Vector $\vec { A }$ has magnitude 2 units and is directed to the north. Vector $\overrightarrow { \mathbf { B } }$ has magnitude $5 \text { units }$ and is directed to the south. Calculate the magnitude and direction of $\overrightarrow { \mathbf { A } } - \overrightarrow { \mathbf { B } }$.
A) 7 units, north
B) 7 units, south
C) 3 units, north
D) 3 units, south

Question

Quizplus · Accepted Answer

To calculate $\overrightarrow { \mathbf { A } } - \overrightarrow { \mathbf { B } }$, we need to subtract the components of $\vec{B}$ from the components of $\vec{A}$. Since $\vec{A}$ is directed to the north, its components are $\begin{pmatrix}0 \ 2\end{pmatrix}$ (where the first entry represents the x-component and the second entry represents the y-component). Since $\vec{B}$ is directed to the south, its components are $\begin{pmatrix}0 \ -5\end{pmatrix}$.

Subtracting the components, we get:

$$\overrightarrow { \mathbf { A } } - \overrightarrow { \mathbf { B } } = \begin{pmatrix}0 \ 2\end{pmatrix} - \begin{pmatrix}0 \ -5\end{pmatrix} = \begin{pmatrix}0 \ 7\end{pmatrix}$$

The magnitude of $\overrightarrow { \mathbf { A } } - \overrightarrow { \mathbf { B } }$ is $\sqrt{0^2 + 7^2} = 7$ units. The direction is north, since the y-component is positive. Therefore, the answer is:

Answer: A

Vector A⃗\vec { A }A Has Magnitude 2 Units and Is Directed to the North

Vector $\vec { A }$ Has Magnitude 2 Units and Is Directed to the North