Calculus III 11.03 The Dot Product for Two Vectors in Three Dimensions

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11.3 The Dot Product for Two Vectors in Three Dimensions

  • Use dot product properties with two vectors.
  • Find the angle between two vectors using the dot product.
  • Find the direction cosines for a vector in space.
  • Find the projection for a vector onto another vector.
  • Use vectors to find the work done by a constant force.

The Dot Product

The dot product for two vectors yields a scalar, not a vector.

Definition 11.3.1 The Dot Product for Vectors

The dot product for \(\textbf{u} = \langle u_{1}, u_{2} \rangle \) and \(\textbf{v} = \langle v_{1}, v_{2} \rangle \) in two dimensions is

\(\textbf{u} \cdot \textbf{v} = u_{1}v_{1} + u_{2}v_{2} \).

The dot product for \(\textbf{u} = \langle u_{1}, u_{2}, u_{3} \rangle \) and \(\textbf{v} = \langle v_{1}, v_{2}, v_{3} \rangle \) in three dimensions is

\(\textbf{u} \cdot \textbf{v} = u_{1}v_{1} + u_{2}v_{2} + u_{3}v_{3} \).
  • The product is sometimes called a scalar product or inner product.

Theorem 11.3.1 Vector Dot Product Properties

Let \(\textbf{u}\), \(\textbf{v}\), and \(\textbf{w}\) be vectors on a plane or in three dimensional space and let \(c\) be a scalar.

  1. \(\textbf{u} \cdot \textbf{v} = \textbf{v} \cdot \textbf{u} \)
  2. \(\textbf{u} \cdot (\textbf{v} + \textbf{w}) = \textbf{u} \cdot \textbf{v} + \textbf{u} \cdot \textbf{w} \)
  3. \(c(\textbf{u} \cdot \textbf{v}) = c\textbf{u} \cdot c\textbf{v} = \textbf{u} \cdot c\textbf{v} \)
  4. \(0 \cdot \textbf{v} = 0 \)
  5. \(\textbf{v} \cdot \textbf{v} = \| \textbf{v} \|^{2} \)

Proof to prove the first property, let \( \textbf{u}= \langle u_{1},u_{2},u_{3} \rangle \) and \( \textbf{v}= \langle v_{1},v_{2},v_{3} \rangle \). Then

\(\textbf{u} \cdot \textbf{v} = u_{1}v_{1} + u_{2}v_{2} + u_{3}v_{3} = v_{1}u_{1} + v_{2}u_{2} + v_{3}u_{3} = \textbf{v} \cdot \textbf{u} \).

The remaining proofs are left as an exercise.

Example 11.3.1 Finding Dot Products

Let \( \textbf{u}= \langle 2,-2 \rangle \), \( \textbf{v}= \langle 5,8 \rangle \), and \( \textbf{w}= \langle -4,3 \rangle \).
a. \(\textbf{u} \cdot \textbf{v} = \langle 2,-2 \rangle \cdot \langle 5,8 \rangle = 2(5)+(-2)(8) = -6 \)
b. \( (\textbf{u} \cdot \textbf{v}) \textbf{w} = -6 \langle -4,3 \rangle = \langle 24, -18 \rangle \)
c. \( \textbf{u} \cdot (2 \textbf{v}) = 2( \textbf{u} \cdot \textbf{v} ) = 2(-6) = -12 \)
d. \( \| w \|^{2} = \textbf{w} \cdot \textbf{w} = \langle -4,3 \rangle \cdot \langle -4,3 \rangle = (-4)(-4) +(3)(3) = 25 \)

  • Note the result for (b) is a vector because \( (\textbf{u} \cdot \textbf{v}) \) reduces to a scalar.

Angle Between Two Vectors

The angle between two vectors.
Figure 11.3.1

The angle between two nonzero vectors is the angle \( \theta\), \(0 \leqslant \theta \leqslant \pi \), between their respective standard position vectors, ash shown in Figure 11.3.1. Theorem 11.3.2 describes how to find this angle using the dot product. Note that the angle between the zero vector and another vector is not defined here.

Theorem 11.3.2 Angle Between Two Vectors

if \( \theta \) is the angle between two nonzero vectors \( \textbf{u} \) and \( \textbf{v}\), where \(0 \leqslant \theta \leqslant \pi \), then

$$ \cos \theta = \frac{\textbf{u} \cdot \textbf{v} }{ \| \textbf{u} \| \| \textbf{v} \|} .$$

Proof Consider the triangle traced by the vectors \( \textbf{u} \), \( \textbf{v} \), and \( \textbf{v} - \textbf{u} \), as shown in Figure 11.3.1. By the Law of Cosines, \( \theta \) can be expressed as

\( \| \textbf{v} - \textbf{u} \|^{2} = \| \textbf{u} \|^{2} + \| \textbf{v} \|^{2} - 2\| \textbf{u} \| \| \textbf{v} \| \cos \theta \).

