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12.2: Vectors in Three Dimensions - Mathematics


Learning Objectives

  • Describe three-dimensional space mathematically.
  • Locate points in space using coordinates.
  • Write the distance formula in three dimensions.
  • Write the equations for simple planes and spheres.
  • Perform vector operations in (mathbb{R}^{3}).

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.

Definition: Three-dimensional Rectangular Coordinate System

The three-dimensional rectangular coordinate system consists of three perpendicular axes: the (x)-axis, the (y)-axis, and the (z)-axis. Because each axis is a number line representing all real numbers in (ℝ), the three-dimensional system is often denoted by (ℝ^3).

In Figure (PageIndex{1a}), 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 this 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 (Figure (PageIndex{1b})). 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), is appended to indicate alignment with the (z)-axis: ((x,y,z)). A point in space is identified by all three coordinates (Figure (PageIndex{2})). To plot the point ((x,y,z)), go (x) units along the (x)-axis, then (y) units in the direction of the (y)-axis, then (z) units in the direction of the (z)-axis.

Example (PageIndex{1}): Locating Points in Space

Sketch the point ((1,−2,3)) in three-dimensional space.

Solution

To sketch a point, start by sketching three sides of a rectangular prism along the coordinate axes: one unit in the positive (x) direction, (2) units in the negative (y) direction, and (3) units in the positive (z) direction. Complete the prism to plot the point (Figure).

Exercise (PageIndex{1})

Sketch the point ((−2,3,−1)) in three-dimensional space.

Hint

Start by sketching the coordinate axes. e.g., Figure (PageIndex{3}). Then sketch a rectangular prism to help find the point in space.

Answer

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 (Figure (PageIndex{3})). We define the (xy)-plane formally as the following set: ({(x,y,0):x,y∈ℝ}.) Similarly, the (xz)-plane and the (yz)-plane are defined as ({(x,0,z):x,z∈ℝ}) and ({(0,y,z):y,z∈ℝ},) respectively.

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) in the same way that quadrants fill (ℝ^2), as shown in Figure (PageIndex{4}).

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) between two points ((x_1,y_1)) and ((x_2,y_2)) in the x(y)-coordinate plane is given by the formula

[d=sqrt{(x_2−x_1)^2+(y_2−y_1)^2}.]

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

The Distance between Two Points in Space

The distance (d) between points ((x_1,y_1,z_1)) and ((x_2,y_2,z_2)) is given by the formula

[d=sqrt{(x_2−x_1)^2+(y_2−y_1)^2+(z_2−z_1)^2}. label{distanceForm}]

The proof of this theorem is left as an exercise. (Hint: First find the distance (d_1) between the points ((x_1,y_1,z_1)) and ((x_2,y_2,z_1)) as shown in Figure (PageIndex{5}).)

Example (PageIndex{2}): Distance in Space

Find the distance between points (P_1=(3,−1,5)) and (P_2=(2,1,−1).)

Solution

Substitute values directly into the distance formula (Equation ef{distanceForm}):

[egin{align*} d(P_1,P_2) &=sqrt{(x_2−x_1)^2+(y_2−y_1)^2+(z_2−z_1)^2} [4pt] &=sqrt{(2−3)^2+(1−(−1))^2+(−1−5)^2} [4pt] &=sqrt{(-1)^2+2^2+(−6)^2} [4pt] &=sqrt{41}. end{align*}]

Exercise (PageIndex{2})

Find the distance between points (P_1=(1,−5,4)) and (P_2=(4,−1,−1)).

Hint

(d=sqrt{(x_2−x_1)^2+(y_2−y_1)^2+(z_2−z_1)^2})

Answer

(5sqrt{2})

Before moving on to the next section, let’s get a feel for how (ℝ^3) differs from (ℝ^2). For example, in (ℝ^2), lines that are not parallel must always intersect. This is not the case in (ℝ^3). For example, consider the line shown in Figure (PageIndex{7}). These two lines are not parallel, nor do they intersect.

Figure (PageIndex{7}): These two lines are not parallel, but still do not intersect.

You can also have circles that are interconnected but have no points in common, as in Figure (PageIndex{8}).

Figure (PageIndex{8}): These circles are interconnected, but have no points in common.

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

Writing Equations in (ℝ^3)

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) in (ℝ), (ℝ^2),and (ℝ^3) (Figure (PageIndex{9})). From these graphs, we can see the same equation can describe a point, a line, or a plane.

In space, the equation (x=0) describes all points ((0,y,z)). This equation defines the (yz)-plane. Similarly, the (xy)-plane contains all points of the form ((x,y,0)). The equation (z=0) defines the (xy)-plane and the equation (y=0) describes the (xz)-plane (Figure (PageIndex{10})).

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.

Equations of Planes Parallel to Coordinate Planes

  1. The plane in space that is parallel to the (xy)-plane and contains point ((a,b,c)) can be represented by the equation (z=c).
  2. The plane in space that is parallel to the (xz)-plane and contains point ((a,b,c)) can be represented by the equation (y=b).
  3. The plane in space that is parallel to the (yz)-plane and contains point ((a,b,c)) can be represented by the equation (x=a).

Example (PageIndex{3}): Writing Equations of Planes Parallel to Coordinate Planes

  1. Write an equation of the plane passing through point ((3,11,7)) that is parallel to the (yz)-plane.
  2. Find an equation of the plane passing through points ((6,−2,9), (0,−2,4),) and ((1,−2,−3).)

