# 4 Vector Geometry

# 4.1 Vectors and Lines

In this chapter we study the geometry of 3-dimensional space. We view a point in 3-space as an arrow from the origin to that point. Doing so provides a “picture” of the point that is truly worth a thousand words.

## Vectors in

Introduce a coordinate system in 3-dimensional space in the usual way. First, choose a point called the , then choose three mutually perpendicular lines through , called the , , and , and establish a number scale on each axis with zero at the origin. Given a point in -space we associate three numbers , , and with , as described in Figure 4.1.1.

These numbers are called the of , and we denote the point as , or to emphasize the label . The result is called a coordinate system for 3-space, and the resulting description of 3-space is called .

As in the plane, we introduce vectors by identifying each point with the vector

in , represented by the from the origin to as in Figure 4.1.1. Informally, we say that the point has vector , and that vector has point . In this way 3-space is identified with , and this identification will be made throughout this chapter, often without comment. In particular, the terms “vector” and “point” are interchangeable. The resulting description of 3-space is called . Note that the origin is .

## Length and direction

We are going to discuss two fundamental geometric properties of vectors in : length and direction. First, if is a vector with point , the of vector is defined to be the distance from the origin to , that is the length of the arrow representing . The following properties of length will be used frequently.

Theorem 4.1.1

Let be a vector.

- .
- if and only if
- for all scalars .

Proof:

Let have point .

- In Figure 4.1.2, is the hypotenuse of the right triangle , and so by Pythagoras’ theorem. But is the hypotenuse of the right triangle , so . Now (1) follows by eliminating and taking positive square roots.
- If = 0, then by (1). Because squares of real numbers are nonnegative, it follows that , and hence that . The converse is because .
- We have so (1) gives
Hence , and we are done because for any real number .

Example 4.1.1

If

then . Similarly if

in 2-space then .

When we view two nonzero vectors as arrows emanating from the origin, it is clear geometrically what we mean by saying that they have the same or opposite . This leads to a fundamental new description of vectors.

Theorem 4.1.2

Let and be vectors in . Then as matrices if and only if and have the same direction and the same length.

Proof:

If , they clearly have the same direction and length. Conversely, let and be vectors with points and respectively. If and have the same length and direction then, geometrically, and must be the same point.

Hence , , and , that is .

Note that a vector’s length and direction do depend on the choice of coordinate system in . Such descriptions are important in applications because physical laws are often stated in terms of vectors, and these laws cannot depend on the particular coordinate system used to describe the situation.

## Geometric Vectors

If and are distinct points in space, the arrow from to has length and direction.

Hence,

Definition 4.1 Geometric vectors

Suppose that and are any two points in . In Figure 4.1.4 the line segment from to is denoted and is called the from to . Point is called the of , is called the and the is denoted .

Note that if is any vector in with point then is itself a geometric vector where is the origin. Referring to as a “vector” seems justified by Theorem 4.1.2 because it has a direction (from to ) and a length . However there appears to be a problem because two geometric vectors can have the same length and direction even if the tips and tails are different.

For example and in Figure 4.1.5 have the same length and the same direction (1 unit left and 2 units up) so, by Theorem 4.1.2, they are the same vector! The best way to understand this apparent paradox is to see and as different of the same underlying vector . Once it is clarified, this phenomenon is a great benefit because, thanks to Theorem 4.1.2, it means that the same geometric vector can be positioned anywhere in space; what is important is the length and direction, not the location of the tip and tail. This ability to move geometric vectors about is very useful.

## The Parallelogram Law

We now give an intrinsic description of the sum of two vectors and in , that is a description that depends only on the lengths and directions of and and not on the choice of coordinate system. Using Theorem 4.1.2 we can think of these vectors as having a common tail . If their tips are and respectively, then they both lie in a plane containing , , and , as shown in Figure 4.1.6. The vectors and create a parallelogram in , shaded in Figure 4.1.6, called the parallelogram by and .

