Vectors in Three Dimensions

Vectors are useful tools for solving two-dimensional problems. Life, however, happens in three dimensions. To expand the use of vectors to more realistic applications, it is necessary to create a framework for describing three-dimensional space. For example, although a two-dimensional map is a useful tool for navigating from one place to another, in some cases the topography of the land is important. Does your planned route go through the mountains? Do you have to cross a river? To appreciate fully the impact of these geographic features, you must use three dimensions. This section presents a natural extension of the two-dimensional Cartesian coordinate plane into three dimensions.

Three-Dimensional Coordinate Systems

As we have learned, the two-dimensional rectangular coordinate system contains two perpendicular axes: the horizontal x-axis and the vertical y-axis. We can add a third dimension, the z-axis, which is perpendicular to both the x-axis and the y-axis. We call this system the three-dimensional rectangular coordinate system. It represents the three dimensions we encounter in real life.

In [link](a), the positive z-axis is shown above the plane containing the x- and y-axes. The positive x-axis appears to the left and the positive y-axis is to the right. A natural question to ask is: How was arrangement determined? The system displayed follows the right-hand rule. If we take our right hand and align the fingers with the positive x-axis, then curl the fingers so they point in the direction of the positive y-axis, our thumb points in the direction of the positive z-axis. In this text, we always work with coordinate systems set up in accordance with the right-hand rule. Some systems do follow a left-hand rule, but the right-hand rule is considered the standard representation.

In two dimensions, we describe a point in the plane with the coordinates

(x, y) .

Each coordinate describes how the point aligns with the corresponding axis. In three dimensions, a new coordinate,

z,

A point in space is identified by all three coordinates ([link]). To plot the point

(x, y, z),

In two-dimensional space, the coordinate plane is defined by a pair of perpendicular axes. These axes allow us to name any location within the plane. In three dimensions, we define coordinate planes by the coordinate axes, just as in two dimensions. There are three axes now, so there are three intersecting pairs of axes. Each pair of axes forms a coordinate plane: the xy-plane, the xz-plane, and the yz-plane ([link]). We define the xy-plane formally as the following set:

{(x, y, 0) : x, y \in ℝ} .

Similarly, the xz-plane and the yz-plane are defined as

{(x, 0, z) : x, z \in ℝ}

To visualize this, imagine you’re building a house and are standing in a room with only two of the four walls finished. (Assume the two finished walls are adjacent to each other.) If you stand with your back to the corner where the two finished walls meet, facing out into the room, the floor is the xy-plane, the wall to your right is the xz-plane, and the wall to your left is the yz-plane.

In two dimensions, the coordinate axes partition the plane into four quadrants. Similarly, the coordinate planes divide space between them into eight regions about the origin, called octants. The octants fill

ℝ^{3}

Most work in three-dimensional space is a comfortable extension of the corresponding concepts in two dimensions. In this section, we use our knowledge of circles to describe spheres, then we expand our understanding of vectors to three dimensions. To accomplish these goals, we begin by adapting the distance formula to three-dimensional space.

If two points lie in the same coordinate plane, then it is straightforward to calculate the distance between them. We that the distance

d

The formula for the distance between two points in space is a natural extension of this formula.

The proof of this theorem is left as an exercise. (Hint: First find the distance

d_{1}

lines that are not parallel must always intersect. This is not the case in

ℝ^{3} .

For example, consider the line shown in [link]. These two lines are not parallel, nor do they intersect.

You can also have circles that are interconnected but have no points in common, as in [link].

We have a lot more flexibility working in three dimensions than we do if we stuck with only two dimensions.

Writing Equations in ℝ³

Now that we can represent points in space and find the distance between them, we can learn how to write equations of geometric objects such as lines, planes, and curved surfaces in

ℝ^{3} .

First, we start with a simple equation. Compare the graphs of the equation

x = 0

([link]). From these graphs, we can see the same equation can describe a point, a line, or a plane.

This equation defines the yz-plane. Similarly, the xy-plane contains all points of the form

(x, y, 0) .

