Precalculus by Richard Wright

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6-05 Dot Products

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 \(\overset{\rightharpoonup}{u} \cdot \overset{\rightharpoonup}{v} = u_x v_x + u_y v_y\).

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.

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.

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 \(\overset{\rightharpoonup}{w_2} = \overset{\rightharpoonup}{u} - \overset{\rightharpoonup}{w_1}\).

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.

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

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

Practice Exercises

1. Evaluate the dot product.
2. ⟨0, 3⟩ · ⟨−2, 5⟩
3. ⟨−2, 5⟩ · ⟨10, 0⟩
4. ⟨6, −3⟩ · ⟨5, 3⟩
5. 10⟨cos 30°, sin 30°⟩ · 12⟨cos 90°, sin 90°⟩
6. Find the angle between the vectors.
7. ⟨0, −4⟩ · ⟨1, 5⟩
8. ⟨1, 2⟩ · ⟨−2, 3⟩
9. ⟨−3, 4⟩ · ⟨4, −2⟩
10. Are vectors parallel, orthogonal, or neither?
11. ⟨2, 6⟩, ⟨−9, 3⟩
12. ⟨6, −3⟩, ⟨2, 4⟩
13. ⟨12, 3⟩, ⟨−4, −1⟩
14. Find orthogonal components of \(\overset{\rightharpoonup}{u}\) where one component is orthogonal to \(\overset{\rightharpoonup}{v}\).
15. \(\overset{\rightharpoonup}{u} = \langle 3, 2 \rangle, \overset{\rightharpoonup}{v} = \langle 8, 10 \rangle\)
16. \(\overset{\rightharpoonup}{u} = \langle -1, 3 \rangle, \overset{\rightharpoonup}{v} = \langle -4, 5 \rangle\)
17. \(\overset{\rightharpoonup}{u} = \langle -4, -7 \rangle, \overset{\rightharpoonup}{v} = \langle -10, -10 \rangle\)
18. Problem Solving: In physics, work is the dot product of force and distance. Calculate the work for each situation.
19. A mother applies 200 N of force at angle of 20° below horizontal while pushing a baby stroller 300 m along a horizontal path.
20. A student uses 18 lbs of force at an angle of 40° above the horizontal to pull a backpack 75 ft down the hall.
21. Mixed Review
22. (6-04) Add: (20 m at E 30° N) + (25 m E 60° S).
23. (6-04) Write \(\langle 8\sqrt{3}, 8 \rangle\) in trigonometric form.
24. (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
25. (6-02) In ∆ABC where A = 82°, b = 41, and c = 28, find a.
26. (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