If we now choose a coordinate system in the plane with as origin, then the parallelogram law in the plane shows that their sum is the diagonal of the parallelogram they determine with tail . This is an intrinsic description of the sum because it makes no reference to coordinates. This discussion proves:

The Parallelogram Law

In the parallelogram determined by two vectors and , the vector is the diagonal with the same tail as and .

Because a vector can be positioned with its tail at any point, the parallelogram law leads to another way to view vector addition. In Figure 4.1.7 (a) the sum of two vectors and is shown as given by the parallelogram law. If is moved so its tail coincides with the tip of (shown in (b)) then the sum is seen as “first and then . Similarly, moving the tail of to the tip of shows in (c) that is “first and then .” This will be referred to as the , and it gives a graphic illustration of why .

Since denotes the vector from a point to a point , the tip-to-tail rule takes the easily remembered form

for any points , , and .

One reason for the importance of the tip-to-tail rule is that it means two or more vectors can be added by placing them tip-to-tail in sequence. This gives a useful “picture” of the sum of several vectors, and is illustrated for three vectors in Figure 4.1.8 where is viewed as first , then , then .

There is a simple geometrical way to visualize the (matrix) of two vectors. If and are positioned so that they have a common tail , and if and are their respective tips, then the tip-to-tail rule gives . Hence is the vector from the tip of to the tip of . Thus both and appear as diagonals in the parallelogram determined by and (see Figure 4.1.9.

Theorem 4.1.3

If and have a common tail, then is the vector from the tip of to the tip of .

One of the most useful applications of vector subtraction is that it gives a simple formula for the vector from one point to another, and for the distance between the points.

Theorem 4.1.4

Let and be two points. Then:

- .
- The distance between and is

Can you prove these results?

Example 4.1.3

The distance between and is , and the vector from to is

.

The next theorem tells us what happens to the length and direction of a scalar multiple of a given vector.

Scalar Multiple Law

If a is a real number and is a vector then:

- The length of is .
- If , the direction of is the same as if ; opposite to if

Proof:

The first statement is true due to Theorem 4.1.1.

To prove the second statement, let denote the origin in Let have point , and choose any plane containing and . If we set up a coordinate system in this plane with as origin, then so the result follows from the scalar multiple law in the plane.

A vector is called a if . Then

, , and

are unit vectors, called the vectors.

Example 4.1.4

If show that is the unique unit vector in the same direction as

Solution:

The vectors in the same direction as are the scalar multiples where . But when , so is a unit vector if and only if .

Definition 4.2 Parallel vectors in

Two nonzero vectors are called if they have the same or opposite direction.

Theorem 4.1.5

Two nonzero vectors and are parallel if and only if one is a scalar multiple of the other.

Example 4.1.5

Given points , , , and , determine if and are parallel.

Solution:

By Theorem 4.1.3, and . If

then , so and , which is impossible. Hence is a scalar multiple of , so these vectors are not parallel by Theorem 4.1.5.

## Lines in Space

These vector techniques can be used to give a very simple way of describing straight lines in space. In order to do this, we first need a way to

specify the orientation of such a line.

Definition 4.3 Direction Vector of a Line

We call a nonzero vector a **direction vector **for the line if it is parallel to for some pair of distinct points and on the line.

Note that any nonzero scalar multiple of would also serve as a direction vector of the line.

We use the fact that there is exactly one line that passes through a particular point and has a given direction vector

. We want to describe this line by giving a condition on , , and that the point lies on this line. Let

and denote the vectors of and , respectively.

Then

Hence lies on the line if and only if is parallel to —that is, if and only if for some scalar by Theorem 4.1.5. Thus is the vector of a point on the line if and only if for some scalar .

Vector Equation of a line

The line parallel to through the point with vector is given by

In other words, the point with vector is on this line if and only if a real number t exists such that .

In component form the vector equation becomes

Equating components gives a different description of the line.

Parametric Equations of a line

The line through with direction vector

is given by

In other words, the point is on this line if and only if a real number exists such that , , and .

Example 4.1.6

Find the equations of the line through the points and .

