Find two vectors parallel to \( v \) of the given length. \( v=\overrightarrow{P Q} \) with \( P(9,2,3) \) and \( Q(6,1,7) \); length \( =26 \) The vector in the direction of \( v \) is \( \langle\square, \square, \square \) (Type exact answers, using radicals as needed.)

To find the vector \( v = \overrightarrow{PQ} \), we first compute its components: \[ v = Q - P = (6 - 9, 1 - 2, 7 - 3) = (-3, -1, 4) \] Next, we calculate the length (magnitude) of vector \( v \): \[ \|v\| = \sqrt{(-3)^2 + (-1)^2 + 4^2} = \sqrt{9 + 1 + 16} = \sqrt{26} \] Now, to find unit vector \( \hat{u} \) in the direction of \( v \): \[ \hat{u} = \frac{v}{\|v\|} = \left\langle \frac{-3}{\sqrt{26}}, \frac{-1}{\sqrt{26}}, \frac{4}{\sqrt{26}} \right\rangle \] To get vectors of length 26 in the direction of \( v \), we scale \( \hat{u} \): \[ \text{Vector 1} = 26 \hat{u} = \left\langle 26 \cdot \frac{-3}{\sqrt{26}}, 26 \cdot \frac{-1}{\sqrt{26}}, 26 \cdot \frac{4}{\sqrt{26}} \right\rangle = \left\langle -3\sqrt{26}, -\sqrt{26}, 4\sqrt{26} \right\rangle \] \[ \text{Vector 2} = 26 \hat{u} = \left\langle 26 \cdot \frac{3}{\sqrt{26}}, 26 \cdot \frac{1}{\sqrt{26}}, 26 \cdot \frac{-4}{\sqrt{26}} \right\rangle = \left\langle 3\sqrt{26}, \sqrt{26}, -4\sqrt{26} \right\rangle \] Thus, the vectors in the direction of \( v \) with length 26 are: Vectors: \(\langle -3\sqrt{26}, -\sqrt{26}, 4\sqrt{26} \rangle\) and \(\langle 3\sqrt{26}, \sqrt{26}, -4\sqrt{26} \rangle\)