Solution

  1. When a plane is parallel to the (yz)-plane, only the (y)- and (z)-coordinates may vary. The (x)-coordinate has the same constant value for all points in this plane, so this plane can be represented by the equation (x=3).
  2. Each of the points ((6,−2,9), (0,−2,4),) and ((1,−2,−3)) has the same (y)-coordinate. This plane can be represented by the equation (y=−2).

Exercise (PageIndex{3})

Write an equation of the plane passing through point ((1,−6,−4)) that is parallel to the (xy)-plane.

Hint

If a plane is parallel to the (xy)-plane, the z-coordinates of the points in that plane do not vary.

Answer

(z=−4)

As we have seen, in (ℝ^2) the equation (x=5) describes the vertical line passing through point ((5,0)). This line is parallel to the (y)-axis. In a natural extension, the equation (x=5) in (ℝ^3) describes the plane passing through point ((5,0,0)), which is parallel to the (yz)-plane. Another natural extension of a familiar equation is found in the equation of a sphere.

Definition: Sphere

A sphere is the set of all points in space equidistant from a fixed point, the center of the sphere (Figure (PageIndex{11})), just as the set of all points in a plane that are equidistant from the center represents a circle. In a sphere, as in a circle, the distance from the center to a point on the sphere is called the radius.

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.

Standard Equation of a Sphere

The sphere with center ((a,b,c)) and radius (r) can be represented by the equation

[(x−a)^2+(y−b)^2+(z−c)^2=r^2.]

This equation is known as the standard equation of a sphere.

Example (PageIndex{4}): Finding an Equation of a Sphere

Find the standard equation of the sphere with center ((10,7,4)) and point ((−1,3,−2)), as shown in Figure (PageIndex{12}).

Figure (PageIndex{12}): The sphere centered at ((10,7,4)) containing point ((−1,3,−2).)

Solution

Use the distance formula to find the radius (r) of the sphere:

[egin{align*} r &=sqrt{(−1−10)^2+(3−7)^2+(−2−4)^2} [4pt] &=sqrt{(−11)^2+(−4)^2+(−6)^2} [4pt] &=sqrt{173} end{align*} ]

The standard equation of the sphere is

[(x−10)^2+(y−7)^2+(z−4)^2=173. onumber]

Exercise (PageIndex{4})

Find the standard equation of the sphere with center ((−2,4,−5)) containing point ((4,4,−1).)

Hint

First use the distance formula to find the radius of the sphere.

Answer

[(x+2)^2+(y−4)^2+(z+5)^2=52 onumber]

Example (PageIndex{5}): Finding the Equation of a Sphere

Let (P=(−5,2,3)) and (Q=(3,4,−1)), and suppose line segment (overline{PQ}) forms the diameter of a sphere (Figure (PageIndex{13})). Find the equation of the sphere.

Solution:

Since (overline{PQ}) is a diameter of the sphere, we know the center of the sphere is the midpoint of (overline{PQ}).Then,

[C=left(dfrac{−5+3}{2},dfrac{2+4}{2},dfrac{3+(−1)}{2} ight)=(−1,3,1). onumber]

Furthermore, we know the radius of the sphere is half the length of the diameter. This gives

[egin{align*} r &=dfrac{1}{2}sqrt{(−5−3)^2+(2−4)^2+(3−(−1))^2} [4pt] &=dfrac{1}{2}sqrt{64+4+16} [4pt] &=sqrt{21} end{align*}]

Then, the equation of the sphere is ((x+1)^2+(y−3)^2+(z−1)^2=21.)

Exercise (PageIndex{5})

Find the equation of the sphere with diameter (overline{PQ}), where (P=(2,−1,−3)) and (Q=(−2,5,−1).)

Hint

Find the midpoint of the diameter first.

Answer

[x^2+(y−2)^2+(z+2)^2=14 onumber]

Example (PageIndex{6}): Graphing Other Equations in Three Dimensions

Describe the set of points that satisfies ((x−4)(z−2)=0,) and graph the set.

Solution

We must have either (x−4=0) or (z−2=0), so the set of points forms the two planes (x=4) and (z=2) (Figure (PageIndex{14})).

Exercise (PageIndex{6})

Describe the set of points that satisfies ((y+2)(z−3)=0,) and graph the set.

Hint

One of the factors must be zero.

Answer

The set of points forms the two planes (y=−2) and (z=3).

Example (PageIndex{7}): Graphing Other Equations in Three Dimensions

Describe the set of points in three-dimensional space that satisfies ((x−2)^2+(y−1)^2=4,) and graph the set.

Solution

The (x)- and (y)-coordinates form a circle in the (xy)-plane of radius (2), centered at ((2,1)). Since there is no restriction on the (z)-coordinate, the three-dimensional result is a circular cylinder of radius (2) centered on the line with (x=2) and (y=1). The cylinder extends indefinitely in the (z)-direction (Figure (PageIndex{15})).

Exercise (PageIndex{7})

Describe the set of points in three dimensional space that satisfies (x^2+(z−2)^2=16), and graph the surface.

Hint

Think about what happens if you plot this equation in two dimensions in the (xz)-plane.

