Precalculus by Richard Wright

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“He himself bore our sins” in his body on the cross, so that we might die to sins and live for righteousness; “by his wounds you have been healed.” 1 Peter‬ ‭2‬:‭24‬ ‭NIV‬‬‬

# 11-02 Vectors in space

Summary: In this section, you will:

• Use vector operations in three dimensions.
• Find the angle between vectors.

SDA NAD Content Standards (2018): PC.5.3, PC.6.4

In physics, some measurements have direction as an important part of the measurement, such as forces. It is important which direction a force is applied. Measurements with direction are called vectors, and forces are vectors. To solve force problems, the forces acting on an object are drawn on a freebody diagram as in figure 1. The force vectors are then added together. This lesson will explore how to do vector operations such as vector addition.

Vectors in two dimensions are written in component form like

$$\overset{\rightharpoonup}{v} = \langle v_1, v_2 \rangle$$

Vectors in three dimensions are written similarly with z included.

$$\overset{\rightharpoonup}{v} = \langle v_1, v_2, v_3 \rangle$$

To find the component form of a vector from the initial point (p1, p2, p3) to the terminal point (q1, q2, q3), subtract the corresponding coordinates. This will give the distance in each component direction.

$$\overset{\rightharpoonup}{v} = \langle q_1 - p_1, q_2 - p_2, q_3 - p_3 \rangle$$

###### Vectors in 3-D

Component form: $$\overset{\rightharpoonup}{v} = \langle v_1, v_2, v_3 \rangle$$

Find component form from initial point, P to terminal point Q: $$\overset{\rightharpoonup}{v} = \langle q_1 - p_1, q_2 - p_2, q_3 - p_3 \rangle$$

## Vector Operations

Vector operations in three dimensions are essentially the same as in two dimension, just include z. See lesson 6-03 for indepth explanations.

###### Vector Operations

Let $$\overset{\rightharpoonup}{v} = \langle v_1, v_2, v_3 \rangle$$ and $$\overset{\rightharpoonup}{u} = \langle u_1, u_2, u_3 \rangle$$.

$$\overset{\rightharpoonup}{v} + \overset{\rightharpoonup}{u} = \langle v_1 + u_1, v_2 + u_2, v_3 + u_3 \rangle$$

Scalar Multiplication: Distribute the scalar to all the components.

$$c\overset{\rightharpoonup}{v} = \langle cv_1, cv_2, cv_3 \rangle$$

Dot Product: Add products of corresponding elements.

$$\overset{\rightharpoonup}{v} \cdot \overset{\rightharpoonup}{u} = v_1 u_1 + v_2 u_2 + v_3 u_3$$

Magnitude: Distance between the initial and terminal points.

$$\| \overset{\rightharpoonup}{v} \| = \sqrt{v_1^2 + v_2^2 + v_3^2}$$

Unit Vector: Divide vector by its magnitude to get a vector in direction of $$\overset{\rightharpoonup}{v}$$ but length 1.

$$\frac{\overset{\rightharpoonup}{v}}{\| \overset{\rightharpoonup}{v} \|}$$

The dot product can be used to find the angle between vectors. Dot product can be found by adding the products of corresponding components

$$\overset{\rightharpoonup}{v} \cdot \overset{\rightharpoonup}{u} = v_1 u_1 + v_2 u_2 + v_3 u_3$$

The dot product can also be found by multiplying the magnitudes of the vectors with the cosine of the angle between the vectors.

$$\overset{\rightharpoonup}{v} \cdot \overset{\rightharpoonup}{u} = \| \overset{\rightharpoonup}{v} \| \| \overset{\rightharpoonup}{u} \| \cos θ$$

So, if the dot product and magnitudes are known, or can be calculated, then the angle between the vectors can be found. The angle will always be between 0 and π radians (0 and 180˚). Because cosine is zero at $$\frac{π}{2}$$, the dot product equals zero if the vectors are orthogonal (perpendicular).

###### Angle Between Vectors

$$\overset{\rightharpoonup}{v} \cdot \overset{\rightharpoonup}{u} = \| \overset{\rightharpoonup}{v} \| \| \overset{\rightharpoonup}{u} \| \cos θ$$

If $$\overset{\rightharpoonup}{v} \cdot \overset{\rightharpoonup}{u} = 0$$, then the vectors are orthogonal, or perpendicular.

If $$\overset{\rightharpoonup}{v} = c\overset{\rightharpoonup}{u}$$, then the vectors are parallel.