Understanding the equations of the coordinate planes allows us to write an equation for any plane that is parallel to one of the coordinate planes. When a plane is parallel to the xy-plane, for example, the z-coordinate of each point in the plane has the same constant value. Only the x- and y-coordinates of points in that plane vary from point to point.

This line is parallel to the y-axis. In a natural extension, the equation

x = 5

which is parallel to the yz-plane. Another natural extension of a familiar equation is found in the equation of a sphere.

The equation of a circle is derived using the distance formula in two dimensions. In the same way, the equation of a sphere is based on the three-dimensional formula for distance.

Working with Vectors in ℝ³

Just like two-dimensional vectors, three-dimensional vectors are quantities with both magnitude and direction, and they are represented by directed line segments (arrows). With a three-dimensional vector, we use a three-dimensional arrow.

Three-dimensional vectors can also be represented in component form. The notation

v = 〈 x, y, z 〉

is a natural extension of the two-dimensional case, representing a vector with the initial point at the origin,

(0, 0, 0),

Vector addition and scalar multiplication are defined analogously to the two-dimensional case. If

v = 〈 x_{1}, y_{1}, z_{1} 〉

The standard unit vectors extend easily into three dimensions as well—

i = 〈 1, 0, 0 〉,

—and we use them in the same way we used the standard unit vectors in two dimensions. Thus, we can represent a vector in

ℝ^{3}

As described earlier, vectors in three dimensions behave in the same way as vectors in a plane. The geometric interpretation of vector addition, for example, is the same in both two- and three-dimensional space ([link]).

We have already seen how some of the algebraic properties of vectors, such as vector addition and scalar multiplication, can be extended to three dimensions. Other properties can be extended in similar fashion. They are summarized here for our reference.

We have seen that vector addition in two dimensions satisfies the commutative, associative, and additive inverse properties. These properties of vector operations are valid for three-dimensional vectors as well. Scalar multiplication of vectors satisfies the distributive property, and the zero vector acts as an additive identity. The proofs to verify these properties in three dimensions are straightforward extensions of the proofs in two dimensions.

Key Concepts

Key Equations

Consider a rectangular box with one of the vertices at the origin, as shown in the following figure. If point $A (2, 3, 5)$

is the opposite vertex to the origin, then find

the coordinates of the other six vertices of the box and
the length of the diagonal of the box determined by the vertices $O$
and
$A .$

This figure is the first octant of the 3-dimensional coordinate system. It has a point labeled “A(2, 3, 5)” drawn.

a. $(2, 0, 5), (2, 0, 0), (2, 3, 0), (0, 3, 0), (0, 3, 5), (0, 0, 5);$

b. $\sqrt{38}$

Find the coordinates of point $P$

and determine its distance to the origin.

This figure is the first octant of the 3-dimensional coordinate system. It has a point drawn at (2, 1, 1). The point is labeled “P.”

For the following exercises, describe and graph the set of points that satisfies the given equation.

(y - 5) (z - 6) = 0

A union of two planes: $y = 5$

(a plane parallel to the xz-plane) and $z = 6$

(a plane parallel to the xy-plane)* * *

This figure is the first octant of the 3-dimensional coordinate system. It has two planes drawn. The first plane is parallel to the x y-plane and is at z = 6. The second plane is parallel to the x z-plane and is at y = 5. The planes are perpendicular.

(z - 2) (z - 5) = 0

{(y - 1)}^{2} + {(z - 1)}^{2} = 1

A cylinder of radius $1$

centered on the line $y = 1, z = 1$

This figure is the first octant of the 3-dimensional coordinate system. It has a cylinder drawn. The axis of the cylinder is parallel to the x-axis.

{(x - 2)}^{2} + {(z - 5)}^{2} = 4

Write the equation of the plane passing through point $(1, 1, 1)$

that is parallel to the xy-plane.

z = 1

Write the equation of the plane passing through point $(1, −3, 2)$

that is parallel to the xz-plane.