Answer

A cylinder of radius 4 centered on the line with (x=0) and (z=2).

Working with Vectors in (ℝ^3)

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 (vecs{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)), and terminal point ((x,y,z)). The zero vector is (vecs{0}=⟨0,0,0⟩). So, for example, the three dimensional vector (vecs{v}=⟨2,4,1⟩) is represented by a directed line segment from point ((0,0,0)) to point ((2,4,1)) (Figure (PageIndex{16})).

Vector addition and scalar multiplication are defined analogously to the two-dimensional case. If (vecs{v}=⟨x_1,y_1,z_1⟩) and (vecs{w}=⟨x_2,y_2,z_2⟩) are vectors, and (k) is a scalar, then

[vecs{v}+vecs{w}=⟨x_1+x_2,y_1+y_2,z_1+z_2⟩]

and

[kvecs{v}=⟨kx_1,ky_1,kz_1⟩.]

If (k=−1,) then (kvecs{v}=(−1)vecs{v}) is written as (−vecs{v}), and vector subtraction is defined by (vecs{v}−vecs{w}=vecs{v}+(−vecs{w})=vecs{v}+(−1)vecs{w}).

The standard unit vectors extend easily into three dimensions as well, (hat{mathbf i}=⟨1,0,0⟩), (hat{mathbf j}=⟨0,1,0⟩), and (hat{mathbf k}=⟨0,0,1⟩), 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) in the following ways:

[vecs{v}=⟨x,y,z⟩=xhat{mathbf i}+yhat{mathbf j}+zhat{mathbf k}].

Example (PageIndex{8}): Vector Representations

Let (vecd{PQ}) be the vector with initial point (P=(3,12,6)) and terminal point (Q=(−4,−3,2)) as shown in Figure (PageIndex{17}). Express (vecd{PQ}) in both component form and using standard unit vectors.

Solution

In component form,

[egin{align*} vecd{PQ} =⟨x_2−x_1,y_2−y_1,z_2−z_1⟩ [4pt] =⟨−4−3,−3−12,2−6⟩ [4pt] =⟨−7,−15,−4⟩. end{align*}]

In standard unit form,

[vecd{PQ}=−7hat{mathbf i}−15hat{mathbf j}−4hat{mathbf k}. onumber]

Exercise (PageIndex{8})

Let (S=(3,8,2)) and (T=(2,−1,3)). Express (vec{ST}) in component form and in standard unit form.

Hint

Write (vecd{ST}) in component form first. (T) is the terminal point of (vecd{ST}).

Answer

(vecd{ST}=⟨−1,−9,1⟩=−hat{mathbf i}−9hat{mathbf j}+hat{mathbf k})

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 (Figure (PageIndex{18})).

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.

Properties of Vectors in Space

Let (vecs{v}=⟨x_1,y_1,z_1⟩) and (vecs{w}=⟨x_2,y_2,z_2⟩) be vectors, and let (k) be a scalar.

  • Scalar multiplication: [kvecs{v}=⟨kx_1,ky_1,kz_1⟩]
  • Vector addition: [vecs{v}+vecs{w}=⟨x_1,y_1,z_1⟩+⟨x_2,y_2,z_2⟩=⟨x_1+x_2,y_1+y_2,z_1+z_2⟩]
  • Vector subtraction: [vecs{v}−vecs{w}=⟨x_1,y_1,z_1⟩−⟨x_2,y_2,z_2⟩=⟨x_1−x_2,y_1−y_2,z_1−z_2⟩]
  • Vector magnitude: [|vecs{v}|=sqrt{x_1^2+y_1^2+z_1^2}]
  • Unit vector in the direction of (vecs{v}): [dfrac{1}{|vecs{v}|}vecs{v}=dfrac{1}{|vecs{v}|}⟨x_1,y_1,z_1⟩=⟨dfrac{x_1}{|vecs{v}|},dfrac{y_1}{|vecs{v}|},dfrac{z_1}{|vecs{v}|}⟩, quad ext{if} , vecs{v}≠vecs{0}]

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.

Example (PageIndex{9}): Vector Operations in Three Dimensions

Let (vecs{v}=⟨−2,9,5⟩) and (vecs{w}=⟨1,−1,0⟩) (Figure (PageIndex{19})). Find the following vectors.

  1. (3vecs{v}−2vecs{w})
  2. (5|vecs{w}|)
  3. (|5 vecs{w}|)
  4. A unit vector in the direction of (vecs{v})

Solution

a. First, use scalar multiplication of each vector, then subtract:

[egin{align*} 3vecs{v}−2vecs{w} =3⟨−2,9,5⟩−2⟨1,−1,0⟩ [4pt] =⟨−6,27,15⟩−⟨2,−2,0⟩ [4pt] =⟨−6−2,27−(−2),15−0⟩ [4pt] =⟨−8,29,15⟩. end{align*}]

b. Write the equation for the magnitude of the vector, then use scalar multiplication:

[5|vecs{w}|=5sqrt{1^2+(−1)^2+0^2}=5sqrt{2}. onumber]

c. First, use scalar multiplication, then find the magnitude of the new vector. Note that the result is the same as for part b.:

[|5 vecs{w}|=∥⟨5,−5,0⟩∥=sqrt{5^2+(−5)^2+0^2}=sqrt{50}=5sqrt{2} onumber]

d. Recall that to find a unit vector in two dimensions, we divide a vector by its magnitude. The procedure is the same in three dimensions:

