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13.3 Dot Product

multiply scalars: \( 3 \cdot 5 = 15 \rightarrow a \cdot b = ab \)

multiply vectors are more complicated:

  • dot product (today)
  • cross product (next time)

Dot product:

if \( \vec{u} = \langle a, b \rangle \) and \( \vec{v} = \langle c, d \rangle \)

then dot product of \( \vec{u} \) and \( \vec{v} \) is \( \vec{u} \cdot \vec{v} = ac + bd \)

example: \( \vec{u} = \langle 1, 2 \rangle \), \( \vec{v} = \langle 3, 4 \rangle \)

\[ \vec{u} \cdot \vec{v} = 1 \cdot 3 + 2 \cdot 4 = 11 \]

example: \( \vec{u} = \langle 1, 2, 3 \rangle \), \( \vec{v} = \langle 4, 5, 7 \rangle \)

\[ \vec{u} \cdot \vec{v} = 1 \cdot 4 + 2 \cdot 5 + 3 \cdot 7 = 35 \]

note the dot product is a scalar

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\( \vec{u} = \langle 1, 2 \rangle \)

\[ \begin{aligned} \vec{u} \cdot \vec{u} &= \langle 1, 2 \rangle \cdot \langle 1, 2 \rangle = 1^2 + 2^2 \\ &= (|\vec{u}|)^2 \end{aligned} \]

square of magnitude

recall \( |\vec{u}| = |\langle 1, 2 \rangle| \)

\( = \sqrt{1^2 + 2^2} \)

\( \vec{u} = \langle 1, 2 \rangle \quad \vec{v} = \langle 3, 4 \rangle \)

\[ \vec{v} \cdot \vec{u} = 1 \cdot 3 + 2 \cdot 4 = 11 = \vec{u} \cdot \vec{v} \]

in general, \( \vec{u} \cdot \vec{v} = \vec{v} \cdot \vec{u} \)

order doesn't matter

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The Geometric Definition of Dot Product

Let \(\theta\) be the included angle (angle between vectors \(\vec{u}\) and \(\vec{v}\)).

\[ \vec{u} \cdot \vec{v} = |\vec{u}| |\vec{v}| \cos \theta \]

Diagram showing two vectors u and v with an angle theta between them.

Example

Find the angle between \(\vec{u} = \langle 1, 2, -2 \rangle\) and \(\vec{v} = \langle 6, 0, -8 \rangle\).

\[ \vec{u} \cdot \vec{v} = |\vec{u}| |\vec{v}| \cos \theta \]

\[ \langle 1, 2, -2 \rangle \cdot \langle 6, 0, -8 \rangle = \sqrt{1^2 + 2^2 + (-2)^2} \sqrt{6^2 + 0^2 + (-8)^2} \cos \theta \]

\[ 6 + 0 + 16 = 3 \cdot 10 \cdot \cos \theta \]

\[ 22 = 30 \cos \theta \]

\[ \cos \theta = \frac{22}{30} \]

\[ \theta = \cos^{-1}\left(\frac{22}{30}\right) \approx 43^\circ \]

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Example

What is the angle between \(\vec{a} = \langle 3, 1, 5 \rangle\) and the positive x-axis?

3D coordinate axes x, y, z with vector a = <3,1,5> and angle theta marked from the x-axis.

Pick any vector along the positive x-axis to be the second vector, then find the angle between them.

Let \(\vec{b} = \langle 1, 0, 0 \rangle\) (but any \(\langle k, 0, 0 \rangle\) works, as \(k\) does not affect the angle between the vectors).

\[ \vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta \]

\[ \langle 3, 1, 5 \rangle \cdot \langle 1, 0, 0 \rangle = \sqrt{35} \sqrt{1} \cos \theta \]

\[ 3 = \sqrt{35} \cos \theta \]

\[ \cos \theta = \frac{3}{\sqrt{35}} \quad \text{direction cosine} \]

\[ \theta \approx 60^\circ \quad \text{direction angle} \]

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Vector Projections

The dot product can give us the projection of one vector onto another.

Diagram showing vector b projected onto vector a, forming a right triangle with angle theta.

