The Central Idea
For two-dimensional vectors
\[ \mathbf{u}=\langle a,b\rangle \]and
\[ \mathbf{v}=\langle c,d\rangle, \]the dot product is
\[ \boxed{ \mathbf{u}\cdot\mathbf{v} = ac+bd }. \]Multiply corresponding components, then add the products.
The Dot Product Produces a Scalar
The result of a dot product is not another vector.
It is a scalar.
For example,
\[ \langle 2,3\rangle \cdot \langle 4,5\rangle = 2(4)+3(5) = 8+15 = 23. \]Therefore,
\[ \boxed{ \langle 2,3\rangle \cdot \langle 4,5\rangle = 23 }. \]Why the Name Dot Product?
The centered dot in
\[ \mathbf{u}\cdot\mathbf{v} \]distinguishes this operation from ordinary vector addition and from other vector products.
Because the result is a scalar, the dot product is also called the scalar product or inner product in many contexts.
Example: Dot Product with Negative Components
Let
\[ \mathbf{u}=\langle 3,-2\rangle \]and
\[ \mathbf{v}=\langle -4,5\rangle. \]Then
\[ \mathbf{u}\cdot\mathbf{v} = 3(-4)+(-2)(5). \]Therefore,
\[ \boxed{ \mathbf{u}\cdot\mathbf{v} = -12-10 = -22 }. \]Dot Products in Three Dimensions
If
\[ \mathbf{u} = \langle a,b,c\rangle \]and
\[ \mathbf{v} = \langle d,e,f\rangle, \]then
\[ \boxed{ \mathbf{u}\cdot\mathbf{v} = ad+be+cf }. \]The rule remains the same: multiply corresponding components and add.
Example: Three-Dimensional Dot Product
Let
\[ \mathbf{u}=\langle 1,2,-3\rangle \]and
\[ \mathbf{v}=\langle 4,-1,2\rangle. \]Then
\[ \mathbf{u}\cdot\mathbf{v} = 1(4)+2(-1)+(-3)(2). \]Thus,
\[ \boxed{ \mathbf{u}\cdot\mathbf{v} = 4-2-6 = -4 }. \]The Dot Product of a Vector with Itself
If
\[ \mathbf{v}=\langle a,b\rangle, \]then
\[ \mathbf{v}\cdot\mathbf{v} = a^2+b^2. \]But
\[ \|\mathbf{v}\| = \sqrt{a^2+b^2}. \]Therefore,
\[ \boxed{ \mathbf{v}\cdot\mathbf{v} = \|\mathbf{v}\|^2 }. \]Magnitude from the Dot Product
Since
\[ \mathbf{v}\cdot\mathbf{v} = \|\mathbf{v}\|^2, \]we can write
\[ \boxed{ \|\mathbf{v}\| = \sqrt{\mathbf{v}\cdot\mathbf{v}} }. \]This gives a second way to view the magnitude formula.
Example: Magnitude through the Dot Product
Let
\[ \mathbf{v}=\langle 5,12\rangle. \]Then
\[ \mathbf{v}\cdot\mathbf{v} = 5^2+12^2 = 25+144 = 169. \]Therefore,
\[ \|\mathbf{v}\| = \sqrt{169} = \boxed{13}. \]The Geometric Dot Product Formula
For nonzero vectors \(\mathbf{u}\) and \(\mathbf{v}\), let \(\theta\) be the angle between them.
Then
\[ \boxed{ \mathbf{u}\cdot\mathbf{v} = \|\mathbf{u}\|\, \|\mathbf{v}\| \cos\theta }. \]This is the central bridge between vector algebra and angular geometry.
What the Geometric Formula Tells Us
The dot product depends on three things:
- the magnitude of \(\mathbf{u}\);
- the magnitude of \(\mathbf{v}\);
- the cosine of the angle between them.
Thus the dot product measures both size and directional alignment.
Vectors Pointing in the Same Direction
If two nonzero vectors point in the same direction, then
\[ \theta=0^\circ. \]Since
\[ \cos 0^\circ=1, \]we have
\[ \boxed{ \mathbf{u}\cdot\mathbf{v} = \|\mathbf{u}\|\|\mathbf{v}\| }. \]The dot product is as positive as possible for those magnitudes.
Vectors Pointing in Opposite Directions
If two nonzero vectors point in opposite directions, then
\[ \theta=180^\circ. \]Since
\[ \cos 180^\circ=-1, \]we obtain
\[ \boxed{ \mathbf{u}\cdot\mathbf{v} = -\|\mathbf{u}\|\|\mathbf{v}\| }. \]The dot product is as negative as possible for those magnitudes.