[egin{align*} dfrac{vecs{v}}{|vecs{v}|} =dfrac{1}{|vecs{v}|}⟨−2,9,5⟩ [4pt] =dfrac{1}{sqrt{(−2)^2+9^2+5^2}}⟨−2,9,5⟩ [4pt] =dfrac{1}{sqrt{110}}⟨−2,9,5⟩ [4pt] =⟨dfrac{−2}{sqrt{110}},dfrac{9}{sqrt{110}},dfrac{5}{sqrt{110}}⟩ . end{align*}]

Exercise (PageIndex{9}):

Let (vecs{v}=⟨−1,−1,1⟩) and (vecs{w}=⟨2,0,1⟩). Find a unit vector in the direction of (5vecs{v}+3vecs{w}.)

Hint

Start by writing (5vecs{v}+3vecs{w}) in component form.

Answer

(⟨dfrac{1}{3sqrt{10}},−dfrac{5}{3sqrt{10}},dfrac{8}{3sqrt{10}}⟩)

Example (PageIndex{10}): Throwing a Forward Pass

A quarterback is standing on the football field preparing to throw a pass. His receiver is standing 20 yd down the field and 15 yd to the quarterback’s left. The quarterback throws the ball at a velocity of 60 mph toward the receiver at an upward angle of (30°) (see the following figure). Write the initial velocity vector of the ball, (vecs{v}), in component form.

Solution

The first thing we want to do is find a vector in the same direction as the velocity vector of the ball. We then scale the vector appropriately so that it has the right magnitude. Consider the vector (vecs{w}) extending from the quarterback’s arm to a point directly above the receiver’s head at an angle of (30°) (see the following figure). This vector would have the same direction as (vecs{v}), but it may not have the right magnitude.

The receiver is 20 yd down the field and 15 yd to the quarterback’s left. Therefore, the straight-line distance from the quarterback to the receiver is

Dist from QB to receiver(=sqrt{15^2+20^2}=sqrt{225+400}=sqrt{625}=25) yd.

We have (dfrac{25}{|vecs{w}|}=cos 30°.) Then the magnitude of (vecs{w}) is given by

(|vecs{w}|=dfrac{25}{cos 30°}=dfrac{25⋅2}{sqrt{3}}=dfrac{50}{sqrt{3}}) yd

and the vertical distance from the receiver to the terminal point of (vecs{w}) is

Vert dist from receiver to terminal point of (vecs{w}=|vecs{w}| sin 30°=dfrac{50}{sqrt{3}}⋅dfrac{1}{2}=dfrac{25}{sqrt{3}}) yd.

Then (vecs{w}=⟨20,15,dfrac{25}{sqrt{3}}⟩), and has the same direction as (vecs{v}).

Recall, though, that we calculated the magnitude of (vecs{w}) to be (|vecs{w}|=dfrac{50}{sqrt{3}}), and (vecs{v}) has magnitude (60) mph. So, we need to multiply vector (vecs{w}) by an appropriate constant, (k). We want to find a value of (k) so that (∥kvecs{w}∥=60) mph. We have

(|k vecs{w}|=k|vecs{w}|=kdfrac{50}{sqrt{3}}) mph,

so we want

(kdfrac{50}{sqrt{3}}=60)

(k=dfrac{60sqrt{3}}{50})

(k=dfrac{6sqrt{3}}{5}).

Then

(vecs{v}=kvecs{w}=k⟨20,15,dfrac{25}{sqrt{3}}⟩=dfrac{6sqrt{3}}{5}⟨20,15,dfrac{25}{sqrt{3}}⟩=⟨24sqrt{3},18sqrt{3},30⟩).

Let’s double-check that (|vecs{v}|=60.) We have

(|vecs{v}|=sqrt{(24sqrt{3})^2+(18sqrt{3})^2+(30)^2}=sqrt{1728+972+900}=sqrt{3600}=60) mph.

So, we have found the correct components for (vecs{v}).

Exercise (PageIndex{10})

Assume the quarterback and the receiver are in the same place as in the previous example. This time, however, the quarterback throws the ball at velocity of (40) mph and an angle of (45°). Write the initial velocity vector of the ball, (vecs{v}), in component form.

Hint

Follow the process used in the previous example.

Answer

(v=⟨16sqrt{2},12sqrt{2},20sqrt{2}⟩)

Key Concepts

  • The three-dimensional coordinate system is built around a set of three axes that intersect at right angles at a single point, the origin. Ordered triples ((x,y,z)) are used to describe the location of a point in space.
  • The distance (d) between points ((x_1,y_1,z_1)) and ((x_2,y_2,z_2)) is given by the formula [d=sqrt{(x_2−x_1)^2+(y_2−y_1)^2+(z_2−z_1)^2}. onumber]
  • In three dimensions, the equations (x=a,y=b,) and (z=c) describe planes that are parallel to the coordinate planes.
  • The standard equation of a sphere with center ((a,b,c)) and radius (r) is [(x−a)^2+(y−b)^2+(z−c)^2=r^2. onumber ]
  • In three dimensions, as in two, vectors are commonly expressed in component form, (v=⟨x,y,z⟩), or in terms of the standard unit vectors, (xi+yj+zk.)
  • Properties of vectors in space are a natural extension of the properties for vectors in a plane. Let (v=⟨x_1,y_1,z_1⟩) and (w=⟨x_2,y_2,z_2⟩) be vectors, and let (k) be a scalar.