"shadow" of \( \vec{b} \) on \( \vec{a} \)

The length of this "shadow" is called the scalar projection of \( \vec{b} \) onto \( \vec{a} \).

Written as \( \text{scal}_{\vec{a}} \vec{b} \)

Right triangle with hypotenuse magnitude of vector b, base scalar projection of b onto a, and angle theta.

From basic trig,

\[ \cos \theta = \frac{\text{scal}_{\vec{a}} \vec{b}}{|\vec{b}|} \]
\[ \text{scal}_{\vec{a}} \vec{b} = |\vec{b}| \cos \theta \]
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From \( \vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta \) we get \( \cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|} \)

Replace the \( \cos \theta \) on bottom of last page:

\[ \begin{aligned} \text{scal}_{\vec{a}} \vec{b} &= |\vec{b}| \cos \theta \\ &= |\vec{b}| \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|} \end{aligned} \]
\[ \text{scal}_{\vec{a}} \vec{b} = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}|} \]
Vector diagram showing the vector projection of b onto a as a vector in the same direction as a.

Vector projection of \( \vec{b} \) onto \( \vec{a} \): vector in same direction as \( \vec{a} \) w/ magnitude \( \text{scal}_{\vec{a}} \vec{b} \)

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To give it the direction of \(\vec{a}\) w/o changing magnitude, we use a unit vector pointing in same direction as \(\vec{a}\).

So, the vector projection of \(\vec{b}\) onto \(\vec{a}\) is

\[ \text{proj}_{\vec{a}} \vec{b} = \underbrace{\text{scal}_{\vec{a}} \vec{b}}_{\text{magnitude}} \underbrace{\frac{\vec{a}}{|\vec{a}|}}_{\text{direction (as a unit vector)}} \]
\[ \text{proj}_{\vec{a}} \vec{b} = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}|} \frac{\vec{a}}{|\vec{a}|} = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}|^2} \vec{a} \]

Example

\(\vec{a} = \langle 1, 2 \rangle \quad \vec{b} = \langle 3, -4 \rangle\)

\[ \text{scal}_{\vec{a}} \vec{b} = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}|} = \frac{-5}{\sqrt{5}} = -\sqrt{5} \]
\[ \text{proj}_{\vec{a}} \vec{b} = \text{scal}_{\vec{a}} \vec{b} \frac{\vec{a}}{|\vec{a}|} = -\sqrt{5} \frac{\langle 1, 2 \rangle}{\sqrt{5}} = -\langle 1, 2 \rangle = \langle -1, -2 \rangle \]
Vector diagram showing vector b pointing away from vector a, indicating an obtuse angle and negative projection.
Negative because of the obtuse angle between vectors.
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\[ \vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta \]

Notice if \(\vec{a} \perp \vec{b}\), then \(\theta = 90^\circ = \pi/2\) and \(\cos \theta = 0\).

So, if \(\vec{a} \perp \vec{b}\) then \(\vec{a} \cdot \vec{b} = 0\).

Example

\(\vec{u} = \langle -1, -4, -1 \rangle \quad \vec{v} = \langle 2, -2, 3 \rangle\)

Express \(\vec{u}\) as \(\vec{u} = \vec{p} + \vec{n}\) where \(\vec{p}\) is parallel to \(\vec{v}\) and \(\vec{n}\) is perpendicular to \(\vec{v}\).

Vector decomposition diagram showing vector u as the sum of parallel component p and orthogonal component n.

\(\vec{p}\) is \(\text{proj}_{\vec{v}} \vec{u} = \) (see examples)

\[ = \dots = \left\langle \frac{6}{17}, -\frac{6}{17}, \frac{9}{17} \right\rangle \]

Find \(\vec{n}\): \(\vec{u} = \vec{p} + \vec{n}\) so \(\vec{n} = \vec{u} - \vec{p}\)

\[ \vec{n} = \langle -1, -4, -1 \rangle - \left\langle \frac{6}{17}, -\frac{6}{17}, \frac{9}{17} \right\rangle = \left\langle -\frac{23}{17}, -\frac{62}{17}, -\frac{26}{17} \right\rangle \]