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Define the points \( P(-4,-2) \) and \( Q(3,-4) \). Carry out the following calculation. Find two vectors parallel to \( \overrightarrow{\mathrm{QP}} \) with length 2 . The parallel vector of length 2 with the same direction is (Type exact answers, using radicals as needed.)

Ask by Cox Black. in the United States
Jan 22,2025

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One vector parallel to \(\overrightarrow{\mathrm{QP}}\) with length 2 is \(\left( \frac{-14\sqrt{53}}{53}, \frac{4\sqrt{53}}{53} \right)\). Another vector in the opposite direction is \(\left( \frac{14\sqrt{53}}{53}, \frac{-4\sqrt{53}}{53} \right)\).

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To find the vector \( \overrightarrow{QP} \), we start by calculating it: \[ \overrightarrow{QP} = P - Q = (-4 - 3, -2 - (-4)) = (-7, 2). \] Next, we find the length (magnitude) of this vector: \[ |\overrightarrow{QP}| = \sqrt{(-7)^2 + 2^2} = \sqrt{49 + 4} = \sqrt{53}. \] Now, we can find a unit vector in the direction of \( \overrightarrow{QP} \): \[ \hat{QP} = \frac{\overrightarrow{QP}}{|\overrightarrow{QP}|} = \left( \frac{-7}{\sqrt{53}}, \frac{2}{\sqrt{53}} \right). \] To find the parallel vector of length 2 in the same direction, we multiply the unit vector by 2: \[ \overrightarrow{QP}_{length2} = 2 \cdot \hat{QP} = 2 \left( \frac{-7}{\sqrt{53}}, \frac{2}{\sqrt{53}} \right) = \left( \frac{-14}{\sqrt{53}}, \frac{4}{\sqrt{53}} \right). \] To find another parallel vector of length 2, we can take the negative of this vector: \[ \overrightarrow{QP}_{length2, negative} = -\overrightarrow{QP}_{length2} = \left( \frac{14}{\sqrt{53}}, \frac{-4}{\sqrt{53}} \right). \] Thus, the two vectors parallel to \( \overrightarrow{QP} \) with length 2 are: \[ \left( \frac{-14}{\sqrt{53}}, \frac{4}{\sqrt{53}} \right) \quad \text{and} \quad \left( \frac{14}{\sqrt{53}}, \frac{-4}{\sqrt{53}} \right). \]

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