Precalculus by Richard Wright

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For God has not given us a spirit of fear, but of power and of love and of a sound mind. 2 Timothy‬ ‭1‬:‭7‬ ‭NKJV‬‬

6-05 Dot Products

Mr. Wright teaches the lesson.

Summary: In this section, you will:

SDA NAD Content Standards (2018): PC.6.4

person pulling sled
Figure 1: Researcher doing work by pulling a sled up Mr. Denali. credit (nps.gov/Kent Miller)

In physics, work is found by multiplying force and the distance the force moves an object. However, the force and distance need to be in the same direction. The easiest way to calculate this is by describing both the force and distance as vectors and finding the dot product.

Dot Product

There are two ways to multiply vectors, the dot product and the cross product. The dot product of two vectors is sometimes called the scalar product because the result is a scalar number. The dot product of two vectors \(\overset{\rightharpoonup}{u} = \langle u_x, u_y \rangle\) and \(\overset{\rightharpoonup}{v} = \langle v_x, v_y \rangle\) is \(\overset{\rightharpoonup}{u} \cdot \overset{\rightharpoonup}{v} = u_x v_x + u_y v_y\).

Dot Product of Two Vectors

Let \(\overset{\rightharpoonup}{u} = \langle u_x, u_y \rangle\) and \(\overset{\rightharpoonup}{v} = \langle v_x, v_y \rangle\), then

$$\overset{\rightharpoonup}{u} \cdot \overset{\rightharpoonup}{v} = u_x v_x + u_y v_y$$

Properties of Dot Products

If \(\overset{\rightharpoonup}{u}\), \(\overset{\rightharpoonup}{v}\), and \(\overset{\rightharpoonup}{w}\) are vectors and a is a scalar, then

  1. \(\overset{\rightharpoonup}{u} \cdot \overset{\rightharpoonup}{v} = \overset{\rightharpoonup}{v} \cdot \overset{\rightharpoonup}{u}\)
  2. \(\overset{\rightharpoonup}{v} \cdot \overset{\rightharpoonup}{0} = 0\)
  3. \(\overset{\rightharpoonup}{u} \cdot \left(\overset{\rightharpoonup}{v} + \overset{\rightharpoonup}{w}\right) = \overset{\rightharpoonup}{u} \cdot \overset{\rightharpoonup}{v} + \overset{\rightharpoonup}{u} \cdot \overset{\rightharpoonup}{w}\)
  4. \(\overset{\rightharpoonup}{v} \cdot \overset{\rightharpoonup}{v} = \lVert \overset{\rightharpoonup}{v} \rVert^2\)
  5. \(a\left(\overset{\rightharpoonup}{u} \cdot \overset{\rightharpoonup}{v} \right) = a\overset{\rightharpoonup}{u} \cdot \overset{\rightharpoonup}{v} = \overset{\rightharpoonup}{u} \cdot c\overset{\rightharpoonup}{v}\)

Example 1: Evaluate a Dot Product

Find the dot product of \(\langle 2, -1 \rangle \cdot \langle 7, 3 \rangle\).

Solution

$$\overset{\rightharpoonup}{u} \cdot \overset{\rightharpoonup}{v} = u_x v_x + u_y v_y$$

$$\langle 2, -1 \rangle \cdot \langle 7, 3 \rangle = 2\left(7\right) + -1\left(3\right)$$

$$= 11$$

Try It 1

Find the dot product of \(\langle 2, 9 \rangle \cdot \langle 10, -1 \rangle\).

Answer

11

Angle Between Two Vectors

An alternate way of evaluating the dot product is \(\overset{\rightharpoonup}{u} \cdot \overset{\rightharpoonup}{v} = \lVert \overset{\rightharpoonup}{u} \rVert \lVert \overset{\rightharpoonup}{v} \rVert \cos θ\) where θ is the angle between the vectors. This can be used to find the dot product when the magnitudes and angle are known, or it can be used to find the angle between the vectors.

Angle Between Two Vectors

If \(\overset{\rightharpoonup}{u}\) and \(\overset{\rightharpoonup}{v}\) are vectors, then

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

Example 2: Find the Angle Between Two Vectors

Find the angle between \(\langle 2, -1 \rangle\) and \(\langle 7, 3 \rangle\).

