\textbf{(D) }2.0\text{ m at an angle 30° south of east.}}$$

Vector A⃗\vec { A }A Has a Magnitude of 4 B→\overrightarrow { \mathbf { B } }B

Vector $\vec { A }$ Has a Magnitude of 4 $\overrightarrow { \mathbf { B } }$