Find an equation of the plane passing through points $(1, −3, −2),$

(0, 3, −2),

and $(1, 0, −2) .$

z = −2

Find an equation of the plane passing through points $(1, 9, 2),$

(1, 3, 6),

and $(1, −7, 8) .$

For the following exercises, find the equation of the sphere in standard form that satisfies the given conditions.

Center $C (−1, 7, 4)$

and radius $4$

{(x + 1)}^{2} + {(y - 7)}^{2} + {(z - 4)}^{2} = 16

Center $C (−4, 7, 2)$

and radius $6$

Diameter $P Q,$

where $P (−1, 5, 7)$

and $Q (−5, 2, 9)$

{(x + 3)}^{2} + {(y - 3.5)}^{2} + {(z - 8)}^{2} = \frac{29}{4}

Diameter $P Q,$

where $P (−16, −3, 9)$

and $Q (−2, 3, 5)$

For the following exercises, find the center and radius of the sphere with an equation in general form that is given.

P (1, 2, 3)

x^{2} + y^{2} + z^{2} - 4 z + 3 = 0

Center $C (0, 0, 2)$

and radius $1$

x^{2} + y^{2} + z^{2} - 6 x + 8 y - 10 z + 25 = 0

For the following exercises, express vector $\vec{P Q}$

with the initial point at $P$

and the terminal point at $Q$

in component form and
by using standard unit vectors.

P (3, 0, 2)

and $Q (−1, −1, 4)$

a. $\vec{P Q} = 〈 −4, −1, 2 〉;$

b. $\vec{P Q} = −4 i - j + 2 k$

P (0, 10, 5)

and $Q (1, 1, −3)$

P (−2, 5, −8)

and $M (1, −7, 4),$

where $M$

is the midpoint of the line segment $P Q$

a. $\vec{P Q} = 〈 6, −24, 24 〉;$

b. $\vec{P Q} = 6 i - 24 j + 24 k$

Q (0, 7, −6)

and $M (−1, 3, 2),$

where $M$

is the midpoint of the line segment $P Q$

Find terminal point $Q$

of vector $\vec{P Q} = 〈 7, −1, 3 〉$

with the initial point at $P (−2, 3, 5) .$

Q (5, 2, 8)

Find initial point $P$

of vector $\vec{P Q} = 〈 −9, 1, 2 〉$

with the terminal point at $Q (10, 0, −1) .$

For the following exercises, use the given vectors $a$

and $b$

to find and express the vectors $a + b,$

4 a,

and $−5 a + 3 b$

in component form.

a = 〈 −1, −2, 4 〉,

b = 〈 −5, 6, −7 〉

a + b = 〈 −6, 4, −3 〉,

4 a = 〈 −4, −8, 16 〉,

−5 a + 3 b = 〈 −10, 28, −41 〉

a = 〈 3, −2, 4 〉,

b = 〈 −5, 6, −9 〉

a = − k,

b = − i

a + b = 〈 −1, 0, −1 〉,

4 a = 〈 0, 0, −4 〉,

−5 a + 3 b = 〈 −3, 0, 5 〉

a = i + j + k,

b = 2 i - 3 j + 2 k

For the following exercises, vectors u and v are given. Find the magnitudes of vectors $u - v$

and $−2 u .$

u = 2 i + 3 j + 4 k,

v = − i + 5 j - k

‖ u - v ‖ = \sqrt{38},

‖ −2 u ‖ = 2 \sqrt{29}

u = i + j,

v = j - k

u = 〈 2 cos t, −2 sin t, 3 〉,

v = 〈 0, 0, 3 〉,

where $t$

is a real number.

‖ u - v ‖ = 2,

‖ −2 u ‖ = 2 \sqrt{13}

u = 〈 0, 1, sinh t 〉,

v = 〈 1, 1, 0 〉,

where $t$

is a real number.

