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6.8. The Dot Product

There is an important operation on vectors, known as the dot product. The * dot product* of two vectors

**u**= (u

_{x}, u

_{y}) and

**v**= (v

_{x}, v

_{y}) is a scalar given by

**(6.8.1) u · v =**u_{x}v_{x}+ u_{y}v_{y}

**Example 1**: Compute the dot product of the following pairs of vectors:

- (a)
**u**= (2, 5),**v**= (3, 7) - (b)
**u**= (3, 4),**v**= (2, -3) - (c)
**u**= (2, 5),**v**= (-10, 4)

**Solution:**

- (a) We have
**u · v =**(2)(3) + (5)(7) = 6 + 35 = 41. - (b) We have
**u · v =**(3)(2) + (4)(-3) = 6 - 12 = -6. - (c) We have
**u · v =**(2)(-10) + (5)(4) = -20 + 20 = 0.

Note that the dot product of two vectors may be positive, negative, or zero.

It is easy to see from (6.8.1) that the dot product is * bilinear*, namely, that both of the following equations hold for arbitrary vectors u, v, and w and arbitrary scalars a and b. We leave the proofs as an exercise.

**(6.8.2a)**(a**u**+ b**v**)**·****w**= a**u · w**+ b**v · w****(6.8.2b) u ·**(a**v**+ b**w**) = a**u · v**+ b**u · w**

The dot product has several important geometric applications. Not that by Equations (6.7.6) and (6.8.1), the magnitude of a vector **v** is equal to the square root of **v · v**. Specifically we have

**(6.8.3)**|**v**| = √**v · v**

Perhaps the most important geometric application of the dot product is the following equation, which we state without proof:

**(6.8.4) u · v =**|**u**| |**v**| cos θ

where θ is the angle formed by the vectors **u** and **v** when they are brought tail to tail, as the following diagram illustrates:

It should be noted that Equation (6.8.4) has a nice geometric interpretation. Consider the **projection****v**_{proj} of **v** onto **u**, as shown in the following figure.

Figure 6.8.5: Vector Projection

It is easy to see from the geometry of the figure that **v**_{proj} has magnitude given by

**(6.8.6)**|**v**_{proj}| = |**v**| |cos θ|.

Therefore, Equation (6.8.4) becomes

**(6.8.7) u · v =**±|**u**| |**v**_{proj}|

the minus sign being used only when cos θ is negative, i.e. when θ is obtuse. In other words, the absolute value of the dot product of the vectors u and v is equal to the product of the magnitudes of **u** and **v**_{proj}. Now clearly **v**_{proj} is zero if and only if the vectors **u** and **v** are perperndicular, whence the dot product of two nonzero vectors is zero if and only if the vectors are perperndicular. (Perpendicular vectors are also said to be * orthogonal*.) It is also easy to see this fact directly from Equation (6.8.4).

Perhaps the most important application of Equation (6.8.4) is in determining the angle between two vectors. Solving (6.8.4) for θ, we obtain

- (6.8.8) θ = arccos(
**u · v /**(|**u**| |**v**|))

**Example 1: **Estimate the angle formed by the vectors **u **= (2, 1) and **v ** = (3, 4).

**Solution: ** We have **u · v = **(2)(3) + (1)(4) = 6 + 4 = 10. We also have |**u**| = √2^{2} + 1^{2} = √5 and |**v**| = √3^{2} + 4^{2} = 5 . Thus, by (6.8.8) we have cos θ = 10/(5√5) = 2/√5, whence θ = arccos(2/√5) ≈ 26.6°.

Equation (6.8.8) may also be used to determine the angles of a triangle with given vertices, as the following example shows:

**Example 2:** Estimate the angles of a triangle with vertices at (-4, -2), (1, 3), and (4, -1).

**Solution: **To solve this problem, we construct the vectors u, v, and w as shown below:

From the figure, we see that A = arccos(**u · v / ** (|**u**| |**v**|)), B = arccos((-**v**) **·** (-**w**)** / ** (|**-v**| |**-w**|)) = arccos(**v · w / ** (|**v**| |**w**|)), and C = arccos((-**u**) **·** **w** ** / ** (|**-u**| |**w**|)) = 180° - arccos(**u · w / ** (|**u**| |**w**|)). First we compute the following:

**u** = (4 - (-4), (-1) - (-2)) = (8, 1)

**v** = (1 - (-4), 3 - (-2)) = (5, 5)

**w** = (1 - 4, 3 - (-1)) = (-3, 4)

**u · v **= (8)(5) + (1)(5) = 40 + 5 = 45

**u · w **= (8)(-3) + (1)(4) = -24 + 4 = -20

**v · w **= (5)(-3) + (5)(4) = -15 + 20 = 5

|**u**| = √8^{2} + 1^{2} = √65

|**v**| = √5^{2} + 5^{2} = √50 = 5√2

|**w**| = √(-3)^{2} + 4^{2} = 5

From the above relatioins we then compute

A = arccos(**u · v / ** (|**u**| |**v**|)) = arccos(45/(5√130)) = arccos(9/√130) ≈ 37.9°.

B = arccos(**v · w / ** (|**v**| |**w**|)) = arccos(5/(25√2)) = arccos(1/(5√2)) ≈ 81.9°.

C = 180° - arccos(**u · w / ** (|**u**| |**w**|)) - 180° - arccos((-20 / (5√65)) = 180° - arccos(-4 / √65) ≈ 60.3°.

Finally we check that these angles add up to 180°, as they should. The sum of the angles given is 180.1°, the difference most likely being due to rounding error.

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