Solution

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

$$\langle 2, -1 \rangle \cdot \langle 7, 3 \rangle = \lVert \langle 2, -1 \rangle\ \rVert \lVert \langle 7, 3 \rangle \rVert \cos θ$$

$$2\left(7\right) + -1\left(3\right) = \sqrt{2^2 + \left(-1\right)^2} \sqrt{7^2 + 3^2} \cos θ$$

$$11 = \sqrt{5} \sqrt{58} \cos θ$$

$$0.6459422415 = \cos θ$$

$$θ ≈ 49.8˚$$

Try It 2

Find the angle between \(\langle -3, 4 \rangle\) and \(\langle 12, -5 \rangle\).

Answer

149.5˚

Notice in this form, the dot product depends on cosine. If θ = 90˚, then cos θ = 0. Thus, if the dot product is zero, the vectors are orthogonal or at 90˚ to each other.

Orthogonal Vectors

If \(\overset{\rightharpoonup}{u} \cdot \overset{\rightharpoonup}{v} = 0\), then \(\overset{\rightharpoonup}{u}\) and \(\overset{\rightharpoonup}{v}\) are orthogonal.

Parallel Vectors

If \(\overset{\rightharpoonup}{u} = k\overset{\rightharpoonup}{v}\), then \(\overset{\rightharpoonup}{u}\) and \(\overset{\rightharpoonup}{v}\) are parallel.

Example 3: Orthogonal Vectors

Are the vectors \(\langle 3, 6 \rangle\) and \(\langle -4, 2 \rangle\) parallel, orthogonal, or neither?

Solution

$$\overset{\rightharpoonup}{u} \cdot \overset{\rightharpoonup}{v}$$

$$= \langle 3, 6 \rangle \cdot \langle -4, 2 \rangle$$

$$= 3\left(-4\right) + 6\left(2\right)$$

$$= 0$$

Because \(\overset{\rightharpoonup}{u} \cdot \overset{\rightharpoonup}{v} = 0\), the vectors are orthogonal.

Try It 3

Are the vectors \(\langle 12, -18 \rangle\) and \(\langle -8, 12 \rangle\) orthogonal, parallel, or neither?

Answer

Parallel

Find Orthogonal Vector Components

Sometimes it is useful to be able to write a vector in components that are not simply horizontal and vertical. Vector components are two orthogonal vectors that add to produce the original vector. An example is forces and work on an incline where it is most convenient to have one component parallel to the incline and the other component orthogonal to the incline.

To find orthogonal components of a vector \(\overset{\rightharpoonup}{u}\), you need to find two vectors \(\overset{\rightharpoonup}{w_1}\) and \(\overset{\rightharpoonup}{w_2}\) such that \(\overset{\rightharpoonup}{u} = \overset{\rightharpoonup}{w_1} + \overset{\rightharpoonup}{w_2}\) where \(\overset{\rightharpoonup}{w_1}\) is in the direction of \(\overset{\rightharpoonup}{v}\). \(\overset{\rightharpoonup}{w_1} = \text{proj}_\overset{\rightharpoonup}{v} \overset{\rightharpoonup}{u}\) and \(\overset{\rightharpoonup}{w_2} = \overset{\rightharpoonup}{u} - \overset{\rightharpoonup}{w_1}\).

Figure 2: \(\overset{\rightharpoonup}{w}_1\) is the projection \(\text{proj}_\overset{\rightharpoonup}{v} \overset{\rightharpoonup}{u}\). \(\overset{\rightharpoonup}{w}_1\) and \(\overset{\rightharpoonup}{w}_2\) are orthogonal components of \(\overset{\rightharpoonup}{u}\).
Find orthogonal Components of a Vector

Find the orthogonal components of \(\overset{\rightharpoonup}{u}\) where one component is in the direction of \(\overset{\rightharpoonup}{v}\).