Perpendicular Vectors
If two nonzero vectors are perpendicular, then
\[ \theta=90^\circ. \]Since
\[ \cos90^\circ=0, \]it follows that
\[ \boxed{ \mathbf{u}\cdot\mathbf{v}=0 }. \]This gives a powerful algebraic test for perpendicularity.
Orthogonal Vectors
Vectors whose dot product is zero are called orthogonal.
For nonzero geometric vectors, orthogonal means perpendicular.
Thus,
\[ \boxed{ \mathbf{u}\perp\mathbf{v} \iff \mathbf{u}\cdot\mathbf{v}=0 } \]when both vectors are nonzero.
Example: Test for Perpendicularity
Let
\[ \mathbf{u}=\langle 3,4\rangle \]and
\[ \mathbf{v}=\langle 4,-3\rangle. \]Compute:
\[ \mathbf{u}\cdot\mathbf{v} = 3(4)+4(-3). \]Therefore,
\[ \mathbf{u}\cdot\mathbf{v} = 12-12 = 0. \]So the vectors are perpendicular.
Positive Dot Products
If
\[ \mathbf{u}\cdot\mathbf{v}>0, \]then
\[ \cos\theta>0. \]Therefore, the angle between the nonzero vectors is acute:
\[ \boxed{ 0^\circ<\theta<90^\circ }. \]Negative Dot Products
If
\[ \mathbf{u}\cdot\mathbf{v}<0, \]then
\[ \cos\theta<0. \]Therefore, the angle between the nonzero vectors is obtuse:
\[ \boxed{ 90^\circ<\theta<180^\circ }. \]Zero Dot Products
If two nonzero vectors satisfy
\[ \mathbf{u}\cdot\mathbf{v}=0, \]then
\[ \theta=90^\circ. \]Thus the sign of the dot product gives immediate angular information.
Angular Classification from the Dot Product
| Dot Product | Angle Type |
|---|---|
| \(\mathbf{u}\cdot\mathbf{v}>0\) | Acute |
| \(\mathbf{u}\cdot\mathbf{v}=0\) | Right angle, for nonzero vectors |
| \(\mathbf{u}\cdot\mathbf{v}<0\) | Obtuse |
Finding the Angle Between Two Vectors
Start from
\[ \mathbf{u}\cdot\mathbf{v} = \|\mathbf{u}\| \|\mathbf{v}\| \cos\theta. \]Solve for cosine:
\[ \boxed{ \cos\theta = \frac{ \mathbf{u}\cdot\mathbf{v} }{ \|\mathbf{u}\|\|\mathbf{v}\| } }. \]Then use inverse cosine to find the angle.
Angle Formula
For nonzero vectors,
\[ \boxed{ \theta = \cos^{-1} \left( \frac{ \mathbf{u}\cdot\mathbf{v} }{ \|\mathbf{u}\|\|\mathbf{v}\| } \right) }. \]This gives the smaller angle between the vectors, from \(0^\circ\) through \(180^\circ\).
Example: Find an Angle
Let
\[ \mathbf{u}=\langle 1,0\rangle \]and
\[ \mathbf{v}=\langle 1,1\rangle. \]The dot product is
\[ \mathbf{u}\cdot\mathbf{v} = 1. \]The magnitudes are
\[ \|\mathbf{u}\|=1 \]and
\[ \|\mathbf{v}\|=\sqrt2. \]Completing the Angle Calculation
Substitute into the angle formula:
\[ \cos\theta = \frac{1}{\sqrt2}. \]Therefore,
\[ \boxed{ \theta=45^\circ }. \]Example: An Obtuse Angle
Let
\[ \mathbf{u}=\langle 1,0\rangle \]and
\[ \mathbf{v}=\langle -1,1\rangle. \]Then
\[ \mathbf{u}\cdot\mathbf{v} = -1. \]Since the dot product is negative, the angle must be obtuse.
Completing the Obtuse-Angle Example
The magnitudes are
\[ \|\mathbf{u}\|=1, \qquad \|\mathbf{v}\|=\sqrt2. \]Therefore,
\[ \cos\theta = -\frac{1}{\sqrt2}. \]Hence,
\[ \boxed{ \theta=135^\circ }. \]Unit Vectors Simplify the Geometric Formula
If \(\mathbf{u}\) and \(\mathbf{v}\) are both unit vectors, then
\[ \|\mathbf{u}\|=\|\mathbf{v}\|=1. \]The geometric dot product formula becomes
\[ \boxed{ \mathbf{u}\cdot\mathbf{v} = \cos\theta }. \]For unit vectors, the dot product directly measures angular alignment.