For the following exercises, find the unit vector in the direction of the given vector $a$

and express it using standard unit vectors.

a = 3 i - 4 j

a = \frac{3}{5} i - \frac{4}{5} j

a = 〈 4, −3, 6 〉

a = \vec{P Q},

where $P (−2, 3, 1)$

and $Q (0, −4, 4)$

〈 \frac{2}{\sqrt{62}}, - \frac{7}{\sqrt{62}}, \frac{3}{\sqrt{62}} 〉

a = \vec{O P},

where $P (−1, −1, 1)$

a = u - v + w,

where $u = i - j - k,$

v = 2 i - j + k,

and $w = − i + j + 3 k$

〈 - \frac{2}{\sqrt{6}}, \frac{1}{\sqrt{6}}, \frac{1}{\sqrt{6}} 〉

a = 2 u + v - w,

where $u = i - k,$

v = 2 j,

and $w = i - j$

Determine whether $\vec{A B}$

and $\vec{P Q}$

are equivalent vectors, where $A (1, 1, 1), B (3, 3, 3), P (1, 4, 5),$

and $Q (3, 6, 7) .$

Equivalent vectors

Determine whether the vectors $\vec{A B}$

and $\vec{P Q}$

are equivalent, where $A (1, 4, 1),$

B (−2, 2, 0),

P (2, 5, 7),

and $Q (−3, 2, 1) .$

For the following exercises, find vector $u$

with a magnitude that is given and satisfies the given conditions.

v = 〈 7, −1, 3 〉,

‖ u ‖ = 10,

u

and $v$

have the same direction

u = 〈 \frac{70}{\sqrt{59}}, - \frac{10}{\sqrt{59}}, \frac{30}{\sqrt{59}} 〉

v = 〈 2, 4, 1 〉,

‖ u ‖ = 15,

u

and $v$

have the same direction

v = 〈 2 sin t, 2 cos t, 1 〉,

‖ u ‖ = 2,

u

and $v$

have opposite directions for any $t,$

where $t$

is a real number

u = 〈 - \frac{4}{\sqrt{5}} sin t, - \frac{4}{\sqrt{5}} cos t, - \frac{2}{\sqrt{5}} 〉

v = 〈 3 sinh t, 0, 3 〉,

‖ u ‖ = 5,

u

and $v$

have opposite directions for any $t,$

where $t$

is a real number

Determine a vector of magnitude $5$

in the direction of vector $\vec{A B},$

where $A (2, 1, 5)$

and $B (3, 4, −7) .$

〈 \frac{5}{\sqrt{154}}, \frac{15}{\sqrt{154}}, - \frac{60}{\sqrt{154}} 〉

Find a vector of magnitude $2$

that points in the opposite direction than vector $\vec{A B},$

where $A (−1, −1, 1)$

and $B (0, 1, 1) .$

Express the answer in component form.

Consider the points $A (2, α, 0), B (0, 1, β),$

and $C (1, 1, β),$

where $α$

and $β$

are negative real numbers. Find $α$

and $β$

such that $‖ \vec{O A} - \vec{O B} + \vec{O C} ‖ = ‖ \vec{O B} ‖ = 4 .$

α = − \sqrt{7},

β = − \sqrt{15}

Consider points $A (α, 0, 0), B (0, β, 0),$

and $C (α, β, β),$

where $α$

and $β$

are positive real numbers. Find $α$

and $β$

such that $‖ \overset{—}{O A} + \overset{—}{O B} ‖ = \sqrt{2} and ‖ \overset{—}{O C} ‖ = \sqrt{3} .$

Let $P (x, y, z)$

be a point situated at an equal distance from points $A (1, −1, 0)$

and $B (−1, 2, 1) .$

Show that point $P$

lies on the plane of equation $−2 x + 3 y + z = 2 .$

Let $P (x, y, z)$

be a point situated at an equal distance from the origin and point $A (4, 1, 2) .$

Show that the coordinates of point $P$

satisfy the equation $8 x + 2 y + 4 z = 21 .$

The points $A, B,$

and $C$

are collinear (in this order) if the relation $‖ \vec{A B} ‖ + ‖ \vec{B C} ‖ = ‖ \vec{A C} ‖$

is satisfied. Show that $A (5, 3, −1),$

B (−5, −3, 1),

and $C (−15, −9, 3)$

are collinear points.