Scalar multiplication:

[(kvecs{v}=⟨kx_1,ky_1,kz_1⟩ onumber]

Vector addition:

[vecs{v}+vecs{w}=⟨x_1,y_1,z_1⟩+⟨x_2,y_2,z_2⟩=⟨x_1+x_2,y_1+y_2,z_1+z_2⟩ onumber]

Vector subtraction:

[vecs{v}−vecs{w}=⟨x_1,y_1,z_1⟩−⟨x_2,y_2,z_2⟩=⟨x_1−x_2,y_1−y_2,z_1−z_2⟩ onumber]

Vector magnitude:

[‖vecs{v}‖=sqrt{x_1^2+y_1^2+z_1^2} onumber]

Unit vector in the direction of (vecs{v}):

[dfrac{vecs{v}}{‖vecs{v}‖}=dfrac{1}{‖vecs{v}‖}⟨x_1,y_1,z_1⟩=⟨dfrac{x_1}{‖vecs{v}‖},dfrac{y_1}{‖vecs{v}‖},dfrac{z_1}{‖vecs{v}‖}⟩, vecs{v}≠vecs{0} onumber]

Key Equations

Distance between two points in space:

[d=sqrt{(x_2−x_1)^2+(y_2−y_1)^2+(z_2−z_1)^2}]

Sphere with center ((a,b,c)) and radius (r):

[(x−a)^2+(y−b)^2+(z−c)^2=r^2]

Glossary

coordinate plane
a plane containing two of the three coordinate axes in the three-dimensional coordinate system, named by the axes it contains: the (xy)-plane, (xz)-plane, or the (yz)-plane
right-hand rule
a common way to define the orientation of the three-dimensional coordinate system; when the right hand is curved around the (z)-axis in such a way that the fingers curl from the positive (x)-axis to the positive (y)-axis, the thumb points in the direction of the positive (z)-axis
octants
the eight regions of space created by the coordinate planes
sphere
the set of all points equidistant from a given point known as the center
standard equation of a sphere
((x−a)^2+(y−b)^2+(z−c)^2=r^2) describes a sphere with center ((a,b,c)) and radius (r)
three-dimensional rectangular coordinate system
a coordinate system defined by three lines that intersect at right angles; every point in space is described by an ordered triple ((x,y,z)) that plots its location relative to the defining axes

Dimension (vector space)

In mathematics, the dimension of a vector space V is the cardinality (i.e. the number of vectors) of a basis of V over its base field. [1] [2] It is sometimes called Hamel dimension (after Georg Hamel) or algebraic dimension to distinguish it from other types of dimension.

The dimension of the vector space V over the field F can be written as dimF(V) or as [V : F], read "dimension of V over F". When F can be inferred from context, dim(V) is typically written.


Figures in Three Dimensions

We can still plot 2 dimensional shapes in 3 dimensions, all we need is to set any
of the axes to 0.

The formula for a circle in the xy plane would be

Notice that when z is 0, it is the same equation for a 2 dimensional circle. When
z does not equal 0, the graph will form a sphere.

The graph of a circle in the xy plane forms a cylinder when z can take on any value

Let’s graph x = 3 in one, two, and three dimensions. Here is x = 3 in R 1

Since x is the only dimension, the image is just a point on a number line.

Because y can take on any value when x is 3, the image is a straight line.

Here, y and z can take on any value, so the image is the yz plane at x =
3.

Notice that first we have a point, then a line, and then finally a plane.


NCERT Solutions for Class 11 Maths Chapter 12 Introduction to three Dimensional Geometry Ex 12.2

NCERT Solutions for Class 11 Maths Chapter 12 Introduction to Three Dimensional Geometry Ex 12.2

Question 1.
Find the distance between the following pairs of points:
(i) (2, 3, 5) and (4, 3, 1)
(ii) (-3, 7, 2) and (2, 4, -1)
(iii) (-1, 3, -4) and (1, -3, 4)
(iv) (2, -1, 3) and (-2, 1, 3)
Solution:
(i) The distance PQ between the points P(2, 3, 5) and Q(4, 3, 1) is
(PQ=sqrt < left( 4-2 ight) ^< 2 >+left( 3-3 ight) ^< 2 >left( 1-5 ight) ^ < 2 >> )
= (sqrt < 4+0+16= >sqrt < 20 >=2sqrt < 5 >units.)