  1. \(\overset{\rightharpoonup}{w_1} = \text{proj}_\overset{\rightharpoonup}{v} \overset{\rightharpoonup}{u}\) where \(\text{proj}_\overset{\rightharpoonup}{v} \overset{\rightharpoonup}{u} = \frac{\overset{\rightharpoonup}{u} \cdot \overset{\rightharpoonup}{v}}{\lVert \overset{\rightharpoonup}{v} \rVert^2} \overset{\rightharpoonup}{v}\)
  2. \(\overset{\rightharpoonup}{w_2} = \overset{\rightharpoonup}{u} - \overset{\rightharpoonup}{w_1}\)

The vector projection is \(\overset{\rightharpoonup}{u} \cdot \overset{\rightharpoonup}{v} = \lVert \overset{\rightharpoonup}{u} \rVert \lVert \overset{\rightharpoonup}{v} \rVert \cos θ\) solved for \(\lVert \overset{\rightharpoonup}{u} \rVert \cos θ = \frac{\overset{\rightharpoonup}{u} \cdot \overset{\rightharpoonup}{v}}{\lVert \overset{\rightharpoonup}{v} \rVert}\). This gives the length of \(\overset{\rightharpoonup}{u}\) which covers \(\overset{\rightharpoonup}{v}\). This is multiplied by the unit vector in the direction of \(\overset{\rightharpoonup}{v}\) which is \(\frac{\overset{\rightharpoonup}{v}}{\lVert \overset{\rightharpoonup}{v} \rVert}\) to give a direction to the length.

$$\text{proj}_\overset{\rightharpoonup}{v} \overset{\rightharpoonup}{u} = \frac{\overset{\rightharpoonup}{u} \cdot \overset{\rightharpoonup}{v}}{\lVert \overset{\rightharpoonup}{v} \rVert} \frac{\overset{\rightharpoonup}{v}}{\lVert \overset{\rightharpoonup}{v} \rVert}$$

$$\text{proj}_\overset{\rightharpoonup}{v} \overset{\rightharpoonup}{u} = \frac{\overset{\rightharpoonup}{u} \cdot \overset{\rightharpoonup}{v}}{\lVert \overset{\rightharpoonup}{v} \rVert^2} \overset{\rightharpoonup}{v}$$

Vector Projection

The projection of \(\overset{\rightharpoonup}{u}\) onto \(\overset{\rightharpoonup}{v}\) is

$$\text{proj}_\overset{\rightharpoonup}{v} \overset{\rightharpoonup}{u} = \frac{\overset{\rightharpoonup}{u} \cdot \overset{\rightharpoonup}{v}}{\lVert \overset{\rightharpoonup}{v} \rVert^2} \overset{\rightharpoonup}{v}$$

Example 4: Find orthogonal Vector Components

Find orthogonal components of \(\overset{\rightharpoonup}{u} = \langle 2, 3 \rangle \) where one component is orthogonal to \(\overset{\rightharpoonup}{v} = \langle 5, 4 \rangle \).

Solution

The first component is the projection of \(\overset{\rightharpoonup}{u}\) onto \(\overset{\rightharpoonup}{v}\).

$$\overset{\rightharpoonup}{w_1} = \text{proj}_\overset{\rightharpoonup}{v} \overset{\rightharpoonup}{u}$$

$$= \frac{\overset{\rightharpoonup}{u} \cdot \overset{\rightharpoonup}{v}}{\lVert \overset{\rightharpoonup}{v} \rVert^2} \overset{\rightharpoonup}{v}$$

$$= \frac{\langle 2, 3 \rangle \cdot \langle 5, 4 \rangle}{\lVert \langle 5, 4 \rangle \rVert^2} \langle 5, 4 \rangle$$

$$= \frac{2\left(5\right) + 3\left(4\right)}{\sqrt{5^2 + 4^2}^2} \langle 5, 4 \rangle$$

$$= \frac{22}{41} \langle 5, 4 \rangle$$

$$= \left\langle \frac{110}{41}, \frac{88}{41} \right\rangle$$

The second component is found by subtracting the \(\overset{\rightharpoonup}{w_1}\) from \(\overset{\rightharpoonup}{u}\).

$$\overset{\rightharpoonup}{w_2} = \overset{\rightharpoonup}{u} - \overset{\rightharpoonup}{w_1}$$

$$= \langle 2, 3 \rangle - \left\langle \frac{110}{41}, \frac{88}{41} \right\rangle$$

$$= \left\langle -\frac{28}{41}, \frac{35}{41} \right\rangle$$

Try It 4

Find orthogonal components of \(\overset{\rightharpoonup}{u} = \langle -4, 1 \rangle\) where one component is orthogonal to \(\overset{\rightharpoonup}{v} = \langle -5, 2 \rangle\).

