Scalar Multiplication and Linear Combinations | Malone Math

The Central Idea

A scalar is an ordinary numerical quantity.

Multiplying a vector by a scalar changes the vector according to that number.

\[ \mathbf{v}=\langle a,b\rangle \]

and \(k\) is a scalar, then

\[ \boxed{ k\mathbf{v} = \langle ka,kb\rangle }. \]

Every component is multiplied by the same scalar.

What Scalar Multiplication Does Geometrically

Multiplying a vector by a scalar can change its magnitude and possibly its direction.

For a nonzero vector \(\mathbf{v}\):

if \(k>1\), the vector becomes longer;
if \(0
if \(k=1\), the vector is unchanged;
if \(k=0\), the result is the zero vector;
if \(k<0\), the vector reverses direction and is scaled by \(|k|\).

Example: Positive Scaling

Let

\[ \mathbf{v}=\langle 2,3\rangle. \]

Multiply by \(4\):

\[ 4\mathbf{v} = 4\langle 2,3\rangle. \]

Multiply each component:

\[ 4\mathbf{v} = \langle 8,12\rangle. \]

Therefore,

\[ \boxed{ 4\mathbf{v} = \langle 8,12\rangle }. \]

Positive Scalars Preserve Direction

If \(k>0\), then \(k\mathbf{v}\) points in the same direction as \(\mathbf{v}\).

For example,

\[ \mathbf{v}=\langle 2,1\rangle \]

and

\[ 3\mathbf{v}=\langle 6,3\rangle \]

point in the same direction.

Only the magnitude changes.

Scaling by a Fraction

Suppose

\[ \mathbf{v}=\langle 8,4\rangle. \]

Multiplying by

\[ \frac12 \]

gives

\[ \frac12\mathbf{v} = \left\langle \frac12(8),\frac12(4) \right\rangle. \]

Therefore,

\[ \boxed{ \frac12\mathbf{v} = \langle 4,2\rangle }. \]

The direction is unchanged, while the magnitude is halved.

Negative Scalars Reverse Direction

Let

\[ \mathbf{v}=\langle 3,-2\rangle. \]

Then

\[ -2\mathbf{v} = \langle -6,4\rangle. \]

The factor \(2\) doubles the magnitude, while the negative sign reverses the direction.

Multiplication by \(-1\)

The simplest negative scalar is \(-1\).

For any vector \(\mathbf{v}\),

\[ \boxed{ (-1)\mathbf{v} = -\mathbf{v} }. \]

This produces the opposite vector.

For example,

\[ -\langle 5,-3\rangle = \langle -5,3\rangle. \]

Multiplication by Zero

For every vector \(\mathbf{v}\),

\[ \boxed{ 0\mathbf{v} = \mathbf{0} }. \]

In component form,

\[ 0\langle a,b\rangle = \langle 0,0\rangle. \]

Scaling by zero removes all directed magnitude.

Multiplication by One

The scalar \(1\) leaves a vector unchanged:

\[ \boxed{ 1\mathbf{v} = \mathbf{v} }. \]

Thus \(1\) acts as the multiplicative identity for scalar multiplication of vectors.

Magnitude under Scalar Multiplication

If \(k\) is a scalar, then

\[ \boxed{ \|k\mathbf{v}\| = |k|\,\|\mathbf{v}\| }. \]

The absolute value appears because magnitude cannot be negative.

Example: Magnitude after Scaling

Suppose

\[ \|\mathbf{v}\|=5. \]

Then

\[ \|-3\mathbf{v}\| = |-3|\,\|\mathbf{v}\|. \]

Therefore,

\[ \boxed{ \|-3\mathbf{v}\| = 15 }. \]

The negative sign reverses direction but does not make magnitude negative.

Verifying the Magnitude Rule with Components

Let

\[ \mathbf{v}=\langle a,b\rangle. \]

Then

\[ k\mathbf{v} = \langle ka,kb\rangle. \]

Its magnitude is

\[ \sqrt{(ka)^2+(kb)^2}. \]

Factor \(k^2\):

\[ \sqrt{k^2(a^2+b^2)} = |k|\sqrt{a^2+b^2}. \]

Hence,

\[ \boxed{ \|k\mathbf{v}\| = |k|\,\|\mathbf{v}\| }. \]

Scalar Multiples Are Parallel

\[ \mathbf{u}=k\mathbf{v} \]

for some nonzero scalar \(k\), then \(\mathbf{u}\) and \(\mathbf{v}\) point along the same line of direction.