(ii) The distance PQ between the points P(-3, 7, 2) and Q(2, 4, -1) is
(PQ=sqrt < left[ 2-left( -3 ight) ight] ^< 2 >+left( 4-7 ight) ^< 2 >left( -1-2 ight) ^ < 2 >> )
(=sqrt < left( 2+3 ight) ^< 2 >+left( 4-7 ight) ^< 2 >+left( -1-2 ight) ^ < 2 >> )
(=sqrt < 25+9+9 >=sqrt < 43 >units)

(iii) The distance PQ between the points P(-1, 3, -4) and Q(1, -3, 4) is
(PQ=sqrt < left[ 1-left( -1 ight) ight] ^< 2 >+left( -3-3 ight) ^< 2 >left[ 4-left( -4 ight) ight] ^ < 2 >> )
(=sqrt < 4+36+64 >=sqrt < 104 >=2sqrt < 26 >units)

(iv) The distance PQ between the points P(2, -1, 3) and Q(-2, 1, 3) is
(PQ=sqrt < left( -2-2 ight) ^< 2 >+left[ 1-left( -1 ight) ight] ^< 2 >+left( 3-3 ight) ^ < 2 >> )
(=sqrt < 16+4+0 >=sqrt < 20 >=2sqrt < 5 >units)

Question 2.
Show that the points (-2, 3, 5), (1, 2, 3) and (7, 0, -1) are collinear.
Solution:
Let A(-2, 3, 5), B(1, 2, 3) and C(7, 0, -1) be three given points.

Now AC = AB + BC
Thus, points A, B and C are collinear.

Question 3.
Verify the following:
(i) (0, 7, -10), (1, 6, -6) and (4, 9, -6) are the vertices of an isosceles triangle.
(ii) (0, 7, 10), (-1, 6, 6) and (-4, 9, 6) are the vertices of a right angled triangle.
(iii) (-1, 2, 1), (1, -2, 5), (4, -7,8) and (2, -3,4) are the vertices of a parallelogram.
Solution:
(i) Let A(0, 7, -10), B(l, 6, -6) and C(4, 9, -6) be three vertices of triangle ABC. Then

Now, AB = BC
Thus, ABC is an isosceles triangle.

(ii) Let A(0, 7,10), B(-l, 6, 6) and C(-A, 9, 6) be three vertices of triangle ABC. Then

Now, AC 2 = AB 2 + BC 2
Thus, ABC is a right angled triangle.

(iii) Let A(-1, 2, 1), B(1, -2, 5) and C(4, -7, 8) and D(2, -3,4) be four vertices of quadrilateral ABCD. Then

Now AB = CD, BC = AD and AC ≠ BD
Thus A, B, C and D are vertices of a parallelogram ABCD.

Question 4.
Find the equation of the set of points which are equidistant from the points (1, 2, 3) and (3, 2, -1).
Solution:
Let A(x, y, z) be any point which is equidistant from points B(1, 2, 3) and C(3, 2, -1).

Question 5.
Find the equation of the set of points P, the sum of whose distances from A(4, 0, 0) and B(-4,0,0) is equal to 10.
Solution:
Let P(x, y, z) be any point.


The Scalar Triple Product of Vectors in Three-Dimensional Space

Geometrically, the absolute value of the scalar triple product, $iggr vert vec cdot ( vec imes vec ) iggr vert$ is equal to the volume of the parallelepiped formed from the vectors $vec, vec, vec$ as depicted in the following image.

Furthermore, it is important to note that $vec cdot (vec imes vec) = vec cdot (vec imes vec) = vec cdot (vec imes vec)$ , which can be proven by noting the following determinants result in a two row interchanges which is equivalent to multiplying the determinant $vec cdot (vec imes vec)$ by $(-1)(-1) = 1$ .

We're about to look more into lines, planes, and surfaces in general in $mathbb^3$ . The following definition will be important coming up.

Definition: Three vectors $vec, vec, vec$ are said to be Coplanar if all three vectors lie on the same plane.

For example, the vectors $(0, 1, 2), (0, 2, 3), (0, -1, 3) in mathbb^3$ are coplanar as they all lie on the $yz$ -plane. The following theorem gives us a nice way of checking if three vectors in $mathbb^3$ are coplanar.

  • Proof: $Rightarrow$ Suppose that $vec, vec, vec in mathbb^3$ are coplanar. Then these vectors lie on a plane and thus one of the dimensions of the corresponding parallelepiped formed by these vectors has zero length, and so $vec cdot (vec imes vec) = 0$ .
  • $Leftarrow$ Suppose that $vec cdot (vec imes vec) = 0$ . Then the volume spanned by $vec, vec, vec$ is zero which implies that these vectors must line on the some similar plane, and thus they are coplanar. $lacksquare$

The Cross Product of Vectors in Three-Dimensional Space

Recall that for any two vectors $vec, vec in mathbb^n$ that the dot product $vec cdot vec = u_1v_1 + u_2v_2 + . + u_nv_n$ produces a scalar. We will now look at another type of vector product known as the Cross Product. We emphasize that the cross product between two vectors is only defined in $mathbb^3$ .

Using the latter definition of the cross product of two vectors in terms of a 3 x 3 Determinant:

Thus, this "determinant" form of the cross product can be rather useful instead of memorizing the longer formula for the cross product. We will now look at some properties of the cross product and prove a few of them. The rest will be left for the reader to prove.


12.2: Vectors in Three Dimensions - Mathematics

This is a basic introduction to the mathematics of vectors. Vectors are defined as mathematical expressions possessing magnitude and direction, which add according to the parallelogram law. Forces, displacements, velocities, accelerations, and momenta are examples of physical quantities that may be represented mathematically by vectors. Physical quantities that do not have direction, such as volume, mass or energy, are represented by ordinary number or scalars.

In this article, scalar variables will be denoted as an italicized variable, such as a. Variables that are vectors will be indicated with a boldface variable, such as a, although it is common to see vectors denoted with small arrows above the variable.

