Two vectors are given by overrightarrow (A)=2overrightarrow (i)+3overrightarrow (j) and overrightarrow (B)=4overrightarrow (i)-3overrightarrow (j) Find: (a) the magnitude and direction of the vector sum overrightarrow (R)=overrightarrow (A)+overrightarrow (B) (b) the magnitude and direction of the vector difference overrightarrow (S)=overrightarrow (A)-overrightarrow (B)

(a) To find the magnitude and direction of the vector sum $\overrightarrow{R} = \overrightarrow{A} + \overrightarrow{B}$, we first calculate the sum of the two vectors:<br /><br />$\overrightarrow{R} = \overrightarrow{A} + \overrightarrow{B} = (2\overrightarrow{i} + 3\overrightarrow{j}) + (4\overrightarrow{i} - 3\overrightarrow{j}) = 6\overrightarrow{i}$<br /><br />The magnitude of $\overrightarrow{R}$ is given by the square root of the sum of the squares of its components:<br /><br />$|\overrightarrow{R}| = \sqrt{(6)^2 + (0)^2} = 6$<br /><br />The direction of $\overrightarrow{R}$ is along the positive x-axis since it has no y-component.<br /><br />(b) To find the magnitude and direction of the vector difference $\overrightarrow{S} = \overrightarrow{A} - \overrightarrow{B}$, we calculate the difference of the two vectors:<br /><br />$\overrightarrow{S} = \overrightarrow{A} - \overrightarrow{B} = (2\overrightarrow{i} + 3\overrightarrow{j}) - (4\overrightarrow{i} - 3\overrightarrow{j}) = -2\overrightarrow{i} + 6\overrightarrow{j}$<br /><br />The magnitude of $\overrightarrow{S}$ is given by the square root of the sum of the squares of its components:<br /><br />$|\overrightarrow{S}| = \sqrt{(-2)^2 + (6)^2} = \sqrt{4 + 36} = \sqrt{40} = 2\sqrt{10}$<br /><br />The direction of $\overrightarrow{S}$ is 126.87 degrees counterclockwise from the positive x-axis, which can be calculated using the arctangent function:<br /><br />$\theta = \arctan\left(\frac{6}{-2}\right) = \arctan(-3) \approx -71.57^\circ$<br /><br />Since the direction is negative, we add 360 degrees to get the positive angle:<br /><br />$\theta = 360^\circ - 71.57^\circ = 288.43^\circ$<br /><br />Therefore, the magnitude and direction of the vector sum $\overrightarrow{R}$ are 6 and along the positive x-axis, respectively. The magnitude and direction of the vector difference $\overrightarrow{S}$ are $2\sqrt{10}$ and 126.87 degrees counterclockwise from the positive x-axis, respectively.