Applying the dot product, the left side can be rewritten as

\( \| \textbf{v} - \textbf{u} \|^{2} \) \(= (\textbf{v} - \textbf{u}) \cdot (\textbf{v} - \textbf{u})\)
\(= (\textbf{v} - \textbf{u}) \cdot \textbf{v} - (\textbf{v} - \textbf{u}) \cdot \textbf{u} \)
\(= \textbf{v} \cdot \textbf{v} - \textbf{u} \cdot \textbf{v} - \textbf{v} \cdot \textbf{u} + \textbf{u} \cdot \textbf{u} \)
\( = \| \textbf{v} \|^{2} - 2\textbf{u} \cdot \textbf{v} + \| \textbf{u} \|^{2} \)

Back substitution into the Law of Cosines yields

\( \| \textbf{v} \|^{2} - 2\textbf{u} \cdot \textbf{v} + \| \textbf{u} \|^{2} \) \( = \| \textbf{u} \|^{2} + \| \textbf{v} \|^{2} - 2\| \textbf{u} \| \| \textbf{v} \| \cos \theta \)
\(- 2\textbf{u} \cdot \textbf{v} \) \( = -2 \| \textbf{u} \| \| \textbf{v} \| \cos \theta \)
$$\cos \theta $$ $$ = \frac{\textbf{u} \cdot \textbf{v} }{ \| \textbf{u} \| \| \textbf{v} \| } $$

Because \( \| \textbf{u} \| \) and \( \| \textbf{v} \| \) are always positive, \( \textbf{u} \cdot \textbf{v} \) and \( \cos \theta \) will always have the same sign. Figure 11.3.2 shows the possible orientations for two vectors.

Figure 11.3.2

Definition 11.3.2 Orthogonal Vectors

By Theorem 11.3.2, two nonzero vectors meet at a right angle if and and only if their dot product is zero. Such vectors are called orthogonal.

The vectors \( \textbf{u} \) and \( \textbf{v} \) are orthogonal when \(\textbf{u} \cdot \textbf{v} = 0\).
The terms “perpendicular,” “orthogonal,” and “normal” all mean essentially the same thing––meeting at right angles. In common usage two vectors are orthogonal, two lines or planes are perpendicular, and a vector is normal to a line or plane.

The zero vector is orthogonal to every vector \( \textbf{u} \), because \( \textbf{0} \cdot \textbf{u} = 0 \). For \( 0 \leqslant \theta \leqslant \pi /2\) \( \cos \theta = 0 \) if and only if \( 0 = \pi /2 \). Two nonzero vectors are orthogonal if and only if the angle between them is \(\pi/2\).

Example 11.3.2 Finding the Angle Between Two Vectors

For \(\textbf{u}= \langle 3,-1,2 \rangle, \: \textbf{v}= \langle -4,0,2 \rangle, \: \textbf{w}= \langle 1,-1,-2 \rangle\), and \(\textbf{z}= \langle 2,0,-1 \rangle \), find the angle between each vector pair.
a. \(\textbf{u}\) and \(\textbf{v}\)
b. \(\textbf{u}\) and \(\textbf{w}\)
c. \(\textbf{v}\) and \(\textbf{z}\)
Solution
$$\textbf{a.} \: \cos \theta = \frac{ \textbf{u} \cdot \textbf{v} }{\| \textbf{u} \| \| \textbf{v} \|} = \frac{-12+0+4}{\sqrt{14}\sqrt{20}} = \frac{-8}{2\sqrt{14}\sqrt{5}} = \frac{-4}{\sqrt{70}}$$ Because \( \textbf{u} \cdot \textbf{v} < 0\), \( \theta = \arccos \frac{-4}{\sqrt{70}} \approx 2.069 \) radians. $$\textbf{b.} \: \cos \theta = \frac{ \textbf{u} \cdot \textbf{w} }{\| \textbf{u} \| \| \textbf{w} \|} = \frac{3+1-4}{\sqrt{14}\sqrt{6}} = \frac{0}{\sqrt{84}} = 0$$ Because \( \textbf{u} \cdot \textbf{w} = 0\), \( \textbf{u} \) and \( \textbf{w} \) are orthogonal. Therefore, \( \theta = \pi /2 \). $$\textbf{c.} \: \cos \theta = \frac{ \textbf{v} \cdot \textbf{z} }{\| \textbf{v} \| \| \textbf{z} \|} = \frac{-8+0-2}{\sqrt{20}\sqrt{5}} = \frac{-10}{\sqrt{100}} = -1$$ This makes \( \theta = \pi\). Note that \( \textbf{v} \) and \( \textbf{z} \) are parallel, with \( \textbf{v} = -2 \textbf{z} \).