#### Example 1: Vector Operations

Let $$\overset{\rightharpoonup}{m} = ⟨1, 0, 3⟩$$ and $$\overset{\rightharpoonup}{n} = ⟨−2, 1, −4⟩$$. (a) Find $$\| \overset{\rightharpoonup}{m} \|$$, (b) find the unit vector in the direction of $$\overset{\rightharpoonup}{m}$$, and (c) find $$\overset{\rightharpoonup}{m} + 2\overset{\rightharpoonup}{n}$$.

###### Solutions
1. Use the distance formula with the components to find the magnitude.

$$\| \overset{\rightharpoonup}{m} \| = \sqrt{𝑚_1^2 + 𝑚_2^2 + 𝑚_3^2}$$

$$\qquad = \sqrt{1^2 + 0^2 + 3^2}$$

$$\qquad = \sqrt{10}$$

2. To find a unit vector, divide the vector by its magnitude. The magnitude was evaluated in part a.

$$\frac{\overset{\rightharpoonup}{m}}{\| \overset{\rightharpoonup}{m} \|} = \frac{⟨1, 0, 3⟩}{\sqrt{10}}$$

$$\qquad = \left\langle \frac{1}{\sqrt{10}}, \frac{0}{\sqrt{10}}, \frac{3}{\sqrt{10}} \right\rangle$$

$$\qquad = \left\langle \frac{\sqrt{10}}{10}, 0, \frac{3\sqrt{10}}{10}\right\rangle$$

3. Fill in the component form for the vectors first.

$$\overset{\rightharpoonup}{m} + 2\overset{\rightharpoonup}{n}$$

$$\langle 1, 0, 3 \rangle + 2 \langle −2, 1, −4 \rangle$$

Follow the order of operations, so do the scalar multiplication by distributing the 2.

$$\langle 1, 0, 3 \rangle + \langle −4, 2, −8 \rangle$$

$$\langle −3, 2, −5\rangle$$

##### Try It 1

Let $$\overset{\rightharpoonup}{a} = ⟨-2, 4, 1⟩$$ and $$\overset{\rightharpoonup}{b} = ⟨−1, 2, 0⟩$$. (a) Find $$\| \overset{\rightharpoonup}{b} \|$$, (b) find the unit vector in the direction of $$\overset{\rightharpoonup}{b}$$, and (c) find $$-2\overset{\rightharpoonup}{a} + \overset{\rightharpoonup}{b}$$.

$$\sqrt{5}$$; $$\left\langle -\frac{1}{5}, \frac{2}{5}, 0\right\rangle$$

#### Example 2: Dot Products and Angles Between Vectors

Let $$\overset{\rightharpoonup}{m} = ⟨1, 0, 3⟩$$ and $$\overset{\rightharpoonup}{n} = ⟨−2, 1, −4⟩$$. (a) Find $$\overset{\rightharpoonup}{𝑚}⋅\overset{\rightharpoonup}{n}$$ and (b) find the angle between $$\overset{\rightharpoonup}{m}$$ and $$\overset{\rightharpoonup}{n}$$

###### Solution
1. Multiply the corresponding components and then add the products.

$$\overset{\rightharpoonup}{v} \cdot \overset{\rightharpoonup}{u} = v_1 u_1 + v_2 u_2 + v_3 u_3$$

$$⟨1, 0, 3⟩⋅⟨−2, 1, −4⟩$$

$$1(−2) + 0(1) + 3(−4)$$

–14

2. The dot product was found in part a. Use that with the other dot product formula to find the angle.

$$\overset{\rightharpoonup}{m}⋅\overset{\rightharpoonup}{n} = \| \overset{\rightharpoonup}{m} \| \| \overset{\rightharpoonup}{n} \| \cos θ$$

$$−14 = \sqrt{1^2+0^2+3^2} \sqrt{(−2)^2+1^2+(−4)^2} \cos θ$$

$$−14 = \sqrt{10} \sqrt{21} \cos θ$$

$$\frac{−14}{\sqrt{10} \sqrt{21}} = \cos θ$$

$$θ ≈ 165.0°$$

##### Try It 2

Let $$\overset{\rightharpoonup}{a} = ⟨-2, 4, 1⟩$$ and $$\overset{\rightharpoonup}{b} = ⟨−1, 2, 0⟩$$. (a) Find $$\overset{\rightharpoonup}{a}·\overset{\rightharpoonup}{b}$$ and (b) find the angle between $$\overset{\rightharpoonup}{a}$$ and $$\overset{\rightharpoonup}{b}$$

10; 12.6°

#### Example 3: Parallel or Perpendicular Vectors

Are $$\overset{\rightharpoonup}{p} = ⟨1, 5, −2⟩$$ and $$\overset{\rightharpoonup}{q} = \left\langle −\frac{1}{5}, −1, \frac{2}{5} \right\rangle$$ parallel, orthogonal, or neither?