Answer

\(\overset{\rightharpoonup}{w_1} = \left\langle -\frac{110}{29}, \frac{44}{29} \right\rangle\), \(\overset{\rightharpoonup}{w_2} = \left\langle -\frac{6}{29}, -\frac{15}{29} \right\rangle\)

Lesson Summary

Dot Product of Two Vectors

Let \(\overset{\rightharpoonup}{u} = \langle u_x, u_y \rangle\) and \(\overset{\rightharpoonup}{v} = \langle v_x, v_y \rangle\), then

$$\overset{\rightharpoonup}{u} \cdot \overset{\rightharpoonup}{v} = u_x v_x + u_y v_y$$


Properties of Dot Products

If \(\overset{\rightharpoonup}{u}\), \(\overset{\rightharpoonup}{v}\), and \(\overset{\rightharpoonup}{w}\) are vectors and a is a scalar, then

  1. \(\overset{\rightharpoonup}{u} \cdot \overset{\rightharpoonup}{v} = \overset{\rightharpoonup}{v} \cdot \overset{\rightharpoonup}{u}\)
  2. \(\overset{\rightharpoonup}{v} \cdot \overset{\rightharpoonup}{0} = 0\)
  3. \(\overset{\rightharpoonup}{u} \cdot \left(\overset{\rightharpoonup}{v} + \overset{\rightharpoonup}{w}\right) = \overset{\rightharpoonup}{u} \cdot \overset{\rightharpoonup}{v} + \overset{\rightharpoonup}{u} \cdot \overset{\rightharpoonup}{w}\)
  4. \(\overset{\rightharpoonup}{v} \cdot \overset{\rightharpoonup}{v} = \lVert \overset{\rightharpoonup}{v} \rVert^2\)
  5. \(a\left(\overset{\rightharpoonup}{u} \cdot \overset{\rightharpoonup}{v} \right) = a\overset{\rightharpoonup}{u} \cdot \overset{\rightharpoonup}{v} = \overset{\rightharpoonup}{u} \cdot c\overset{\rightharpoonup}{v}\)

Angle Between Two Vectors

If \(\overset{\rightharpoonup}{u}\) and \(\overset{\rightharpoonup}{v}\) are vectors, then

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


Orthogonal Vectors

If \(\overset{\rightharpoonup}{u} \cdot \overset{\rightharpoonup}{v} = 0\), then \(\overset{\rightharpoonup}{u}\) and \(\overset{\rightharpoonup}{v}\) are orthogonal.


Parallel Vectors

If \(\overset{\rightharpoonup}{u} = k\overset{\rightharpoonup}{v}\), then \(\overset{\rightharpoonup}{u}\) and \(\overset{\rightharpoonup}{v}\) are parallel.


Find orthogonal Components of a Vector

Find the orthogonal components of \(\overset{\rightharpoonup}{u}\) where one component is in the direction of \(\overset{\rightharpoonup}{v}\).

  1. \(\overset{\rightharpoonup}{w_1} = \text{proj}_\overset{\rightharpoonup}{v} \overset{\rightharpoonup}{u}\) where \(\text{proj}_\overset{\rightharpoonup}{v} \overset{\rightharpoonup}{u} = \frac{\overset{\rightharpoonup}{u} \cdot \overset{\rightharpoonup}{v}}{\lVert \overset{\rightharpoonup}{v} \rVert^2} \overset{\rightharpoonup}{v}\)
  2. \(\overset{\rightharpoonup}{w_2} = \overset{\rightharpoonup}{u} - \overset{\rightharpoonup}{w_1}\)

Vector Projection

The projection of \(\overset{\rightharpoonup}{u}\) onto \(\overset{\rightharpoonup}{v}\) is

$$\text{proj}_\overset{\rightharpoonup}{v} \overset{\rightharpoonup}{u} = \frac{\overset{\rightharpoonup}{u} \cdot \overset{\rightharpoonup}{v}}{\lVert \overset{\rightharpoonup}{v} \rVert^2} \overset{\rightharpoonup}{v}$$

Helpful videos about this lesson.

Practice Exercises

    Evaluate the dot product.