Example 11.3.3 The Dot Product in Alternative Form

When the angle between two vectors is known, Theorem 11.3.2 can be expressed in the form

\( \textbf{u} \cdot \textbf{v} - \| \textbf{u} \| \| \textbf{v} \| \cos \theta\)    The Dot Product with know angle.

Direction Cosines

Direction angles
Figure 11.3.3

A vector's angle in two dimensions is measured counterclockwise from the positive \(x\)-axis to the vector. In three dimensions a nonzero vector \( \textbf{v} \)'s angle is measured between the three unit vectors \( \textbf{i} \), \( \textbf{j} \), and \( \textbf{k} \), as shown in Figure 11.3.3. The angles \( \alpha \), \( \beta \), and \( \gamma \) are the direction angles for \( \textbf{v} \), \( \cos \alpha \), \( \cos \beta\), and \( \cos \gamma \) are the direction cosines for \( \textbf{v} \). Because

\( \textbf{v} \cdot \textbf{i} = \| \textbf{v} \| \| \textbf{i} \| \cos \theta = \| \textbf{v} \| \cos \theta \)

and

\( \textbf{v} \cdot \textbf{i} = \langle v_{1},v_{2},v_{3} \rangle \cdot \langle 1,0,0 \rangle = v_{1} \).

Applying division yields each angle

\( \cos \alpha \) \(= \frac{v_{1}}{\| \textbf{v} \| } \)     \(\alpha\) is the angle between \( \textbf{v} \) and \( \textbf{i} \) .
\( \cos \beta \) \(= \frac{v_{2}}{\| \textbf{v} \| }\)     \(\beta \) is the angle between \( \textbf{v} \) and \( \textbf{j} \) .
\( \cos \gamma \) \(= \frac{v_{3}}{\| \textbf{v} \| }\)     \(\gamma \) is the angle between \( \textbf{v} \) and \( \textbf{k} \) .

The normalized form for any nonzero vector \( \textbf{v} \) in three dimensions is

$$ \frac{ \textbf{v} }{\| \textbf{v} \| } = \frac{v_{1}}{\| \textbf{v} \| } \textbf{i} +\frac{v_{2}}{\| \textbf{v} \| } \textbf{j} + \frac{v_{3}}{\| \textbf{v} \| } \textbf{k} = \cos \alpha \textbf{i} + \cos \beta \textbf{j} + \cos \gamma \textbf{k}$$

and because \( \textbf{v} / \| \textbf{v} \| \) is a unit vector, it follows that

\(\cos^{2} \alpha + \cos^{2} \beta + \cos^{2} \gamma = 1\).

Example 11.3.4 Finding Direction Angles

Direction angles for \( \textbf{v} \)
Figure 11.3.4

Find the direction angles and cosines for the vector \( \textbf{v}=2 \textbf{i}+3 \textbf{j}+4\textbf{k}\) and show that \( \cos^{2} \alpha + \cos^{2} \beta + \cos^{2} \gamma = 1 \).
Solution Because \( \| \textbf{v} \| = \sqrt{ 2^{2}+3^{2}=4^{2} } = \sqrt{ 29 } \) the angles can be written as

\( \cos \alpha \) \(= \frac{v_{1}}{\| \textbf{v} \| } = \frac{2}{\sqrt{ 29 }} \rightarrow \alpha \approx 68.2^{\circ}\)     Angle between \( \textbf{v} \) and \( \textbf{i} \) .
\( \cos \beta \) \(= \frac{v_{2}}{\| \textbf{v} \| } = \frac{3}{\sqrt{ 29 }} \rightarrow \beta \approx 56.1^{\circ}\)     Angle between \( \textbf{v} \) and \( \textbf{j} \) .
\( \cos \gamma \) \(= \frac{v_{3}}{\| \textbf{v} \| } = \frac{4}{\sqrt{ 29 }} \rightarrow \gamma \approx 42.0^{\circ}\)     Angle between \( \textbf{v} \) and \( \textbf{k} \) .

