91Ó°ÊÓ

Using the given magnitudes and directions, the x and y components of the vectors can be calculated as follows: \(\overrightarrow{\mathbf{A}}=(0, \sqrt{3})\) \(\overrightarrow{\mathbf{B}}=(-1.0, 0)\)

We can now compute the sum, difference, and difference for the x- and y- components of vector A and B. (a) \(\overrightarrow{\mathbf{A}}+\overrightarrow{\mathbf{B}}=(0, \sqrt{3})+(-1,0)=(-1, \sqrt{3})\) (b) \(\overrightarrow{\mathbf{A}}-\overrightarrow{\mathbf{B}}=(0, \sqrt{3})-(-1,0)=(1, \sqrt{3})\) (c) \(\overrightarrow{\mathbf{B}}-\overrightarrow{\mathbf{A}}=(-1, 0)-(0, \sqrt{3})=(-1,-\sqrt{3})\)

Using the Pythagorean theorem for each of the resulting vectors, we find: (a) Magnitude: \(|\overrightarrow{\mathbf{A}}+\overrightarrow{\mathbf{B}}| = \sqrt{(-1)^2+(\sqrt{3})^2}=\sqrt{4}=2\) (b) Magnitude: \(|\overrightarrow{\mathbf{A}}-\overrightarrow{\mathbf{B}}| = \sqrt{(1)^2+(\sqrt{3})^2}=\sqrt{4}=2\)

To find the direction of the resulting vectors, we use the tangent inverse function: (a) Direction: \(\tan^{-1}\left (\frac{\sqrt{3}}{-1}\right) = -60°\) The angle -60° can be interpreted as an angle of 300° in the trigonometric circle. (b) Direction: \(\tan^{-1}\left (\frac{\sqrt{3}}{1}\right) = 60°\) (c) The x and y components of \(\overrightarrow{\mathbf{B}}-\overrightarrow{\mathbf{A}}\) are already calculated in Step 2: (-1, -\(\sqrt{3}\)) The answers are: (a) The magnitude of \(\overrightarrow{\mathbf{A}}+\overrightarrow{\mathbf{B}}\) is 2 units, and the direction is \(300°\). (b) The magnitude of \(\overrightarrow{\mathbf{A}}-\overrightarrow{\mathbf{B}}\) is 2 units, and the direction is \(60°\). (c) The x-component of \(\overrightarrow{\mathbf{B}}-\overrightarrow{\mathbf{A}}\) is -1, and the y-component is -\(\sqrt{3}\).