Given the points \( A(-3,2), B(7,13), C(2,4), D(5,4), E(\sqrt{5}, \sqrt{5}) \), and \( F(4 \sqrt{5},-3 \sqrt{5}) \), find the position vector equal to the following vectors. \( \begin{array}{lll}\text { a. } \overrightarrow{A B} & \text { b. } \overrightarrow{A C} & \text { c. } \overrightarrow{E F} \\ \text { d. } \overrightarrow{C D}\end{array} \) a. \( \overrightarrow{A B}=\langle\square, \square \) (Type exact answers, using radicals as needed.)

To find the position vectors given the points, we need to perform vector subtraction for each pair of points. **a.** For \(\overrightarrow{A B}\): \[ \overrightarrow{A B} = B - A = (7 - (-3), 13 - 2) = (7 + 3, 13 - 2) = (10, 11). \] So, \(\overrightarrow{A B} = \langle 10, 11 \rangle\). **b.** For \(\overrightarrow{A C}\): \[ \overrightarrow{A C} = C - A = (2 - (-3), 4 - 2) = (2 + 3, 4 - 2) = (5, 2). \] So, \(\overrightarrow{A C} = \langle 5, 2 \rangle\). **c.** For \(\overrightarrow{E F}\): \[ \overrightarrow{E F} = F - E = (4 \sqrt{5} - \sqrt{5}, -3 \sqrt{5} - \sqrt{5}) = (4\sqrt{5} - \sqrt{5}, -3\sqrt{5} - \sqrt{5}) = (3\sqrt{5}, -4\sqrt{5}). \] So, \(\overrightarrow{E F} = \langle 3\sqrt{5}, -4\sqrt{5} \rangle\). **d.** For \(\overrightarrow{C D}\): \[ \overrightarrow{C D} = D - C = (5 - 2, 4 - 4) = (3, 0). \] So, \(\overrightarrow{C D} = \langle 3, 0 \rangle\). In summary: a. \(\overrightarrow{A B} = \langle 10, 11 \rangle\) b. \(\overrightarrow{A C} = \langle 5, 2 \rangle\) c. \(\overrightarrow{E F} = \langle 3\sqrt{5}, -4\sqrt{5} \rangle\) d. \(\overrightarrow{C D} = \langle 3, 0 \rangle\)