Dot Products of Standard Unit Vectors
In the plane,
\[ \mathbf{i}=\langle 1,0\rangle, \qquad \mathbf{j}=\langle 0,1\rangle. \]Then
\[ \mathbf{i}\cdot\mathbf{i}=1, \] \[ \mathbf{j}\cdot\mathbf{j}=1, \]and
\[ \boxed{ \mathbf{i}\cdot\mathbf{j}=0 }. \]The coordinate axes are perpendicular.
Standard Unit Vectors in Three Dimensions
In three dimensions,
\[ \mathbf{i}=\langle 1,0,0\rangle, \] \[ \mathbf{j}=\langle 0,1,0\rangle, \] \[ \mathbf{k}=\langle 0,0,1\rangle. \]Each has magnitude \(1\), and every distinct pair has dot product \(0\).
Orthogonal Unit Directions
The relations
\[ \mathbf{i}\cdot\mathbf{j}=0, \qquad \mathbf{j}\cdot\mathbf{k}=0, \qquad \mathbf{k}\cdot\mathbf{i}=0 \]express the perpendicularity of the three coordinate axes.
A collection of mutually perpendicular unit vectors is called orthonormal.
Commutative Property of the Dot Product
The dot product is commutative:
\[ \boxed{ \mathbf{u}\cdot\mathbf{v} = \mathbf{v}\cdot\mathbf{u} }. \]In component form,
\[ ac+bd = ca+db. \]Distributive Property of the Dot Product
The dot product distributes over vector addition:
\[ \boxed{ \mathbf{u}\cdot(\mathbf{v}+\mathbf{w}) = \mathbf{u}\cdot\mathbf{v} + \mathbf{u}\cdot\mathbf{w} }. \]This makes the dot product compatible with linear combinations.
Scalars Can Move through a Dot Product
If \(k\) is a scalar, then
\[ \boxed{ (k\mathbf{u})\cdot\mathbf{v} = k(\mathbf{u}\cdot\mathbf{v}) }. \]Similarly,
\[ \mathbf{u}\cdot(k\mathbf{v}) = k(\mathbf{u}\cdot\mathbf{v}). \]Dot Product with the Zero Vector
For any vector \(\mathbf{v}\),
\[ \boxed{ \mathbf{v}\cdot\mathbf{0}=0 }. \]Every component product contains a zero.
This does not mean the zero vector has a defined angle of \(90^\circ\) with every vector. The ordinary angle formula requires nonzero vectors.
Directional Alignment
The sign and size of the dot product indicate directional alignment.
- Large positive values indicate strong same-direction alignment.
- Zero indicates perpendicular directions for nonzero vectors.
- Negative values indicate opposing directional tendencies.
The magnitudes of the vectors also affect the numerical size of the dot product.
Separating Alignment from Magnitude
Divide the dot product by the magnitudes:
\[ \frac{ \mathbf{u}\cdot\mathbf{v} }{ \|\mathbf{u}\|\|\mathbf{v}\| }. \]For nonzero vectors, this equals
\[ \boxed{\cos\theta}. \]This normalized quantity reflects direction rather than vector size.
The Range of the Normalized Dot Product
Since
\[ -1\le\cos\theta\le1, \]we have
\[ \boxed{ -1 \le \frac{ \mathbf{u}\cdot\mathbf{v} }{ \|\mathbf{u}\|\|\mathbf{v}\| } \le 1 }. \]This is one geometric form of the Cauchy-Schwarz inequality.
A First Look at Projection
Sometimes we want to know how much of one vector lies in the direction of another.
The dot product provides exactly this information.
If \(\mathbf{u}\) is a unit vector, then
\[ \boxed{ \mathbf{v}\cdot\mathbf{u} } \]is the signed scalar component of \(\mathbf{v}\) in the direction of \(\mathbf{u}\).
Scalar Projection
For nonzero \(\mathbf{u}\), the scalar projection of \(\mathbf{v}\) onto \(\mathbf{u}\) is
\[ \boxed{ \operatorname{comp}_{\mathbf{u}}\mathbf{v} = \frac{ \mathbf{v}\cdot\mathbf{u} }{ \|\mathbf{u}\| } }. \]This measures the signed length of the part of \(\mathbf{v}\) pointing along \(\mathbf{u}\).