Solution:

Let

denote the vector from to . Then is parallel to the line ( and are* on* the line), so serves as a direction vector for the line. Using as the point on the line leads to the parametric equations

Note that if is used (rather than ), the equations are

These are different from the preceding equations, but this is merely the result of a change of parameter. In fact, .

Example 4.1.7

Determine whether the following lines intersect and, if so, find the point of intersection.

Solution:

Suppose with vector lies on both lines. Then

where the first (second) equation is because lies on the first (second) line. Hence the lines intersect if and only if the three equations

have a solution. In this case, and satisfy all three equations, so the lines *do* intersect and the point of intersection is

using . Of course, this point can also be found from

using .

# 4.2 Projections and Planes

Suppose a point and a plane are given and it is desired to find the point that lies in the plane and is closest to , as shown in Figure 4.2.1.

Clearly, what is required is to find the line through that is perpendicular to the plane and then to obtain as the point of intersection of this line with the plane. Finding the line *perpendicular* to the plane requires a way to determine when two vectors are perpendicular. This can be done using the idea of the dot product of two vectors.

## The Dot Product and Angles

Definition 4.4 Dot Product in

Given vectors

and

, their **dot product ** is a number defined

Because is a number, it is sometimes called the **scalar product **of and

Example 4.2.1

If

and , then .

Theorem 4.2.1

Let , , and denote vectors in (or ).

- is a real number.
- .
- .
- .
- for all scalars .

The readers are invited to prove these properties using the definition of dot products.

Example 4.2.2

Verify that when , , and .

Solution:

We apply Theorem 4.2.1 several times:

There is an intrinsic description of the dot product of two nonzero vectors in . To understand it we require the following result from trigonometry.

Laws of Cosine

If a triangle has sides , , and , and if is the interior angle opposite then

Proof:

We prove it when is acute, that is ; the obtuse case is similar. In Figure 4.2.2 we have and .

Hence Pythagoras’ theorem gives

The law of cosines follows because for any angle .

Note that the law of cosines reduces to Pythagoras’ theorem if is a right angle (because ).

Now let and be nonzero vectors positioned with a common tail. Then they determine a unique angle in the range

This angle will be called the** angle between ** and . Clearly and are parallel if is either or . Note that we do not define the angle between and if one of these vectors is .

The next result gives an easy way to compute the angle between two nonzero vectors using the dot product.

Theorem 4.2.2

Let and be nonzero vectors. If is the angle between and , then

Proof:

We calculate in two ways. First apply the law of cosines to the triangle in Figure 4.2.4 to obtain:

On the other hand, we use Theorem 4.2.1:

Comparing these we see that , and the result follows.

If and are nonzero vectors, Theorem 4.2.2 gives an intrinsic description of because , , and the angle between and do not depend on the choice of coordinate system. Moreover, since and are nonzero ( and are nonzero vectors), it gives a formula for the cosine of the angle :

Since , this can be used to find .

Example 4.2.3

Compute the angle between

and

.

Solution:

Compute . Now recall that and are defined so that (, ) is the point on the unit circle determined by the angle (drawn counterclockwise, starting from the positive axis). In the present case, we know that and that . Because , it follows that .

If and are nonzero, the previous example shows that has the same sign as , so

In this last case, the (nonzero) vectors are perpendicular. The following terminology is used in linear algebra:

Definition 4.5 Orthogonal Vectors in

Two vectors and are said to be \textbf{orthogonal}\index{orthogonal vectors}\index{vectors!orthogonal vectors} if or or the angle between them is .

Since if either or , we have the following theorem:

Theorem 4.2.3

Two vectors and are orthogonal if and only if .

Example 4.2.4

Show that the points , , and are the vertices of a right triangle.

Solution:

The vectors along the sides of the triangle are

Evidently , so and are orthogonal vectors. This means sides and are perpendicular—that is, the angle at is a right angle.

## Projections

In applications of vectors, it is frequently useful to write a vector as the sum of two orthogonal vectors.

If a nonzero vector is specified, the key idea is to be able to write an arbitrary vector as a sum of two vectors,

where is parallel to and is orthogonal to . Suppose that and emanate from a common tail (see Figure 4.2.5). Let be the tip of , and let denote the foot of the perpendicular from to the line through parallel to .