If \(k>0\), they point in the same direction.
If \(k<0\), they point in opposite directions.

This provides an algebraic test for parallel nonzero vectors.

Example: Recognizing Scalar Multiples

Compare

\[ \mathbf{u}=\langle 2,5\rangle \]

and

\[ \mathbf{v}=\langle 6,15\rangle. \]

Since

\[ \mathbf{v} = 3\mathbf{u}, \]

the vectors are parallel and point in the same direction.

Example: Opposite Parallel Directions

Compare

\[ \mathbf{u}=\langle 4,-2\rangle \]

and

\[ \mathbf{v}=\langle -6,3\rangle. \]

Since

\[ \mathbf{v} = -\frac32\mathbf{u}, \]

the vectors are parallel but point in opposite directions.

Not Every Pair of Vectors Are Scalar Multiples

Consider

\[ \mathbf{u}=\langle 2,3\rangle \]

and

\[ \mathbf{v}=\langle 4,7\rangle. \]

If \(\mathbf{v}=k\mathbf{u}\), then the first components would require

\[ k=2. \]

But then the second component would have to be

\[ 2(3)=6, \]

not \(7\).

Therefore, the vectors are not scalar multiples.

Distributive Property over Vector Addition

Scalar multiplication distributes over vector addition:

\[ \boxed{ k(\mathbf{u}+\mathbf{v}) = k\mathbf{u}+k\mathbf{v} }. \]

This mirrors the ordinary distributive property of arithmetic and algebra.

Verifying Distribution with Components

Let

\[ \mathbf{u}=\langle a,b\rangle, \qquad \mathbf{v}=\langle c,d\rangle. \]

Then

\[ k(\mathbf{u}+\mathbf{v}) = k\langle a+c,b+d\rangle. \]

Therefore,

\[ \langle k(a+c),k(b+d)\rangle = \langle ka+kc,kb+kd\rangle. \]

This equals

\[ k\mathbf{u}+k\mathbf{v}. \]

Distributive Property over Scalar Addition

If \(a\) and \(b\) are scalars, then

\[ \boxed{ (a+b)\mathbf{v} = a\mathbf{v}+b\mathbf{v} }. \]

One vector may therefore be scaled by a sum or by the corresponding separate scalar contributions.

Associativity of Scalar Multiplication

If \(a\) and \(b\) are scalars, then

\[ \boxed{ a(b\mathbf{v}) = (ab)\mathbf{v} }. \]

Scaling a vector and then scaling again has the same effect as multiplying the scalars first.

What Is a Linear Combination?

A linear combination is a sum of scalar multiples of vectors.

With two vectors,

\[ \boxed{ a\mathbf{u}+b\mathbf{v} }. \]

Here \(a\) and \(b\) are scalars.

The scalars are often called coefficients or weights.

Example: A Simple Linear Combination

Let

\[ \mathbf{u}=\langle 1,2\rangle, \qquad \mathbf{v}=\langle 3,-1\rangle. \]

Find

\[ 2\mathbf{u}+3\mathbf{v}. \]

First scale:

\[ 2\mathbf{u} = \langle 2,4\rangle, \] \[ 3\mathbf{v} = \langle 9,-3\rangle. \]

Then add:

\[ 2\mathbf{u}+3\mathbf{v} = \boxed{\langle 11,1\rangle}. \]

Linear Combinations Are Built in Two Stages

A linear combination combines two familiar operations:

scale each vector;
add the scaled vectors.

No new arithmetic operation is required.

The power comes from using these two operations together.

Linear Combinations of More Than Two Vectors

With vectors

\[ \mathbf{v}_1,\mathbf{v}_2,\ldots,\mathbf{v}_n, \]

a general linear combination has the form

\[ \boxed{ c_1\mathbf{v}_1 + c_2\mathbf{v}_2 + \cdots + c_n\mathbf{v}_n }. \]

Each \(c_i\) is a scalar coefficient.

The Standard Unit Vectors Build Every Plane Vector

Recall

\[ \mathbf{i}=\langle 1,0\rangle, \qquad \mathbf{j}=\langle 0,1\rangle. \]

Any vector

\[ \mathbf{v}=\langle a,b\rangle \]

can be written

\[ \boxed{ \mathbf{v} = a\mathbf{i}+b\mathbf{j} }. \]

Thus every ordinary plane vector is a linear combination of \(\mathbf{i}\) and \(\mathbf{j}\).

