Define the points \( P(-4,-2) \) and \( Q(3,-4) \). Carry out the following calculation. Find two vectors parallel to \( \overrightarrow{Q P} \) with length 2 . The parallel vector of length 2 with the same direction is \( \left\langle\frac{-14 \sqrt{53}}{53}, \frac{4 \sqrt{53}}{53}\right) \). (Type exact answers, using radicals as needed.) The parallel vector of length 2 with the opposite direction is \( \langle\square, \square \). (Type exact answers, using radicals as needed.)

To find the vector \(\overrightarrow{QP}\), we start by calculating the coordinates of this vector: \[ \overrightarrow{QP} = P - Q = (-4, -2) - (3, -4) = (-4 - 3, -2 + 4) = (-7, 2). \] Next, we find the magnitude of \(\overrightarrow{QP}\): \[ \|\overrightarrow{QP}\| = \sqrt{(-7)^2 + 2^2} = \sqrt{49 + 4} = \sqrt{53}. \] To find a unit vector in the direction of \(\overrightarrow{QP}\), we divide each component of \(\overrightarrow{QP}\) by its magnitude: \[ \frac{\overrightarrow{QP}}{\|\overrightarrow{QP}\|} = \left\langle \frac{-7}{\sqrt{53}}, \frac{2}{\sqrt{53}} \right\rangle. \] Next, we multiply this unit vector by 2 to get the vector in the same direction with a length of 2: \[ \mathbf{v}_{\text{same}} = 2 \cdot \left\langle \frac{-7}{\sqrt{53}}, \frac{2}{\sqrt{53}} \right\rangle = \left\langle \frac{-14}{\sqrt{53}}, \frac{4}{\sqrt{53}} \right\rangle. \] The parallel vector of length 2 with the same direction is given as \[ \left\langle \frac{-14 \sqrt{53}}{53}, \frac{4 \sqrt{53}}{53} \right\rangle. \] Now, for the parallel vector of length 2 with the opposite direction, we simply reverse the signs: \[ \mathbf{v}_{\text{opposite}} = 2 \cdot \left\langle \frac{7}{\sqrt{53}}, \frac{-2}{\sqrt{53}} \right\rangle = \left\langle \frac{14}{\sqrt{53}}, \frac{-4}{\sqrt{53}} \right\rangle. \] Putting it in the requested format, we obtain: \[ \langle \frac{14 \sqrt{53}}{53}, \frac{-4 \sqrt{53}}{53} \rangle. \] Thus, the parallel vector of length 2 with the opposite direction is \[ \langle \frac{14 \sqrt{53}}{53}, \frac{-4 \sqrt{53}}{53} \rangle. \]