Then has the required properties:

1. *is parallel to* .

2. *is orthogonal to* .

3. .

Definition 4.6 Projection in

The vector in Figure 4.2.6 is called *the projection of on . *

It is denoted

In Figure 4.2.5 (a) the vector has the same direction as ; however, and have opposite directions if the angle between and is greater than (see Figure 4.2.5 (b)). Note that the projection is zero if and only if and are orthogonal.

Calculating the projection of on is remarkably easy.

Theorem 4.2.4

Let and be vectors.

- The projection of on is given by .
- The vector is orthogonal to .

Proof:

The vector is parallel to and so has the form for some scalar . The requirement that and are orthogonal determines . In fact, it means that by Theorem 4.2.3. If is substituted here, the condition is

It follows that , where the assumption that guarantees that .

Example 4.2.5

Find the projection of

on

and express where is parallel to and is orthogonal to .

Solution:

The projection of on is

Hence , and this is orthogonal to by Theorem 4.2.4 (alternatively, observe that ). Since , we are done.

Note that the idea of projections can be used to find the shortest distance from a point to a straight line in which is the length of the vector that’s orthogonal to the direction vector of the line.

## Planes

Definition 4.7 Normal vector in a plane

A nonzero vector is called a **normal** for a plane if it is orthogonal to every vector in the plane.

For example, the unit vector is a normal vector for plane.

Given a point and a nonzero vector , there is a unique plane through with normal , shaded in Figure 4.2.6. A point lies on this plane if and only if the vector is orthogonal to —that is, if and only if . Because this gives the following result:

Scalar equation of a plane

The plane through with normal

as a normal vector is given by

In other words, a point is on this plane if and only if , , and satisfy this equation.

Example 4.2.8

Find an equation of the plane through with

as normal.

Solution:

Here the general scalar equation becomes

This simplifies to .

If we write , the scalar equation shows that every plane with normal

has a linear equation of the form

(4.2)

for some constant . Conversely, the graph of this equation is a plane with as a normal vector (assuming that , , and are not all zero).

Example 4.2.9

Find an equation of the plane through that is parallel to the plane with equation .

Solution:

The plane with equation has normal . Because the two planes are parallel, serves as a normal for the plane we seek, so the equation is for some according to (4.2). Insisting that lies on the plane determines ; that is, . Hence, the equation is .

Consider points and with vectors

and

.

Given a nonzero vector , the scalar equation of the plane through with normal takes the vector form:

Vector Equation of a Plane

The plane with normal through the point with vector is given by

In other words, the point with vector is on the plane if and only if satisfies this condition.

Moreover, Equation (4.2) translates as follows:

Every plane with normal has vector equation for some number .

Example 4.2.10

Find the shortest distance from the point to the plane with equation . Also find the point on this plane closest to .

Solution:

The plane in question has normal . Choose any point on the plane—say —and let be the point on the plane closest to (see the diagram). The vector from to is . Now erect with its tail at . Then and is the projection of on :

Hence the distance is . To calculate the point , let

and

be the vectors of and . Then

This gives the coordinates of .

## The Cross Product

If , , and are three distinct points in that are not all on some line, it is clear geometrically that there is a unique plane containing all three. The vectors and both lie in this plane, so finding a normal amounts to finding a nonzero vector orthogonal to both and . The cross product provides a systematic way to do this.

Definition 4.8 Cross Product

Given vectors and , define the **cross product** by

Because it is a vector, is often called the **vector product.** There is an easy way to remember this definition using the **coordinate vectors:**

They are vectors of length pointing along the positive , , and axes. The reason for the name is that any vector can be written as

With this, the cross product can be described as follows:

Determinant form of the cross product

If and are two vectors, then

where the determinant is expanded along the first column.

Example 4.2.11

If and , then

Observe that is orthogonal to both and in Example 4.2.11. This holds in general as can be verified directly by computing and , and is recorded as the first part of the following theorem. It will follow from a more general result which, together with the second part, will be proved later on.