Example: Build a Vector from \(\mathbf{i}\) and \(\mathbf{j}\)

The vector

\[ \langle -4,7\rangle \]

\[ -4\langle 1,0\rangle + 7\langle 0,1\rangle. \]

Therefore,

\[ \boxed{ \langle -4,7\rangle = -4\mathbf{i}+7\mathbf{j} }. \]

Building Vectors from Other Directions

The standard unit vectors are convenient, but linear combinations can use any available vectors.

Suppose

\[ \mathbf{u}=\langle 1,1\rangle, \qquad \mathbf{v}=\langle 1,-1\rangle. \]

Then different choices of scalars \(a\) and \(b\) produce vectors of the form

\[ a\mathbf{u}+b\mathbf{v}. \]

Example: Construct a Target Vector

Using

\[ \mathbf{u}=\langle 1,1\rangle, \qquad \mathbf{v}=\langle 1,-1\rangle, \]

build

\[ \langle 6,2\rangle. \]

We want

\[ a\mathbf{u}+b\mathbf{v} = \langle 6,2\rangle. \]

Component-wise,

\[ \langle a+b,\;a-b\rangle = \langle 6,2\rangle. \]

Solving for the Coefficients

From

\[ a+b=6 \]

and

\[ a-b=2, \]

add the equations:

\[ 2a=8. \]

Thus,

\[ a=4. \]

Then

\[ b=2. \]

Therefore,

\[ \boxed{ \langle 6,2\rangle = 4\mathbf{u}+2\mathbf{v} }. \]

Linear Combination Problems Become Systems of Equations

Suppose

\[ a\mathbf{u}+b\mathbf{v} = \mathbf{w}. \]

Writing all vectors in component form creates one scalar equation for each component.

In two dimensions, this usually produces a system of two equations.

This connects vector algebra directly to systems of linear equations.

Example: Determine Whether a Target Can Be Built

Let

\[ \mathbf{u}=\langle 2,1\rangle, \qquad \mathbf{v}=\langle 1,3\rangle. \]

Can

\[ \mathbf{w}=\langle 5,5\rangle \]

be written as

\[ a\mathbf{u}+b\mathbf{v}? \]

We require

\[ 2a+b=5 \]

and

\[ a+3b=5. \]

Solving the Construction Problem

From

\[ 2a+b=5, \]

we have

\[ b=5-2a. \]

Substitute into

\[ a+3b=5: \] \[ a+3(5-2a)=5. \]

Then

\[ -5a=-10, \qquad a=2. \]

Therefore,

\[ b=1. \]

\[ \boxed{ \mathbf{w} = 2\mathbf{u}+\mathbf{v} }. \]

One Vector Can Only Build a Line of Directions

Fix a nonzero vector \(\mathbf{v}\).

All scalar multiples

\[ k\mathbf{v} \]

lie along the same line through the origin.

Positive \(k\) values produce one direction, negative values the opposite direction, and \(k=0\) gives the origin.

The Idea of Span

The collection of all linear combinations of a set of vectors is called their span.

For one vector,

\[ \operatorname{span}\{\mathbf{v}\} \]

consists of all scalar multiples of \(\mathbf{v}\).

In the plane, this is usually a line through the origin when \(\mathbf{v}\neq\mathbf{0}\).

Two Nonparallel Plane Vectors Can Build the Plane

If two vectors in the plane are not parallel, their directions provide two independent ways to move.

By choosing suitable scalar coefficients, every vector in the plane can be expressed as a linear combination of them.

This is an introductory glimpse of the ideas of span, basis, and independence.

Two Parallel Vectors Do Not Add a New Direction

Suppose

\[ \mathbf{v}=3\mathbf{u}. \]

Then

\[ a\mathbf{u}+b\mathbf{v} = a\mathbf{u}+3b\mathbf{u}. \]

Therefore,

\[ a\mathbf{u}+b\mathbf{v} = (a+3b)\mathbf{u}. \]

Every linear combination remains on the same line.

Linear Dependence: A First Look

When one vector can already be built from the others, it does not add a new independent direction.

For example, if

\[ \mathbf{v}=2\mathbf{u}, \]

then the pair contains directional redundancy.

Later linear algebra formalizes this idea as linear dependence.