Show that points $A (1, 0, 1),$

B (0, 1, 1),

and $C (1, 1, 1)$

are not collinear.

[T] A force $F$

of $50 N$

acts on a particle in the direction of the vector $\vec{O P},$

where $P (3, 4, 0) .$

Express the force as a vector in component form.
Find the angle between force $F$
and the positive direction of the x-axis. Express the answer in degrees rounded to the nearest integer.

a. $F = 〈 30, 40, 0 〉;$

b. $53 °$

[T] A force $F$

of $40 N$

acts on a box in the direction of the vector $\vec{O P},$

where $P (1, 0, 2) .$

Express the force as a vector by using standard unit vectors.
Find the angle between force $F$
and the positive direction of the x-axis.

If $F$

is a force that moves an object from point $P_{1} (x_{1}, y_{1}, z_{1})$

to another point $P_{2} (x_{2}, y_{2}, z_{2}),$

then the displacement vector is defined as $D = (x_{2} - x_{1}) i + (y_{2} - y_{1}) j + (z_{2} - z_{1}) k .$

A metal container is lifted $10$

m vertically by a constant force $F .$

Express the displacement vector $D$

by using standard unit vectors.

D = 10 k

A box is pulled $4$

yd horizontally in the x-direction by a constant force $F .$

Find the displacement vector in component form.

The sum of the forces acting on an object is called the resultant or net force. An object is said to be in static equilibrium if the resultant force of the forces that act on it is zero. Let $F_{1} = 〈 10, 6, 3 〉,$

F_{2} = 〈 0, 4, 9 〉,

and $F_{3} = 〈 10, −3, −9 〉$

be three forces acting on a box. Find the force $F_{4}$

acting on the box such that the box is in static equilibrium. Express the answer in component form.

F_{4} = 〈 −20, −7, −3 〉

[T] Let $F_{k} = 〈 1, k, k^{2} 〉,$

k = 1,..., n

be $n$

forces acting on a particle, with $n \geq 2 .$

Find the net force $F = \sum_{k = 1}^{n} F_{k} .$
Express the answer using standard unit vectors.
Use a computer algebra system (CAS) to find n such that $‖ F ‖ < 100 .$

The force of gravity $F$

acting on an object is given by $F = m g,$

where m is the mass of the object (expressed in kilograms) and $g$

is acceleration resulting from gravity, with $‖ g ‖ = 9.8$

N/kg .

A 2-kg disco ball hangs by a chain from the ceiling of a room.

Find the force of gravity $F$
acting on the disco ball and find its magnitude.
Find the force of tension $T$
in the chain and its magnitude.

Express the answers using standard unit vectors.

This figure shows a disco ball suspended from a ceiling.

a. $F = −19.6 k,$

‖ F ‖ = 19.6

N; b. $T = 19.6 k,$

‖ T ‖ = 19.6

A 5-kg pendant chandelier is designed such that the alabaster bowl is held by four chains of equal length, as shown in the following figure.

Find the magnitude of the force of gravity acting on the chandelier.
Find the magnitudes of the forces of tension for each of the four chains (assume chains are essentially vertical).

This figure shows a light fixture hung from a ceiling, supported by 4 chains from the same point on the ceiling to four points spread evenly around the light fixture.

[T] A 30-kg block of cement is suspended by three cables of equal length that are anchored at points $P (−2, 0, 0),$

Q (1, \sqrt{3}, 0),

and $R (1, − \sqrt{3}, 0) .$

The load is located at $S (0, 0, −2 \sqrt{3}),$

as shown in the following figure. Let $F_{1},$

F_{2},

and $F_{3}$

be the forces of tension resulting from the load in cables $R S, Q S,$

and $P S,$

respectively.

Find the gravitational force $F$
acting on the block of cement that counterbalances the sum
$F_{1} + F_{2} + F_{3}$
of the forces of tension in the cables.
Find forces $F_{1},$ $F_{2},$
and
$F_{3} .$
Express the answer in component form.