###### Solution

Check to see is they are orthogonal. Vectors are orthogonal if $$\overset{\rightharpoonup}{p} ⋅ \overset{\rightharpoonup}{q} = 0$$

$$⟨1, 5, −2⟩⋅ \left\langle −\frac{1}{5}, −1, \frac{2}{5} \right\rangle$$

$$1\left(−\frac{1}{5}\right) + 5(−1) + (−2)\left(\frac{2}{5}\right)$$

$$−\frac{1}{5} − 5 − \frac{4}{5} = −6$$

This is not 0, so the vectors are not orthogonal.

The vectors are parallel if $$\overset{\rightharpoonup}{p} = c \overset{\rightharpoonup}{q}$$.

$$⟨1, 5, −2⟩ = c \left\langle −\frac{1}{5}, −1, \frac{2}{5} \right\rangle$$

Check the x-components.

$$1 = c\left(−\frac{1}{5}\right)$$

$$c = −5$$

Check the y-components.

$$5 = c(−1)$$

$$c = −5$$

Check the z-components.

$$−2 = c\left(\frac{2}{5}\right)$$

$$c = −5$$

Notice that c is always the same, so they are parallel.

##### Try It 3

Are $$\overset{\rightharpoonup}{p} = ⟨1, 0, −2⟩$$ and $$\overset{\rightharpoonup}{q} = \left\langle 2, 5, 1 \right\rangle$$ parallel, orthogonal, or neither?

orthogonal

#### Example 4: Determine If Points Are Collinear

Are P(1, −1, 3), Q(0, 4, −2), and R(6, 13, −5) collinear?

###### Solution

If two vectors from the same point are parallel or antiparallel, then they would go in the same or opposite directions. Antiparallel means the vectors go in opposite directions. Either way, they would form a "line". So start by finding $$\overset{\rightharpoonup}{PQ}$$ and $$\overset{\rightharpoonup}{QR}$$ and see if they are parallel.

Find the vector by subtraction corresponding elements from the points. Terminal point - initial point.

$$\overset{\rightharpoonup}{PQ} = ⟨0−1, 4−(−1), −2−3⟩ = ⟨−1, 5, −5⟩$$

$$\overset{\rightharpoonup}{QR} = ⟨6−0, 13−4, −5−(−2)⟩ = ⟨6, 9, −3⟩$$

They would be parallel or antiparallel if $$\overset{\rightharpoonup}{PQ} = c \overset{\rightharpoonup}{QR}$$.

$$\langle -1, 5, -5 \rangle = c \langle 6, 9, -3 \rangle$$

Check the x-components.

$$-1 = c(6)$$

$$-\frac{1}{6} = c$$

Check the y-components.

$$5 = c(9)$$

$$\frac{5}{9} = c$$

The c is not the same, so the vectors are not parallel or antiparallel. So, they are not going same direction. Therefore, the vectors are not collinear.

##### Try It 4

Are A(2, −1, 4), B(1, 4, 3), and C(3, −15, 3) collinear?

Yes, they are collinear.

##### Lesson Summary

###### Vectors in 3-D

Component form: $$\overset{\rightharpoonup}{v} = \langle v_1, v_2, v_3 \rangle$$

Find component form from initial point, P to terminal point Q: $$\overset{\rightharpoonup}{v} = \langle q_1 - p_1, q_2 - p_2, q_3 - p_3 \rangle$$

###### Vector Operations

Let $$\overset{\rightharpoonup}{v} = \langle v_1, v_2, v_3 \rangle$$ and $$\overset{\rightharpoonup}{u} = \langle u_1, u_2, u_3 \rangle$$.

$$\overset{\rightharpoonup}{v} + \overset{\rightharpoonup}{u} = \langle v_1 + u_1, v_2 + u_2, v_3 + u_3 \rangle$$

Scalar Multiplication: Distribute the scalar to all the components.

$$c\overset{\rightharpoonup}{v} = \langle cv_1, cv_2, cv_3 \rangle$$

Dot Product: Add products of corresponding elements.

$$\overset{\rightharpoonup}{v} \cdot \overset{\rightharpoonup}{u} = v_1 u_1 + v_2 u_2 + v_3 u_3$$

Magnitude: Distance between the initial and terminal points.

