Question
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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.)

Ask by Griffiths Turnbull. in the United States
Jan 22,2025

Real Tutor Solution

Tutor-Verified Answer

Answer

The two unit vectors parallel to \( \overrightarrow{P Q} \) are: \[ \left\langle \frac{3}{\sqrt{19}}, \frac{1}{\sqrt{19}}, \frac{3}{\sqrt{19}} \right\rangle \quad \text{and} \quad \left\langle -\frac{3}{\sqrt{19}}, -\frac{1}{\sqrt{19}}, -\frac{3}{\sqrt{19}} \right\rangle \]

Solution

**Problem c. Find two unit vectors parallel to \( \overrightarrow{P Q} \).** To find unit vectors parallel to \( \overrightarrow{P Q} \), we first need to determine the unit vector in the direction of \( \overrightarrow{P Q} \) and then consider its opposite direction. 1. **Calculate the Unit Vector in the Direction of \( \overrightarrow{P Q} \):** The vector \( \overrightarrow{P Q} \) is given by: \[ \overrightarrow{P Q} = \langle 6, 2, 6 \rangle \] The magnitude of \( \overrightarrow{P Q} \) is: \[ \| \overrightarrow{P Q} \| = 2\sqrt{19} \] The unit vector \( \mathbf{u} \) in the direction of \( \overrightarrow{P Q} \) is: \[ \mathbf{u} = \frac{\overrightarrow{P Q}}{\| \overrightarrow{P Q} \|} = \left\langle \frac{6}{2\sqrt{19}}, \frac{2}{2\sqrt{19}}, \frac{6}{2\sqrt{19}} \right\rangle = \left\langle \frac{3}{\sqrt{19}}, \frac{1}{\sqrt{19}}, \frac{3}{\sqrt{19}} \right\rangle \] Alternatively, expressed with radicals in the denominator rationalized: \[ \mathbf{u} = \left\langle \frac{3\sqrt{19}}{19}, \frac{\sqrt{19}}{19}, \frac{3\sqrt{19}}{19} \right\rangle = \frac{\sqrt{19}}{19} \langle 3, 1, 3 \rangle \] 2. **Determine the Two Unit Vectors Parallel to \( \overrightarrow{P Q} \):** - **First Unit Vector (Same Direction):** \[ \mathbf{u} = \left\langle \frac{3}{\sqrt{19}}, \frac{1}{\sqrt{19}}, \frac{3}{\sqrt{19}} \right\rangle \] - **Second Unit Vector (Opposite Direction):** \[ -\mathbf{u} = \left\langle -\frac{3}{\sqrt{19}}, -\frac{1}{\sqrt{19}}, -\frac{3}{\sqrt{19}} \right\rangle \] **Final Answer:** The two unit vectors parallel to \( \overrightarrow{P Q} \) are \[ \boxed{ \left\langle\, \dfrac{3}{\,\sqrt{19\,}},\;\dfrac{1}{\,\sqrt{19\,}},\;\dfrac{3}{\,\sqrt{19\,}}\,\right\rangle \quad \text{and} \quad \left\langle\, \dfrac{-3}{\,\sqrt{19\,}},\;\dfrac{-1}{\,\sqrt{19\,}},\;\dfrac{-3}{\,\sqrt{19\,}}\,\right\rangle } \] **Alternatively, expressed using the \( i, j, k \) notation:** \[ \boxed{ \left(\dfrac{3}{\,\sqrt{19\,}}\,i + \dfrac{1}{\,\sqrt{19\,}}\,j + \dfrac{3}{\,\sqrt{19\,}}\,k\right) \quad \text{and} \quad \left(-\dfrac{3}{\,\sqrt{19\,}}\,i - \dfrac{1}{\,\sqrt{19\,}}\,j - \dfrac{3}{\,\sqrt{19\,}}\,k\right) } \] These unit vectors have a magnitude of 1 and are parallel to \( \overrightarrow{P Q} \).

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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: \[ \left\langle \frac{3}{\sqrt{19}}, \frac{1}{\sqrt{19}}, \frac{3}{\sqrt{19}} \right\rangle \text{ and } \left\langle -\frac{3}{\sqrt{19}}, -\frac{1}{\sqrt{19}}, -\frac{3}{\sqrt{19}} \right\rangle \]

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