Linear Independence: A First Look

Two nonzero, nonparallel vectors in the plane provide distinct directions.

Neither is a scalar multiple of the other.

This is the basic geometric picture behind linear independence in two dimensions.

A later lesson develops independence more formally.

Weighted Combinations

A linear combination can be viewed as a weighted sum.

\[ a\mathbf{u}+b\mathbf{v}, \]

the scalars \(a\) and \(b\) determine how much of each vector contributes to the result.

The weights may be positive, negative, zero, fractional, or larger than \(1\).

Equal-Weight Averaging

Consider

\[ \frac12\mathbf{u} + \frac12\mathbf{v}. \]

This gives the vector average of \(\mathbf{u}\) and \(\mathbf{v}\).

In component form, if

\[ \mathbf{u}=\langle a,b\rangle, \qquad \mathbf{v}=\langle c,d\rangle, \]

then

\[ \frac12\mathbf{u} + \frac12\mathbf{v} = \left\langle \frac{a+c}{2}, \frac{b+d}{2} \right\rangle. \]

Connection to Midpoints

If \(\mathbf{a}\) and \(\mathbf{b}\) are position vectors of points \(A\) and \(B\), then the midpoint has position vector

\[ \boxed{ \frac12\mathbf{a} + \frac12\mathbf{b} }. \]

Thus the familiar midpoint formula is a linear combination of two position vectors.

More General Weighted Points

Expressions such as

\[ (1-t)\mathbf{a}+t\mathbf{b} \]

create points along the line determined by position vectors \(\mathbf{a}\) and \(\mathbf{b}\).

When

\[ 0\le t\le 1, \]

the resulting point lies between the two endpoints.

Example: A Point Halfway Between Two Position Vectors

Let

\[ \mathbf{a}=\langle 2,4\rangle, \qquad \mathbf{b}=\langle 8,10\rangle. \]

The halfway position is

\[ \frac12\mathbf{a} + \frac12\mathbf{b}. \]

Therefore,

\[ \left\langle 1,2\right\rangle + \left\langle 4,5\right\rangle = \boxed{\langle 5,7\rangle}. \]

Three-Dimensional Scalar Multiplication

\[ \mathbf{v}=\langle a,b,c\rangle, \]

then

\[ \boxed{ k\mathbf{v} = \langle ka,kb,kc\rangle }. \]

The rule is exactly the same as in two dimensions.

Example: Scale a Three-Dimensional Vector

Let

\[ \mathbf{v}=\langle 2,-1,4\rangle. \]

Then

\[ -3\mathbf{v} = \langle -6,3,-12\rangle. \]

Therefore,

\[ \boxed{ -3\mathbf{v} = \langle -6,3,-12\rangle }. \]

Three-Dimensional Linear Combinations

Suppose

\[ \mathbf{u}=\langle 1,0,2\rangle \]

and

\[ \mathbf{v}=\langle 0,3,-1\rangle. \]

Then

\[ 2\mathbf{u}-\mathbf{v} = \langle 2,0,4\rangle - \langle 0,3,-1\rangle. \]

Therefore,

\[ \boxed{ 2\mathbf{u}-\mathbf{v} = \langle 2,-3,5\rangle }. \]

The Standard Basis in Three Dimensions

Recall

\[ \mathbf{i}=\langle 1,0,0\rangle, \qquad \mathbf{j}=\langle 0,1,0\rangle, \qquad \mathbf{k}=\langle 0,0,1\rangle. \]

Any vector

\[ \langle a,b,c\rangle \]

can be written

\[ \boxed{ a\mathbf{i} + b\mathbf{j} + c\mathbf{k} }. \]

Scalar Multiplication in Higher Dimensions

For

\[ \mathbf{v} = \langle x_1,x_2,\ldots,x_n\rangle, \]

scalar multiplication is

\[ \boxed{ k\mathbf{v} = \langle kx_1,kx_2,\ldots,kx_n \rangle }. \]

The algebraic rule does not depend on our ability to visualize the space.

Why Linear Combinations Matter

Linear combinations are a basic construction throughout mathematics.

They are used to:

build vectors from coordinate directions;
describe net effects from weighted inputs;
represent points along lines and planes;
study span and dimension;
solve systems of linear equations;
describe transformations;
analyze data and models.