In many problems it will be found desirable to resolve a force into two or three components that are perpendicular to each other. These component vectors are generally oriented on a coordinate system, the most popular of which is the two-dimensional Cartesian plane. The Cartesian plane has a horizontal axis which is labeled x and a vertical axis labeled y. One can use the same principle to specify the position of any point in three-dimensional space by three Cartesian coordinates, in which the axes are x, y, and z.

Vectors in multiple-dimension coordinate systems can be resolved into their component vectors. In the three-dimensional case, this results in an x-component, a y-component and a z-component. Figure 1 to the right is an example of a vector, a, resolved into its components, ax, ay and az. When breaking a vector into its components, the vector is a sum of the components,

Addition and Subtraction

Adding scalar quantities ignores all information about the directions however, vectors are manipulated somewhat differently &ndash the direction must always be considered. Adding two or more vectors is synonymous to placing the vectors end to end, and creating a new vector running from the starting point to the end point, as demonstrated in Figure 2. If the vectors have the same direction, then this simply means adding the magnitudes, but if they have different directions, the addition is more complex. Vectors are added by resolving them into their components and then adding the components, as shown below,

The subtraction of a vector is defined as the addition of the corresponding negative vector, that is, by a &ndash b we mean a + (&ndashb).

Scalar Multiplication

The operation of multiplying a vector by a scalar is called scalar multiplication. If we define the product na of a scalar n and a vector a as a vector having the same direction as a (if n positive), or a direction opposite to that of a (if n negative), and a magnitude equal to the product of a and of the absolute value of n. The resulting vector is:

n a = n ax + n ay + n az

Scalar multiplication is distributive over vector addition in the following sense:

n (a + b) = n a + n b

We shall, at this point, introduce three vectors of magnitude 1, directed respectively along the positive x-, y- and z-axes. These vectors are called unit vectors and are denoted by i, j and k, respectively. Recalling the definition of the product of a scalar and a vector, we note that the rectangular components ax, ay and az of a vector a may be obtained by multiplying respectively the unit vectors i, j and k by appropriate scalars. We write

a = ax i + ay j + az k

While the scalars ax, ay and az may be positive or negative, depending on the sense of ax, ay and az, their absolute values are respectively equal to the magnitudes of the component vectors ax, ay and az. The scalars ax, ay and az are called the scalar components of the vector a, while the actual components ax, ay and az should be referred to as the vector components of a. However, when there exists no possibility of confusion, the vector as well as the scalar components of a may be referred to simply as the components of a. We note that the scalar component of ax is positive when the vector component ax has the same sense as the unit vector i (i.e. the same sense as the positive x-axis) and negative when ax has the opposite sense. A similar conclusion may be drawn regarding the sign of the scalar components ay and az.

The magnitude or length of the vector a, represented a = |a|, may be obtained by applying the Pythagorean theorem,

a = ( ax 2 + ay 2 + az 2 ) 1/2

The dot product of two vectors a and b is defined as the product of the magnitudes of a and b and the cosine of the angle formed by a and b. The dot product of a and b is denoted by a &bull b and is defined as:

a &bull b = a b cos

Note that the expression just defined is not a vector, but a scalar, which explains why the dot product is also referred to as the scalar product of a and b.

The dot product can also be defined as the sum of the products of the components of each vector,

Combining the above equations, we see that

  • The line of action of a × b is perpendicular to the plane containing a and b (see Figure 4).
  • The magnitude of a × b is the product of the magnitudes of a and b and the sine of the angle formed by a and b.
  • The sense of a × b is such that a person located at the tip of a × b will observe as counterclockwise the rotation through that brings a in line with b. The three vectors a, b and a × b &ndash taken in that order &ndash are said to form a right-handed triad.

The cross product differs from the dot product primarily in that the result of the cross product of two vectors is a vector, thus it is also referred to as the vector product. The cross product is defined as

a × b = a b sin() n

where n is a unit vector perpendicular to both a and b.

The cross product can also be written as

Line of Intersection between two Planes

If we are given two vectors, n1 and n2, that are normal to Plane 1 and Plane 2, then by simple geometrical reasoning, the line of intersection of the two planes is perpendicular to both normals (see Figure 5).

To find the position vector, r, of any point on the line of intersection, we (1) find a vector, v, to which the line is parallel, and (2) find the position vector, a, of a specific point on the line.

As we found above, the cross product of two vectors gives a third vector that is perpendicular to both vectors. We know that v is perpendicular to both n1 and n2 therefore, v is parallel to n1 × n2.

To find a, we find any point on the line of intersection that satisfies the equations of both planes. If we let the point have an x-coordinate of zero, then we have two equations in two unknowns, y and z. Solving for y and z and setting x = 0, we have a point on the line of intersection who's position vector is a.

Substituting v and a into the following yields an equation for the line of intersection:

r = a + t v

Transformation of Coordinates

In astrodynamics it is common to give position vectors in polar coordinates of longitude, l , latitude, b , and radial distance, r. These coordinates are converted to XYZ rectangular coordinates by the following formulae:

x = r cos b cos l
y = r cos b sin l
z = r sin b

Given XYZ coordinates, longitude and latitude are derived as follows:

l = arctan[ y / x ]
b = arctan[ z / (x 2 + y 2 ) 1/2 ] = asin[ z / r ]

where r is simply the vector magnitude. It is important that special care be taken to place l in the correct quadrant.