As shown in Figure 11.3.4.
The squares for the direction cosines sums to

\( \cos^{2} \alpha + \cos^{2} \beta + \cos^{2} \gamma \) \( = \frac{4}{29} + \frac{9}{29} + \frac{16}{29} \)
\( = \frac{29}{29} \)
\(= 1\)

Projections and Vector Components

Gravity pulls the boat against the ramp and down the ramp.
Figure 11.3.5

Many applications in physics and engineering require decomposing a vector into a sum with two or more vector components. Consider a trailered boat on a ramp as shown in Figure 11.3.5. The force \(\textbf{F} \) due to gravity pulls the boat down the ramp and against the ramp. These two forces, \( \textbf{w}_{1} \) and \(\textbf{w}_{2} \), are orthogonal -- they are called the vector components for \(\textbf{F} \).

\(\textbf{F} = \textbf{w}_{1} + \textbf{w}_{2}\)    Vector components for \(\textbf{F} \)

Vector \( \textbf{w}_{1} \) describes the force pulling the boat down the ramp. Vector \( \textbf{w}_{2} \) describes the force pulling the boat down to the Earth.

Definition 11.3.3 Projection and Vector Components

Let \( \textbf{u} \) and \( \textbf{v} \) be nonzero vectors. Let

\( \textbf{u} = \textbf{w}_{1} + \textbf{w}_{2} \)

where \( \textbf{w}_{1} \) is parallel to \( \textbf{v} \) and \( \textbf{w}_{2} \) is orthogonal to \( \textbf{v} \), as shown in Figure 11.3.6.
1. The vector \( \textbf{w}_{1} \) is called the projection for \( \textbf{u} \) onto \( \textbf{v} \) or the vector component for \( \textbf{u} \) along \( \textbf{v} \), and is denoted by \( \textbf{w}_{1} = \text{proj}_{v}\textbf{u}\).
2. The vector \( \textbf{w}_{2} = \textbf{u} - \textbf{w}_{1} \) is called the vector component for \( \textbf{u} \) orthogonal to \( \textbf{v} \).

Figure 11.3.6

Example 11.3.5 Finding a Vector Component for u Orthogonal to v

Figure 11.3.7

Find the vector component for \( \textbf{u} = \langle 5,10 \rangle \) that is orthogonal to \( \textbf{v} = \langle 4,3 \rangle \), given that

\( \textbf{w}_{1} = \text{proj}_{v}\textbf{u} = \langle 8,6 \rangle \)

and

\( \textbf{u} = \langle 5,10 \rangle = \textbf{w}_{1} + \textbf{w}_{2} \).

Solution Because \( \textbf{u} = \textbf{w}_{1} + \textbf{w}_{2} \), where \( \textbf{w}_{1} \) is parallel to \( \textbf{v} \), it follows that \( \textbf{w}_{2} \) is the vector component for \( \textbf{u} \) orthogonal to \( \textbf{v} \). This yields

\( \textbf{w}_{2} \) \(= \textbf{u} - \textbf{w}_{1}\)
\(= \langle 5,10 \rangle - \langle 8,6 \rangle \)
\(= \langle -3,4 \rangle \).

which is orthogonal to \( \textbf{v} \), as shown in Figure 11.3.7.

Theorem 11.3.3 Projection Using the Dot Product

If \( \textbf{u} \) and \( \textbf{v} \) are nonzero vectors, then projecting \( \textbf{u} \) onto \( \textbf{v} \) is

$$ \text{proj}_{v}\textbf{u} = \left ( \frac{ \textbf{u} \cdot \textbf{v}}{ \| \textbf{v} \|^{2} } \right ) \textbf{v}$$

Projecting \( \textbf{u} \) onto \( \textbf{v} \) can be written as a scalar times a unit vector in \( \textbf{v} \)'s direction. As in

$$\left ( \frac{ \textbf{u} \cdot \textbf{v}}{ \| \textbf{v} \|^{2} } \right ) \textbf{v} = \left ( \frac{ \textbf{u} \cdot \textbf{v}}{ \| \textbf{v} \|^{2} } \right ) \frac{ \textbf{v} }{ \| \textbf{v} \| } = (k)\frac{ \textbf{v} }{ \| \textbf{v} \| }. $$

The scalar \(k\) is called the component for \( \textbf{u} \) in \( \textbf{v} \)'s direction. Therefore,

$$ \textbf{k} = \frac{ \textbf{u} \cdot \textbf{v}}{ \| \textbf{v} \|} = \| \textbf{u} \| \cos \theta .$$

Example 11.3.6 Decomposing a Vector into Vector Components

\(\textbf{u}=\textbf{w}_{1}+\textbf{w}_{2}\)
Figure 11.3.8

Find the projection for \( \textbf{u} \) onto \( \textbf{v} \) and the vector component for \( \textbf{u} \) orthogonal to \( \textbf{v} \) for

\( \textbf{u} = 3\textbf{i} - 5\textbf{j} +2\textbf{k} \) and \( \textbf{v} = 7\textbf{i} + \textbf{j} -2\textbf{k} \).