  1. \(\langle 0, 3 \rangle \cdot \langle -2, 5 \rangle\)
  2. \(\langle -2, 5 \rangle \cdot \langle 10, 0 \rangle\)
  3. \(\langle 6, -3 \rangle \cdot \langle 5, 3 \rangle\)
  4. \(10\langle \cos 30˚, \sin 30˚ \rangle \cdot 12\langle \cos 90˚, \sin 90˚ \rangle\)
  5. Find the angle between the vectors.

  6. \(\langle 0, -4 \rangle \cdot \langle 1, 5 \rangle\)
  7. \(\langle 1, 2 \rangle \cdot \langle -2, 3 \rangle\)
  8. \(\langle -3, 4 \rangle \cdot \langle 4, -2 \rangle\)
  9. Are vectors parallel, orthogonal, or neither?

  10. \(\langle 2, 6 \rangle, \langle -9, 3 \rangle\)
  11. \(\langle 6, -3 \rangle, \langle 2, 4 \rangle\)
  12. \(\langle 12, 3 \rangle, \langle -4, -1 \rangle\)
  13. Find orthogonal components of \(\overset{\rightharpoonup}{u}\) where one component is orthogonal to \(\overset{\rightharpoonup}{v}\).

  14. \(\overset{\rightharpoonup}{u} = \langle 3, 2 \rangle, \overset{\rightharpoonup}{v} = \langle 8, 10 \rangle\)
  15. \(\overset{\rightharpoonup}{u} = \langle -1, 3 \rangle, \overset{\rightharpoonup}{v} = \langle -4, 5 \rangle\)
  16. \(\overset{\rightharpoonup}{u} = \langle -4, -7 \rangle, \overset{\rightharpoonup}{v} = \langle -10, -10 \rangle\)
  17. Problem Solving: In physics, work is the dot product of force and distance. Calculate the work for each situation.

  18. A mother applies 200 N of force at angle of 20˚ below horizontal while pushing a baby stroller 300 m along a horizontal path.
  19. A student uses 18 lbs of force at an angle of 40˚ above the horizontal to pull a backpack 75 ft down the hall.
  20. Mixed Review

  21. (6-04) Add: (20 m at E 30˚ N) + (25 m E 60˚ S).
  22. (6-04) Write \(\langle 8\sqrt{3}, 8 \rangle\) in trigonometric form.
  23. (6-03) Given \(\overset{\rightharpoonup}{u} = 2\hat{i} + \hat{j}\) and \(\overset{\rightharpoonup}{v} = -3\hat{i} - 2\hat{j}\), evaluate the \(-\overset{\rightharpoonup}{u} + \overset{\rightharpoonup}{v}\) both (a) graphically and (b) algebraically
  24. (6-02) In ∆ABC where A = 82°, b = 41, and c = 28, find a.
  25. (6-01) In ∆RST where R = 80°, r = 5, s = 7, find S.

Answers

  1. 15
  2. −20
  3. 21
  4. 60
  5. 168.7°
  6. 60.3°
  7. 153.4°
  8. orthogonal
  9. orthogonal
  10. parallel
  11. \(\overset{\rightharpoonup}{w_1} = \left\langle \frac{88}{41}, \frac{110}{41} \right\rangle\), \(\overset{\rightharpoonup}{w_2} = \left\langle \frac{35}{41}, -\frac{28}{41} \right\rangle\)
  12. \(\overset{\rightharpoonup}{w_1} = \left\langle -\frac{76}{41}, \frac{95}{41} \right\rangle\), \(\overset{\rightharpoonup}{w_2} = \left\langle \frac{35}{41}, \frac{28}{41} \right\rangle\)
  13. \(\overset{\rightharpoonup}{w_1} = \left\langle -\frac{11}{2}, -\frac{11}{2} \right\rangle\), \(\overset{\rightharpoonup}{w_2} = \left\langle \frac{3}{2}, -\frac{3}{2} \right\rangle\)
  14. 56382 J
  15. 1034 ft·lbs
  16. 32.0 m at E 21.3° S
  17. \(16\langle \cos \frac{π}{6}, \sin \frac{π}{6} \rangle\)
  18. ; \(-5\hat{i} - 3\hat{j}\)
  19. 46.32
  20. No triangle