Theorem 4.2.5

Let and be vectors in :

- is a vector orthogonal to both and .
- If and are nonzero, then if and only if and are parallel.

Recall that

Example 4.2.12

Find the equation of the plane through , , and .

Solution:

The vectors

and

lie in the plane, so

is a normal for the plane (being orthogonal to both and ). Hence the plane has equation

Since lies in the plane we have . Hence and the equation is . Can you verify that he same equation can be obtained if and , or and , are used as the vectors in the plane?

# 4.3 More on the Cross Product

The cross product of two -vectors and

was defined in Section 4.2 where we observed that it can be best remembered using a determinant:

(4.3)

Here , , and

are the coordinate vectors, and the determinant is expanded along the first column. We observed (but did not prove) in Theorem 4.2.5 that is orthogonal to both and . This follows easily from the next result.

Theorem 4.3.1

If , , and , then .

Proof:

Recall that is computed by multiplying corresponding components of and and then adding. Using equation (4.3), the result is:

where the last determinant is expanded along column 1.

The result in Theorem 4.3.1 can be succinctly stated as follows: If , , and are three vectors in , then

where denotes the matrix with , , and as its columns. Now it is clear that is orthogonal to both and because the determinant of a matrix is zero if two columns are identical.

Because of (4.3) and Theorem 4.3.1, several of the following properties of the cross product follow from

properties of determinants (they can also be verified directly).

Theorem 4.3.2

Let , , and denote arbitrary vectors in .

- is a vector.
- is orthogonal to both and .
- .
- .
- .
- for any scalar .
- .
- .

We have seen some of these results in the past; can you prove 6,7, and 8?

We now come to a fundamental relationship between the dot and cross products.

Theorem 4.3.3 Lagrange Identity

If and are any two vectors in , then

Proof:

Given and , introduce a coordinate system and write

and

in component form. Then all the terms in the identity can be computed in terms of the components.

An expression for the magnitude of the vector can be easily obtained from the Lagrange identity. If is the angle between and , substituting into the Lagrange identity gives

using the fact that . But is nonnegative on the range , so taking the positive square root of both sides gives

This expression for makes no reference to a coordinate system and, moreover, it has a nice geometrical interpretation. The parallelogram determined by the vectors and has base length and altitude . Hence the area of the parallelogram formed by and is

Theorem 4.3.4

If and are two nonzero vectors and is the angle between and , then:

- the area of the parallelogram determined by and .
- and are parallel if and only if .

Proof of 2:

By (1), if and only if the area of the parallelogram is zero. The area vanishes if and only if and have the same or opposite direction—that is, if and only if they are parallel.

Example 4.3.1

Find the area of the triangle with vertices , , and .

Solution:

We have

and . The area of the triangle is half the area of the parallelogram formed by these vectors, and so equals . We have

so the area of the triangle is

If three vectors , , and are given, they determine a “squashed” rectangular solid called a **parallelepiped** (Figure 4.3.2), and it is often useful to be able to find the volume of such a solid. The base of the solid is the parallelogram determined by and , so it has area . The height of the solid is the length of the projection of on . Hence

Thus the volume of the parallelepiped is . This proves

Theorem 4.3.5

The volume of the parallelepiped determined by three vectors , , and is given by .

Example 4.3.2

Find the volume of the parallelepiped determined by the vectors

Solution:

By Theorem 4.3.1, .

Hence the volume is by Theorem 4.3.5.

We can now give an intrinsic description of the cross product .

Right-hand Rule

If the vector is grasped in the right hand and the fingers curl around from to through the angle , the thumb points in the direction for

To indicate why this is true, introduce coordinates in as follows: Let and have a common tail , choose the origin at , choose the axis so that points in the positive direction, and then choose the axis so that is in the – plane and the positive axis is on the same side of the axis as . Then, in this system, and have component form

and

where and . Can you draw a graph based on the description here?

The right-hand rule asserts that should point in the positive direction. But our definition of gives

and has the positive direction because .