Why the Scalar Projection Formula Works
From the geometric dot product formula,
\[ \mathbf{v}\cdot\mathbf{u} = \|\mathbf{v}\|\|\mathbf{u}\|\cos\theta. \]Divide by \(\|\mathbf{u}\|\):
\[ \frac{ \mathbf{v}\cdot\mathbf{u} }{ \|\mathbf{u}\| } = \|\mathbf{v}\|\cos\theta. \]This is the signed length of \(\mathbf{v}\) in the direction of \(\mathbf{u}\).
Vector Projection
The vector projection of \(\mathbf{v}\) onto a nonzero vector \(\mathbf{u}\) is
\[ \boxed{ \operatorname{proj}_{\mathbf{u}}\mathbf{v} = \frac{ \mathbf{v}\cdot\mathbf{u} }{ \mathbf{u}\cdot\mathbf{u} } \mathbf{u} }. \]The result is a vector parallel to \(\mathbf{u}\).
Projection onto a Unit Vector
If \(\mathbf{u}\) is a unit vector, then
\[ \mathbf{u}\cdot\mathbf{u}=1. \]Therefore,
\[ \boxed{ \operatorname{proj}_{\mathbf{u}}\mathbf{v} = (\mathbf{v}\cdot\mathbf{u})\mathbf{u} }. \]The dot product supplies the signed amount, and \(\mathbf{u}\) supplies the direction.
Example: Projection onto the \(x\)-Axis
Let
\[ \mathbf{v}=\langle 5,3\rangle \]and
\[ \mathbf{i}=\langle 1,0\rangle. \]Since
\[ \mathbf{v}\cdot\mathbf{i}=5, \]the projection onto the \(x\)-axis is
\[ \boxed{ 5\mathbf{i} = \langle 5,0\rangle }. \]Perpendicular Decomposition
A vector can be separated into a part parallel to a chosen direction and a part perpendicular to it.
If
\[ \mathbf{p} = \operatorname{proj}_{\mathbf{u}}\mathbf{v}, \]then the remaining part is
\[ \mathbf{v}-\mathbf{p}. \]That remainder is perpendicular to \(\mathbf{u}\).
Why Projection Matters
Projection lets us separate a vector according to chosen directions.
This idea appears in:
- resolving forces;
- finding motion along an axis;
- measuring directional contribution;
- least-squares methods;
- orthogonal decomposition;
- geometry and linear algebra.
Dot Product and Work
In physics, work done by a constant force \(\mathbf{F}\) through a displacement \(\mathbf{d}\) is modeled by
\[ \boxed{ W = \mathbf{F}\cdot\mathbf{d} }. \]Only the component of the force acting along the displacement contributes to this dot product.
Why Perpendicular Force Does No Work in the Idealized Model
If
\[ \mathbf{F}\perp\mathbf{d}, \]then
\[ \mathbf{F}\cdot\mathbf{d}=0. \]Thus the idealized work contribution from that perpendicular force is zero.
Dot Product and Coordinate Geometry
The dot product gives coordinate geometry an efficient test for right angles.
If two line-direction vectors have dot product zero, the corresponding directions are perpendicular.
This provides a vector alternative to slope-based perpendicularity tests.
Example: Perpendicular Segments from Coordinates
Suppose
\[ A=(0,0),\qquad B=(3,4),\qquad C=(4,-3). \]Then
\[ \overrightarrow{AB} = \langle 3,4\rangle \]and
\[ \overrightarrow{AC} = \langle 4,-3\rangle. \]Their dot product is \(0\), so the segments meet at a right angle at \(A\).
The Dot Product Is Not Vector Multiplication in the Ordinary Sense
The notation
\[ \mathbf{u}\cdot\mathbf{v} \]defines a particular vector operation.
It should not be interpreted as multiplying vector symbols exactly as if they were ordinary scalar variables.
Its meaning is fixed by the dot-product definition.
Common Error: Adding Components Instead of Multiplying Them
For
\[ \langle 2,3\rangle \cdot \langle 4,5\rangle, \]do not add all four components.
Multiply corresponding components and then add:
\[ \boxed{ 2(4)+3(5)=23 }. \]Common Error: Returning a Vector
A dot product returns a scalar.
For example,
\[ \langle 2,3\rangle \cdot \langle 4,5\rangle \]is
\[ 23, \]not \(\langle 8,15\rangle\).
Common Error: Forgetting a Component
In three dimensions,
\[ \langle a,b,c\rangle \cdot \langle d,e,f\rangle \]requires all three component products:
\[ \boxed{ ad+be+cf }. \]Omitting the third term changes the result.
