Consider the points \( P(1,5,5) \) and \( Q(7,7,11) \). a. Find \( \overrightarrow{P Q} \) and state your answer in two forms: \( \langle a, b, c\rangle \) and \( a i+b j+c k \). b. Find the magnitude of \( \overrightarrow{P Q} \). c. Find two unit vectors parallel to \( \overrightarrow{P Q} \). a. Find \( \overrightarrow{P Q} \). \( \overrightarrow{P Q}=\langle 6,2,6\rangle=(6) i+(2) \mathbf{j}+(6) \mathbf{k} \) b. Find the magnitude of \( \overrightarrow{P Q} \). The magnitude of \( \overrightarrow{P Q} \) is \( 2 \sqrt{19} \). (Type an exact answer, using radicals as needed.) c. Find two unit vectors parallel to \( \overrightarrow{P Q} \). The unit vector in the direction of \( \overrightarrow{P Q} \) is \( \left\langle\frac{3}{\sqrt{19}}, \frac{1}{\sqrt{19}}, \frac{3}{\sqrt{19}}\right| \) (Type exact answers, using radicals as needed.) The unit vector in the opposite direction of \( \overrightarrow{P Q} \) is \( \left\langle\begin{array}{|}\square \\ \text { (Type exact answers, using radicals as needed.) } \\ \square\end{array}\right. \)

Let's complete part **c** of your problem by finding the two unit vectors parallel to \( \overrightarrow{PQ} \). ### c. Find Two Unit Vectors Parallel to \( \overrightarrow{PQ} \) Given the vector \( \overrightarrow{PQ} = \langle 6, 2, 6 \rangle \), we first found its magnitude to be \( 2\sqrt{19} \). To find a unit vector in the direction of \( \overrightarrow{PQ} \), we divide each component of \( \overrightarrow{PQ} \) by its magnitude: \[ \text{Unit vector in the direction of } \overrightarrow{PQ} = \left\langle \frac{6}{2\sqrt{19}}, \frac{2}{2\sqrt{19}}, \frac{6}{2\sqrt{19}} \right\rangle = \left\langle \frac{3}{\sqrt{19}}, \frac{1}{\sqrt{19}}, \frac{3}{\sqrt{19}} \right\rangle \] In terms of the unit vectors \( \mathbf{i}, \mathbf{j}, \mathbf{k} \), this is: \[ \frac{3}{\sqrt{19}}\,\mathbf{i} + \frac{1}{\sqrt{19}}\,\mathbf{j} + \frac{3}{\sqrt{19}}\,\mathbf{k} \] To find the unit vector in the **opposite** direction of \( \overrightarrow{PQ} \), we simply take the negative of each component of the unit vector we just found: \[ \text{Unit vector in the opposite direction of } \overrightarrow{PQ} = \left\langle -\frac{3}{\sqrt{19}}, -\frac{1}{\sqrt{19}}, -\frac{3}{\sqrt{19}} \right\rangle \] In terms of the unit vectors \( \mathbf{i}, \mathbf{j}, \mathbf{k} \), this is: \[ -\frac{3}{\sqrt{19}}\,\mathbf{i} - \frac{1}{\sqrt{19}}\,\mathbf{j} - \frac{3}{\sqrt{19}}\,\mathbf{k} \] ### **Final Answer for Part c:** - **Unit vector in the direction of \( \overrightarrow{PQ} \):** \[ \left\langle \frac{3}{\sqrt{19}}, \frac{1}{\sqrt{19}}, \frac{3}{\sqrt{19}} \right\rangle \quad \text{or} \quad \frac{3}{\sqrt{19}}\,\mathbf{i} + \frac{1}{\sqrt{19}}\,\mathbf{j} + \frac{3}{\sqrt{19}}\,\mathbf{k} \] - **Unit vector in the opposite direction of \( \overrightarrow{PQ} \):** \[ \left\langle -\frac{3}{\sqrt{19}}, -\frac{1}{\sqrt{19}}, -\frac{3}{\sqrt{19}} \right\rangle \quad \text{or} \quad -\frac{3}{\sqrt{19}}\,\mathbf{i} - \frac{1}{\sqrt{19}}\,\mathbf{j} - \frac{3}{\sqrt{19}}\,\mathbf{k} \]