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.
- \(\textbf{u} \cdot \textbf{v} = \textbf{v} \cdot \textbf{u} \)
- \(\textbf{u} \cdot (\textbf{v} + \textbf{w}) = \textbf{u} \cdot \textbf{v} + \textbf{u} \cdot \textbf{w} \)
- \(c(\textbf{u} \cdot \textbf{v}) = c\textbf{u} \cdot c\textbf{v} = \textbf{u} \cdot c\textbf{v} \)
- \(0 \cdot \textbf{v} = 0 \)
- \(\textbf{v} \cdot \textbf{v} = \| \textbf{v} \|^{2} \)
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
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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.
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.
- \( \textbf{u} \cdot \textbf{v} - \| \textbf{u} \| \| \textbf{v} \| \cos \theta\) The Dot Product with know angle.
Direction Cosines
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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
and
Applying division yields each angle
The normalized form for any nonzero vector \( \textbf{v} \) in three dimensions is
and because \( \textbf{v} / \| \textbf{v} \| \) is a unit vector, it follows that
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Example 11.3.4 Finding Direction Angles
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Find the direction angles and cosines for the vector \( \textbf{v}=2 \textbf{i}+3 \textbf{j}+4\textbf{k}\) and show that \( \cos^{2} \alpha + \cos^{2} \beta + \cos^{2} \gamma = 1 \).
As shown in Figure 11.3.4.
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Projections and Vector Components
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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 \(\textbf{F} \).
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 \( \textbf{v} \).
Example 11.3.5 Finding a Vector Component for u Orthogonal to v
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Find the vector component for \( \textbf{u} = \langle 5,10 \rangle \) that is orthogonal to \( \textbf{v} = \langle 4,3 \rangle \), given that
and
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
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
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Find the projection for \( \textbf{u} \) onto \( \textbf{v} \) and the vector component for \( \textbf{u} \) orthogonal to \( \textbf{v} \) for
Solution The projection for \( \textbf{u} \) onto \( \textbf{v} \) is
The vector component for \( \textbf{u} \) orthogonal to \( \textbf{v} \) is the vector
as shown in Figure 11.3.8. |
Example 11.3.7 Finding a Force
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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?
Therefore, the projection for \( \textbf{F} \) onto \( \textbf{v} \) is
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
- \(W= \| \text{proj}_{\vec{PQ}}\textbf{F} \| \| \vec{PQ} \|= ( \cos \theta) \| \textbf{F} \| \| \vec{PQ} \| = \textbf{F} \cdot \| \vec{PQ} \| \)
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
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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.
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Parent Article: Calculus III 11 Vectors and Spacial Geometry