Three Dimensional Geometry Class 12 Maths Important Questions

CBSE, NCERT, JEE Main, NEET-UG, NDA, Exam Papers, Question Bank, NCERT Solutions, Exemplars, Revision Notes, Free Videos, MCQ Tests & more.

Three Dimensional Geometry Class 12 Maths Important Questions. myCBSEguide has just released Chapter Wise Question Answers for class 12 Maths. There chapter wise Practice Questions with complete solutions are available for download in myCBSEguide website and mobile app. These Questions with solution are prepared by our team of expert teachers who are teaching grade in CBSE schools for years. There are around 4-5 set of solved Chapter 11 Dimensional Geometry Mathematics Extra Questions from each and every chapter. The students will not miss any concept in these Chapter wise question that are specially designed to tackle Board Exam. We have taken care of every single concept given in CBSE Class 12 Mathematics syllabus and questions are framed as per the latest marking scheme and blue print issued by CBSE for class 12.


Resultant Vector In Three Dimensions

A vector in three-dimensional space. A set of three mutually orthogonal unit vectors Right handed system.

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Rectangular component of a Vector.

Resultant vector in three dimensions. Base vectors for a rectangular coordinate system. So a three-dimensional vector field is given by a function a certain multi-variable function that has a three-dimensional input given with coordinates x y and z and then a three-dimensional vector output that has expressions that are somehow dependent on x y and z Ill just put dots in here for now but well fill this in with an example in just a moment. Scalar multiplication of vectors satisfies the distributive property and the zero vector acts as an additive identity.

To find resultant of forces in 3D We can use R Fx2Fy2Fz2 R 450022250211002 R 2025000050625001210000 R 26522500 R 5150 N. The notation is a natural extension of the two-dimensional case representing a vector with the initial point at the origin and terminal point The zero vector is So for example the three dimensional vector is represented by a directed line segment from point to point Figure. We can draw the vector OP as follows.

Just like two-dimensional vectors three-dimensional vectors are quantities with both magnitude and direction and they are represented by directed line segments arrows. For a vector in three dimensions r xr yr z the magnitude is When multiplying numbers there are three different ways to show that multiplication should be performed. The proofs to verify these properties in three dimensions are straightforward extensions of the proofs in two dimensions.

Three-dimensional vectors can also be represented in component form. If displacement vectors A B and C are added together the result will be vector R. A is the angle between u and the x-axis.

Suitable for high school physics. Three-dimensional vectors can also be represented in component form. It is the result of adding two or more vectors together.

The 3D vectors are using the x-y-z axes. Now since Resultant is a 3D vector you will not get one angle theta as in case of 2D vector. G is the angle between u and the z-axis.

X or no symbol. With a three-dimensional vector we use a three-dimensional arrow. Vectors can also be multiplied but there are different kinds.

OP sqrt22 32 52 616 units. For example a multiplied by b can be written a x b a. Calculate the angle of three dimensional vectors 3D Vectors with entered vector coordinates.

As shown in the diagram vector R can be determined by the use of an accurately drawn scaled vector addition diagram. The vector vca is drawn as a green arrow with tail fixed at the origin. I review how to find the resultant graphically and then show how to do it algebraically.

Students will be able to solve 3-D particle equilibrium problems by a Drawing a 3-D free body diagram and b Applying the three scalar equations based on one vector equation of equilibrium. A coordinate system represented by base vectors which follow the right-hand rule. The resultant is the vector sum of two or more vectors.

THREE-DIMENSIONAL FORCE SYSTEMS Todays Objectives. The projections of vector A along the x y and z directions are A x A y and A z respectively. V is the angle between u and the y-axis.

Vectors in Two and Three Dimensions Algebra and Trigonometry - James Stewart Lothar Redlin Saleem Watson All the textbook answers and step-by-step explana. These properties of vector operations are valid for three-dimensional vectors as well. Magnitude of a 3-Dimensional Vector.

A vector of unit length. With a three-dimensional vector we use a three-dimensional arrow. You can drag the head of the green arrow with your mouse to change the vector.

A representation of a vector vcaa_1a_2a_3 in the three-dimensional Cartesian coordinate system. We saw earlier that the distance between 2 points in 3-dimensional space is distance AB sqrt x_2-x_12 y_2-y_12 z_2-z_12 For the vector OP above the magnitude of the vector is given by.

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12.2: Vectors in Three Dimensions - Mathematics

Values used in physics are either scalar or vector.

Scalar values indicate positive magnitudes such as mass, speed, work, charge, or energy, and have no directional component.

Many values do, however, have a directional component. These includes position, velocity, acceleration, and force, which all require a vector to be described properly. See interactive diagram above.

For a three-dimensional vector, we can use three numbers to describe the end-point of a directed line segment that starts at the origin, with each number indicating the displacement along each of the three axis, the x axis, y axis, and z axis.

We can determine the length of such a vector with the equation:

length = sqrt(Rx2 + Ry2 + Rz2)

We can describe the direction of the vector in terms of two angles, where:

We can think of the vector R in the diagram as the sum of the unit vectors, each multiplied by the magnitude of R in each respective dimension (see green dots), or


Watch the video: : Three-Dimensional Coordinate Systems (December 2021).