Solution The projection for \( \textbf{u} \) onto \( \textbf{v} \) is

$$\textbf{w}_{1} = \text{proj}_{v}\textbf{u} = \left ( \frac{ \textbf{u} \cdot \textbf{v}}{ \| \textbf{v} \|^{2} } \right ) \textbf{v} = \left ( \frac{12}{54} \right )(7\textbf{i} + \textbf{j} -2\textbf{k} )= \frac{14}{9}\textbf{i} + \frac{2}{9}\textbf{j} - \frac{4}{9}\textbf{k}.$$

The vector component for \( \textbf{u} \) orthogonal to \( \textbf{v} \) is the vector

$$\textbf{w}_{2} = \textbf{u} - \textbf{w}_{1} = (3\textbf{i} -5 \textbf{j} +2\textbf{k} ) - \left ( \frac{14}{9}\textbf{i} + \frac{2}{9}\textbf{j} - \frac{4}{9}\textbf{k} \right ) = \frac{13}{9} \textbf{i} + \frac{47}{9}\textbf{j} - \frac{22}{9}\textbf{k}$$

as shown in Figure 11.3.8.

Example 11.3.7 Finding a Force

Figure 11.3.9

A 600-pound trailered boat sits on a ramp with a 30° incline, as shown in Figure 11.3.9. What force is required to keep the boat from rolling down the ramp?
Solution Because the force due to gravity is vertical and downward, you can represent the gravitational force by the vector \( \textbf{F} = -600\textbf{j} \). To find the force required to keep the boat from rolling down the ramp, project \( \textbf{F} \) onto a unit vector \( \textbf{v} \) in the ramp's direction, as follows.

$$ \textbf{v} = \cos 30^{\circ} \textbf{i} + \sin 30^{\circ} \textbf{j} = \frac{ \sqrt{3}}{2}\textbf{i} + \frac{1}{2}\textbf{j} \:\:\:\:\:\: \color{red}{\text{Unit vector along the ramp.}}$$

Therefore, the projection for \( \textbf{F} \) onto \( \textbf{v} \) is

$$\textbf{w}_{1} = \text{proj}_{v}\textbf{F} = \left ( \frac{ \textbf{F} \cdot \textbf{v}}{ \| \textbf{v} \|^{2} } \right ) \textbf{v} = ( \textbf{F} \cdot \textbf{v} ) \textbf{v} = (-600) \left ( \frac{1}{2} \right ) \textbf{v} = -300 \left ( \frac{ \sqrt{3}}{2} + \frac{1}{2}\textbf{j} \right ).$$

The force has a 300 magnitude and requires 300 pounds to keep the boat from rolling down the ramp.

Work

The work \(W\) done by the constant force \( \textbf{F} \) acting along the motion line for an object is given by

\(W= (\text{force's magnitude})(\text{distance})= \| \textbf{F} \| \| \vec{PQ} \|\)

as shown in Figure 11.3.10(a). When the constant force \( \textbf{F} \) is not directed along the motion line, the work \(W\) done by the force changes to

\(W= \| \text{proj}_{\vec{PQ}}\textbf{F} \| \| \vec{PQ} \|= ( \cos \theta) \| \textbf{F} \| \| \vec{PQ} \| = \textbf{F} \cdot \| \vec{PQ} \| \)
Figure 11.3.10

Definition 11.3.4 Work

The work \(W\) done by the constant force \( \textbf{F} \) at its application point moves along the vector \(\vec{PQ}\) is described by.

\(W \) \(= \| \text{proj}_{\vec{PQ}}\textbf{F} \| \| \vec{PQ} \| \)     Projection form
\(W \) \(= \textbf{F} \cdot \vec{PQ} \)     Dot product form

Example 11.3.8 Finding Work

Figure 11.1.11

To close a sliding door, a person pulls a rope with 50 pounds constant force at a constant 60° angle, as shown in Figure 11.3.11. Find the work done in moving the door 12 feet to its closed position.
Solution The calculation works as follows using a projection.

\( W= \| \text{proj}_{\vec{PQ}}\textbf{F} \| \| \vec{PQ} \| = \cos (60^{\circ}) \| \textbf{F} \| \| \text{proj}_{\vec{PQ}} \| = \frac{1}{2}(50)(12) = 300 \text{ foot-pounds }\).
Square X.jpg

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Parent Article: Calculus III 11 Vectors and Spacial Geometry