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 \( \overrightarrow{P Q} \). \( \overrightarrow{P Q}=\langle 6,2,6\rangle=(6) i+(2) j+(6) k \) 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 \( \langle\square \), \( \square \). (Type exact answers, using radicals as needed.)
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The Deep Dive
The vector \( \overrightarrow{P Q} \) is found by subtracting the coordinates of point \( P \) from point \( Q \): \[ \overrightarrow{P Q} = Q - P = (7 - 1, 7 - 5, 11 - 5) = \langle 6, 2, 6 \rangle \] In vector notation, it can be expressed as \( 6i + 2j + 6k \). To find the magnitude of \( \overrightarrow{P Q} \), we use the formula: \[ ||\overrightarrow{P Q}|| = \sqrt{6^2 + 2^2 + 6^2} = \sqrt{36 + 4 + 36} = \sqrt{76} = 2\sqrt{19} \] Now, to find the unit vector, we first calculate the unit vector \( \hat{u} \) in the direction of \( \overrightarrow{P Q} \): \[ \hat{u} = \frac{\overrightarrow{P Q}}{||\overrightarrow{P Q}||} = \frac{\langle 6, 2, 6 \rangle}{2\sqrt{19}} = \left\langle \frac{6}{2\sqrt{19}}, \frac{2}{2\sqrt{19}}, \frac{6}{2\sqrt{19}} \right\rangle = \left\langle \frac{3}{\sqrt{19}}, \frac{1}{\sqrt{19}}, \frac{3}{\sqrt{19}} \right\rangle \] To find a second unit vector in the opposite direction, we simply negate the components: \[ -\hat{u} = \left\langle -\frac{3}{\sqrt{19}}, -\frac{1}{\sqrt{19}}, -\frac{3}{\sqrt{19}} \right\rangle \] Thus, the two unit vectors parallel to \( \overrightarrow{P Q} \) are: \[ \left\langle \frac{3}{\sqrt{19}}, \frac{1}{\sqrt{19}}, \frac{3}{\sqrt{19}} \right\rangle \text{ and } \left\langle -\frac{3}{\sqrt{19}}, -\frac{1}{\sqrt{19}}, -\frac{3}{\sqrt{19}} \right\rangle \]