Common Error: Multiplying Only One Component

\[ \mathbf{v}=\langle 2,5\rangle, \]

then

\[ 3\mathbf{v} \]

is not

\[ \langle 6,5\rangle. \]

The scalar multiplies every component:

\[ \boxed{ 3\mathbf{v} = \langle 6,15\rangle }. \]

Common Error: Forgetting Direction Reversal

Multiplying by a negative scalar does more than change component signs algebraically.

It reverses the geometric direction of the vector.

For example,

\[ -2\langle 3,1\rangle = \langle -6,-2\rangle. \]

The new vector points exactly opposite the original.

Common Error: Making Magnitude Negative

\[ \|\mathbf{v}\|=4, \]

then

\[ \|-3\mathbf{v}\| \]

is not \(-12\).

Magnitude is nonnegative:

\[ \boxed{ \|-3\mathbf{v}\|=12 }. \]

Common Error: Treating a Linear Combination as Ordinary Multiplication

The expression

\[ 2\mathbf{u}+3\mathbf{v} \]

means scale \(\mathbf{u}\) by \(2\), scale \(\mathbf{v}\) by \(3\), and then add.

It is not a product of the two vectors.

Common Error: Using Different Coefficients for Components of One Vector

A scalar multiplies an entire vector uniformly.

For example,

\[ 2\langle 3,4\rangle = \langle 6,8\rangle. \]

Applying one multiplier to the first component and another to the second is not ordinary scalar multiplication.

Common Error: Assuming Any Two Vectors Span the Plane

Two vectors do not automatically provide two independent directions.

If one is a scalar multiple of the other, all of their linear combinations remain on one line.

Nonparallel directions are required to span the whole plane.

Common Error: Solving Only One Component Equation

To verify

\[ a\mathbf{u}+b\mathbf{v} = \mathbf{w}, \]

every component must match.

A coefficient choice that works for only one component does not establish vector equality.

A Reliable Procedure for Scalar Multiplication

Identify the scalar.
Multiply every component by that scalar.
Simplify each component.
If the scalar is negative, recognize that direction reverses.
If magnitude is needed, multiply the original magnitude by the scalar's absolute value.

A Reliable Procedure for a Linear Combination

Identify each vector and its scalar coefficient.
Scale each vector separately.
Add the resulting vectors component by component.
Simplify the final components.
Check the result by expanding the calculation directly if needed.

A Reliable Procedure for Building a Target Vector

Write the target as a proposed linear combination.
Replace each vector with component form.
Match corresponding components.
Form the resulting system of scalar equations.
Solve for the coefficients.
Substitute the coefficients back into the vector combination.
Verify that every target component is recovered.

Example: Full Linear-Combination Workflow

Let

\[ \mathbf{u}=\langle 2,-1\rangle, \qquad \mathbf{v}=\langle 3,4\rangle. \]

Find

\[ -2\mathbf{u}+3\mathbf{v}. \]

First scale:

\[ -2\mathbf{u} = \langle -4,2\rangle, \] \[ 3\mathbf{v} = \langle 9,12\rangle. \]

Then add:

\[ \langle -4,2\rangle + \langle 9,12\rangle = \boxed{\langle 5,14\rangle}. \]

Example: Full Target-Construction Workflow

Let

\[ \mathbf{u}=\langle 1,2\rangle, \qquad \mathbf{v}=\langle 2,-1\rangle. \]

Find \(a\) and \(b\) such that

\[ a\mathbf{u}+b\mathbf{v} = \langle 8,1\rangle. \]

Component matching gives

\[ a+2b=8 \]

and

\[ 2a-b=1. \]

Solving the Target-Construction System

From

\[ 2a-b=1, \]

we obtain

\[ b=2a-1. \]

Substitute into

\[ a+2b=8: \] \[ a+2(2a-1)=8. \]

Therefore,

\[ 5a=10, \qquad a=2. \]

Then

\[ b=3. \]

Hence,

\[ \boxed{ \langle 8,1\rangle = 2\mathbf{u}+3\mathbf{v} }. \]

Practice 1: Positive Scalar

Let

\[ \mathbf{v}=\langle 3,-2\rangle. \]

Find

\[ 4\mathbf{v}. \]

Practice 2: Negative Scalar

Let

\[ \mathbf{v}=\langle -5,1\rangle. \]

Find

\[ -2\mathbf{v}. \]

Practice 3: Fractional Scaling

Find

\[ \frac13\langle 9,-6\rangle. \]

Practice 4: Magnitude after Scaling

\[ \|\mathbf{v}\|=7, \]

find

\[ \|-4\mathbf{v}\|. \]

Practice 5: Recognize Scalar Multiples

Determine whether

\[ \mathbf{u}=\langle 3,5\rangle \]

and

\[ \mathbf{v}=\langle -6,-10\rangle \]

are scalar multiples.