$$\| \overset{\rightharpoonup}{v} \| = \sqrt{v_1^2 + v_2^2 + v_3^2}$$

Unit Vector: Divide vector by its magnitude to get a vector in direction of $$\overset{\rightharpoonup}{v}$$ but length 1.

$$\frac{\overset{\rightharpoonup}{v}}{\| \overset{\rightharpoonup}{v} \|}$$

###### Angle Between Vectors

$$\overset{\rightharpoonup}{v} \cdot \overset{\rightharpoonup}{u} = \| \overset{\rightharpoonup}{v} \| \| \overset{\rightharpoonup}{u} \| \cos θ$$

If $$\overset{\rightharpoonup}{v} \cdot \overset{\rightharpoonup}{u} = 0$$, then the vectors are orthogonal, or perpendicular.

If $$\overset{\rightharpoonup}{v} = c\overset{\rightharpoonup}{u}$$, then the vectors are parallel.

## Practice Exercises

Let $$\overset{\rightharpoonup}{r} = \langle 2, 4, -1 \rangle$$, $$\overset{\rightharpoonup}{s} = \langle -3, 0, -2 \rangle$$, and $$\overset{\rightharpoonup}{t} = \langle 1, -5, 4 \rangle$$. Evaluate the following.

1. $$\overset{\rightharpoonup}{r} + \overset{\rightharpoonup}{s}$$
2. $$\overset{\rightharpoonup}{t} - \overset{\rightharpoonup}{r}$$
3. $$3\overset{\rightharpoonup}{r}$$
4. $$2\overset{\rightharpoonup}{s} + 3\overset{\rightharpoonup}{t} - \overset{\rightharpoonup}{r}$$
5. $$\| \overset{\rightharpoonup}{r} \|$$
6. $$\| \overset{\rightharpoonup}{t} \|$$
7. Unit vector in direction of $$\overset{\rightharpoonup}{r}$$
8. Unit vector in direction of $$\overset{\rightharpoonup}{t}$$
9. $$\overset{\rightharpoonup}{r} \cdot \overset{\rightharpoonup}{s}$$
10. $$\overset{\rightharpoonup}{t} \cdot \overset{\rightharpoonup}{r}$$
11. Find the angle between $$\overset{\rightharpoonup}{r}$$ and $$\overset{\rightharpoonup}{s}$$
12. Find the angle between $$\overset{\rightharpoonup}{t}$$ and $$\overset{\rightharpoonup}{s}$$
13. Determine if the vectors are parallel, perpendicular, or neither.

14. $$\left\langle 3, -1, \frac{1}{2}\right\rangle$$ and $$\left\langle -2, -3, 6\right\rangle$$
15. $$\left\langle 36, -1, \frac{1}{2}\right\rangle$$ and $$\left\langle 12, -\frac{1}{3}, \frac{1}{6}\right\rangle$$
16. Determine whether the points G(2, −1, 3), H(4, 1, 5), and J(8, 5, 9) are collinear.
17. Mixed Review

18. (11-01) Find the distance between (3, 1, −2) and (−1, 0, 0).
19. (11-01) Find the equation of the sphere with center (1, 3, −1) and radius of 5.
20. (10-06) Use the binomial theorem to expand (x – 2y)4.
21. (9-05) Find the determinant of $$\left[\begin{matrix} 2 & 1 & 0 \\ -2 & 3 & -1 \\ 0 & 4 & -3 \end{matrix}\right]$$.
22. (7-09) Find the polar equation of a parabola with focus at the origin and directrix y = −4.

1. $$\langle -1, 4, -3 \rangle$$
2. $$\langle -1, -9, 5\rangle$$
3. $$\langle 6, 12, -3 \rangle$$
4. $$\langle -5, -19, 9 \rangle$$
5. $$\sqrt{21}$$
6. $$\sqrt{42}$$
7. $$\left\langle \frac{2\sqrt{21}}{21}, \frac{4\sqrt{21}}{21}, -\frac{\sqrt{21}}{21}\right\rangle$$
8. $$\left\langle \frac{\sqrt{42}}{42}, -\frac{5\sqrt{42}}{42}, \frac{2\sqrt{42}}{21}\right\rangle$$
9. −4
10. −22
11. 104˚
12. 118˚
13. Perpendicular
14. Parallel
15. Collinear
16. $$\sqrt{21}$$
17. $$(x - 1)^2 + (y - 3)^2 + (z + 1)^2 = 25$$
18. x4 – 8x3y + 24x2y2 – 32xy3 + 16y4
19. −16
20. $$r = \frac{4}{1 - \sin θ}$$