This figure is the 3-dimensional coordinate system. It has 4 points drawn. The first point is labeled “P(-2, 0, 0).” The second point is labeled “R(1, -squareroot of 3, 0).” The third point is labeled “S(0, 0, -2squareroots of 3).” The fourth point is labeled “Q(1, squareroot of 3, 0).” There are line segments from P to S, from R to S, and from Q to S. At point S there is a box labeled “30 k g.”

a. $F = −294 k$

N; b. $F_{1} = 〈 - \frac{49 \sqrt{3}}{3}, 49, −98 〉,$

F_{2} = 〈 - \frac{49 \sqrt{3}}{3}, −49, −98 〉,

and $F_{3} = 〈 \frac{98 \sqrt{3}}{3}, 0, −98 〉$

(each component is expressed in newtons)

Two soccer players are practicing for an upcoming game. One of them runs 10 m from point A to point B. She then turns left at $90 °$

and runs 10 m until she reaches point C. Then she kicks the ball with a speed of 10 m/sec at an upward angle of $45 °$

to her teammate, who is located at point A. Write the velocity of the ball in component form.

This figure is the image of two soccer players. The first soccer player is at point A. The second player is at point C. There is a line segment from A to C. Ther is a vector from player C upwards labeled “v.” There is a vector from player A to the bottom of the image. The point at the bottom is labeled “B.” This vector is labeled “10m.” There is a vector from C to B labeled “10m.”

Let $r (t) = 〈 x (t), y (t), z (t) 〉$

be the position vector of a particle at the time $t \in [0, T],$

where $x, y,$

and $z$

are smooth functions on $[0, T] .$

The instantaneous velocity of the particle at time $t$

is defined by vector $v (t) = 〈 x ′ (t), y ′ (t), z ′ (t) 〉,$

with components that are the derivatives with respect to $t,$

of the functions x, y, and z, respectively. The magnitude $‖ v (t) ‖$

of the instantaneous velocity vector is called the speed of the particle at time t. Vector $a (t) = 〈 x ″ (t), y ″ (t), z ″ (t) 〉,$

with components that are the second derivatives with respect to $t,$

of the functions $x, y,$

and $z,$

respectively, gives the acceleration of the particle at time $t .$

Consider $r (t) = 〈 cos t, sin t, 2 t 〉$

the position vector of a particle at time $t \in [0, 30],$

where the components of $r$

are expressed in centimeters and time is expressed in seconds.

Find the instantaneous velocity, speed, and acceleration of the particle after the first second. Round your answer to two decimal places.
Use a CAS to visualize the path of the particle—that is, the set of all points of coordinates $(cos t, sin t, 2 t),$
where
$t \in [0, 30] .$

a. $v (1) = 〈 −0.84, 0.54, 2 〉$

(each component is expressed in centimeters per second); $‖ v (1) ‖ = 2.24$

(expressed in centimeters per second); $a (1) = 〈 −0.54, −0.84, 0 〉$

(each component expressed in centimeters per second squared);

b.* * *

This figure is of the 3-dimensional coordinate system above the xy-plane. It has a spiral drawn resembling a spring. The spiral is around the z-axis. The spiral starts on the x-axis at x = 1.

[T] Let $r (t) = 〈 t, 2 t^{2}, 4 t^{2} 〉$

be the position vector of a particle at time $t$

(in seconds), where $t \in [0, 10]$

(here the components of $r$

are expressed in centimeters).

Find the instantaneous velocity, speed, and acceleration of the particle after the first two seconds. Round your answer to two decimal places.
Use a CAS to visualize the path of the particle defined by the points $(t, 2 t^{2}, 4 t^{2}),$
where
$t \in [0, 60] .$

Vectors in Three Dimensions

Three-Dimensional Coordinate Systems

Writing Equations in ℝ3

Working with Vectors in ℝ3

Key Concepts

Key Equations

Glossary

Writing Equations in ℝ³

Working with Vectors in ℝ³