Practice 6: Linear Combination

Let

\[ \mathbf{u}=\langle 2,1\rangle, \qquad \mathbf{v}=\langle -1,3\rangle. \]

Find

\[ 3\mathbf{u}+2\mathbf{v}. \]

Practice 7: Build with Unit Vectors

Write

\[ \langle 6,-4\rangle \]

as a linear combination of \(\mathbf{i}\) and \(\mathbf{j}\).

Practice 8: Three-Dimensional Combination

Let

\[ \mathbf{u}=\langle 1,2,0\rangle, \qquad \mathbf{v}=\langle 3,-1,4\rangle. \]

Find

\[ 2\mathbf{u}-\mathbf{v}. \]

Practice 9: Find Coefficients

Let

\[ \mathbf{u}=\langle 1,1\rangle, \qquad \mathbf{v}=\langle 1,-1\rangle. \]

Find \(a\) and \(b\) such that

\[ a\mathbf{u}+b\mathbf{v} = \langle 10,4\rangle. \]

Practice 10: Identify the Error

A student writes

\[ -3\langle 2,-4\rangle = \langle -6,-12\rangle. \]

Identify and correct the error.

Practice Answers

Practice 1

\[ 4\langle 3,-2\rangle = \boxed{\langle 12,-8\rangle}. \]

Practice 2

\[ -2\langle -5,1\rangle = \boxed{\langle 10,-2\rangle}. \]

Practice 3

\[ \frac13\langle 9,-6\rangle = \boxed{\langle 3,-2\rangle}. \]

Practice 4

\[ \|-4\mathbf{v}\| = |-4|\|\mathbf{v}\| = 4(7) = \boxed{28}. \]

Practice 5

Yes. Since

\[ \mathbf{v} = -2\mathbf{u}, \]

the vectors are scalar multiples. They are parallel and point in opposite directions.

Practice 6

\[ 3\mathbf{u} = \langle 6,3\rangle, \] \[ 2\mathbf{v} = \langle -2,6\rangle. \]

Therefore,

\[ \boxed{ 3\mathbf{u}+2\mathbf{v} = \langle 4,9\rangle }. \]

Practice 7

\[ \boxed{ \langle 6,-4\rangle = 6\mathbf{i}-4\mathbf{j} }. \]

Practice 8

\[ 2\mathbf{u} = \langle 2,4,0\rangle. \]

Therefore,

\[ 2\mathbf{u}-\mathbf{v} = \langle 2-3,\;4-(-1),\;0-4\rangle = \boxed{\langle -1,5,-4\rangle}. \]

Practice 9

Matching components gives

\[ a+b=10 \]

and

\[ a-b=4. \]

Adding the equations gives

\[ 2a=14, \qquad a=7. \]

Then

\[ b=3. \]

Therefore,

\[ \boxed{ a=7,\qquad b=3 }. \]

Practice 10

The second component was multiplied incorrectly.

\[ (-3)(-4)=12, \]

not \(-12\).

Therefore,

\[ \boxed{ -3\langle 2,-4\rangle = \langle -6,12\rangle }. \]

What to Remember

Scalar multiplication multiplies every vector component by the same scalar.
Positive scalars preserve direction.
Negative scalars reverse direction.
Multiplication by zero produces the zero vector.
\(\|k\mathbf{v}\|=|k|\,\|\mathbf{v}\|\).
Nonzero scalar multiples are parallel.
A linear combination is a sum of scalar multiples of vectors.
Component equations can be used to find unknown linear-combination coefficients.
The span of vectors is the collection of all their linear combinations.
Two nonparallel vectors span the ordinary coordinate plane.
The standard unit vectors provide the usual coordinate directions.

Where We Go Next

Scalar multiplication and linear combinations let us build and compare vectors using their directions and weights. The next step is to introduce a new operation that combines two vectors and produces a scalar.

The next lesson, Dot Products and Angular Relationships, develops the dot product, connects it to vector magnitude and the angle between vectors, and provides an algebraic test for perpendicularity.