Common Error: Treating Zero Dot Product as Evidence of Zero Vectors
A dot product of zero does not imply that both vectors are zero.
Nonzero perpendicular vectors also have dot product zero.
For example,
\[ \langle 1,0\rangle \cdot \langle 0,1\rangle = 0. \]Common Error: Assigning an Angle to the Zero Vector
The angle formula contains
\[ \|\mathbf{u}\|\|\mathbf{v}\| \]in the denominator.
If either vector is zero, the denominator is zero.
Therefore, the ordinary angle-between-vectors formula applies only to nonzero vectors.
Common Error: Forgetting the Magnitudes in the Angle Formula
The expression
\[ \mathbf{u}\cdot\mathbf{v} \]by itself is not generally equal to \(\cos\theta\).
Instead,
\[ \boxed{ \cos\theta = \frac{ \mathbf{u}\cdot\mathbf{v} }{ \|\mathbf{u}\|\|\mathbf{v}\| } } \]unless both vectors are already unit vectors.
Common Error: Using Inverse Sine or Inverse Tangent
The dot-product angle formula naturally gives cosine:
\[ \cos\theta = \frac{ \mathbf{u}\cdot\mathbf{v} }{ \|\mathbf{u}\|\|\mathbf{v}\| }. \]Therefore, the direct inverse operation is
\[ \cos^{-1}. \]Common Error: Confusing Scalar and Vector Projection
The scalar projection is a number:
\[ \operatorname{comp}_{\mathbf{u}}\mathbf{v}. \]The vector projection is a vector:
\[ \operatorname{proj}_{\mathbf{u}}\mathbf{v}. \]They answer related but different questions.
A Reliable Procedure for Computing a Dot Product
- Verify that the vectors have the same number of components.
- Multiply corresponding components.
- Add all component products.
- Report the result as a scalar.
- Interpret its sign if angular information is relevant.
A Reliable Procedure for Testing Perpendicularity
- Write both direction vectors in component form.
- Compute their dot product.
- If the dot product is zero and both vectors are nonzero, conclude that they are perpendicular.
- If the dot product is nonzero, they are not perpendicular.
A Reliable Procedure for Finding an Angle
- Verify that both vectors are nonzero.
- Compute the dot product.
- Compute each vector's magnitude.
- Divide the dot product by the product of the magnitudes.
- Apply inverse cosine.
- State the angle in the requested unit.
- Check whether the dot-product sign agrees with the angle type.
A Reliable Procedure for Projection
- Identify the vector being projected and the nonzero target direction.
- Compute the required dot product.
- Compute the target vector's magnitude or squared magnitude.
- Use the scalar- or vector-projection formula as required.
- Check that the vector projection is parallel to the target direction.
Example: Full Angle Workflow
Let
\[ \mathbf{u}=\langle 2,1\rangle, \qquad \mathbf{v}=\langle 1,3\rangle. \]First compute the dot product:
\[ \mathbf{u}\cdot\mathbf{v} = 2(1)+1(3) = 5. \]Next compute the magnitudes:
\[ \|\mathbf{u}\|=\sqrt5, \qquad \|\mathbf{v}\|=\sqrt{10}. \]Completing the Full Angle Workflow
Substitute:
\[ \cos\theta = \frac{5}{\sqrt5\sqrt{10}} = \frac{1}{\sqrt2}. \]Therefore,
\[ \boxed{ \theta=45^\circ }. \]The positive dot product correctly predicts an acute angle.
Example: Full Projection Workflow
Let
\[ \mathbf{v}=\langle 5,4\rangle \]and
\[ \mathbf{u}=\langle 2,0\rangle. \]Compute
\[ \mathbf{v}\cdot\mathbf{u} = 10 \]and
\[ \mathbf{u}\cdot\mathbf{u} = 4. \]Therefore,
\[ \operatorname{proj}_{\mathbf{u}}\mathbf{v} = \frac{10}{4}\langle 2,0\rangle. \]Completing the Projection Example
Simplify:
\[ \frac{10}{4}\langle 2,0\rangle = \left\langle 5,0 \right\rangle. \]Thus,
\[ \boxed{ \operatorname{proj}_{\mathbf{u}}\mathbf{v} = \langle 5,0\rangle }. \]The projection isolates the horizontal contribution.
Practice 1: Compute a Dot Product
Find
\[ \langle 3,4\rangle \cdot \langle 2,-1\rangle. \]Practice 2: Three-Dimensional Dot Product
Find
\[ \langle 1,-2,3\rangle \cdot \langle 4,5,-1\rangle. \]Practice 3: Test Perpendicularity
Determine whether
\[ \mathbf{u}=\langle 2,5\rangle \]and
\[ \mathbf{v}=\langle 5,-2\rangle \]are perpendicular.
Practice 4: Determine Angle Type
Without finding the exact angle, determine whether the angle between
\[ \mathbf{u}=\langle 1,2\rangle \]and
\[ \mathbf{v}=\langle 3,1\rangle \]is acute, right, or obtuse.
Practice 5: Determine an Obtuse Relationship
Determine whether the angle between
\[ \mathbf{u}=\langle 1,0\rangle \]and
\[ \mathbf{v}=\langle -2,3\rangle \]is acute, right, or obtuse.
Practice 6: Find an Angle
Find the angle between
\[ \mathbf{u}=\langle 1,0\rangle \]and
\[ \mathbf{v}=\langle 0,1\rangle. \]Practice 7: Magnitude from a Dot Product
If
\[ \mathbf{v}\cdot\mathbf{v}=81, \]find
\[ \|\mathbf{v}\|. \]Practice 8: Unit-Vector Dot Product
Two unit vectors form an angle of \(60^\circ\).
Find their dot product.
Practice 9: Projection onto an Axis
Find the projection of
\[ \mathbf{v}=\langle 7,-3\rangle \]onto
\[ \mathbf{i}=\langle 1,0\rangle. \]Practice 10: Identify the Error
A student writes
\[ \langle 2,3\rangle \cdot \langle 4,-1\rangle = \langle 8,-3\rangle. \]Identify and correct the error.
Practice Answers
Practice 1
\[ \langle 3,4\rangle \cdot \langle 2,-1\rangle = 3(2)+4(-1) = \boxed{2}. \]Practice 2
\[ 1(4)+(-2)(5)+3(-1) = 4-10-3 = \boxed{-9}. \]Practice 3
\[ \mathbf{u}\cdot\mathbf{v} = 2(5)+5(-2) = 10-10 = 0. \]Therefore, the vectors are perpendicular.
Practice 4
\[ \mathbf{u}\cdot\mathbf{v} = 1(3)+2(1) = 5. \]Since the dot product is positive, the angle is acute.
Practice 5
\[ \mathbf{u}\cdot\mathbf{v} = 1(-2)+0(3) = -2. \]Since the dot product is negative, the angle is obtuse.
Practice 6
\[ \mathbf{u}\cdot\mathbf{v} = 0. \]Therefore,
\[ \boxed{\theta=90^\circ}. \]Practice 7
\[ \|\mathbf{v}\| = \sqrt{\mathbf{v}\cdot\mathbf{v}} = \sqrt{81} = \boxed{9}. \]Practice 8
Since both vectors are unit vectors,
\[ \mathbf{u}\cdot\mathbf{v} = \cos60^\circ = \boxed{\frac12}. \]Practice 9
\[ \mathbf{v}\cdot\mathbf{i} = 7. \]Since \(\mathbf{i}\) is a unit vector,
\[ \operatorname{proj}_{\mathbf{i}}\mathbf{v} = 7\mathbf{i} = \boxed{\langle 7,0\rangle}. \]Practice 10
The student stopped after multiplying corresponding components.
A dot product requires those products to be added:
\[ 2(4)+3(-1) = 8-3 = \boxed{5}. \]The result is a scalar, not a vector.
What to Remember
- The dot product multiplies corresponding components and adds the results.
- The dot product produces a scalar.
- \(\mathbf{v}\cdot\mathbf{v}=\|\mathbf{v}\|^2\).
- \(\mathbf{u}\cdot\mathbf{v}=\|\mathbf{u}\|\|\mathbf{v}\|\cos\theta\).
- A positive dot product corresponds to an acute angle between nonzero vectors.
- A zero dot product corresponds to perpendicular nonzero vectors.
- A negative dot product corresponds to an obtuse angle.
- The dot product can be used to calculate angles.
- Projection measures how much of one vector lies along another direction.
- The zero vector does not have a defined angle with other vectors.
Where We Go Next
The dot product measures alignment and angular relationships by producing a scalar. In three-dimensional geometry, another vector operation can produce a new vector perpendicular to two given vectors.
The next lesson, Introductory Cross Products, introduces this three-dimensional construction, its geometric meaning, orientation, and applications to perpendicular directions and area.