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Vectors in Three Dimensions

Describe three-dimensional space mathematically.

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 (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, \(z,\) is appended to indicate alignment with the z-axis: \((x,y,z).\) A point in space is identified by all three coordinates (). 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

Try it.

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 ().

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 .

\[d=\sqrt{{({x}_{2}-{x}_{1})}^{2}+{({y}_{2}-{y}_{1})}^{2}}.\]

Condensed — the full section is in OpenStax Calculus Volume 3.

Writing Equations in ℝ

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 \(ℝ,{ℝ}^{2},\ \text{and}\ {ℝ}^{3}\) (). 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 ().

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.

Example

Try it.

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

Solution

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 \(x=3.\)
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 \(y=-2.\)

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.

Condensed — the full section is in OpenStax Calculus Volume 3.

Working with Vectors in ℝ

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 \(\text{v}=〈x,y,z〉\) 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 \(0=〈0,0,0〉.\) So, for example, the three dimensional vector \(\text{v}=〈2,4,1〉\) is represented by a directed line segment from point \((0,0,0)\) to point \((2,4,1)\) ().

Vector addition and scalar multiplication are defined analogously to the two-dimensional case. If \(\text{v}=〈{x}_{1},{y}_{1},{z}_{1}〉\) and \(\text{w}=〈{x}_{2},{y}_{2},{z}_{2}〉\) are vectors, and \(k\) is a scalar, then

\[\text{v}+\text{w}=〈{x}_{1}+{x}_{2},{y}_{1}+{y}_{2},{z}_{1}+{z}_{2}〉\ \text{and}\ k\text{v}=〈k{x}_{1},k{y}_{1},k{z}_{1}〉.\]

If \(k=-1,\) then \(k\text{v}=(-1)\text{v}\) is written as \(\text{-}\text{v},\) and vector subtraction is defined by \(\text{v}-w=v+(\text{-}\text{w})=v+(-1)\text{w}.\)

The standard unit vectors extend easily into three dimensions as well—\(\text{i}=〈1,0,0〉,\) \(\text{j}=〈0,1,0〉,\) and \(\text{k}=〈0,0,1〉\)—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:

\[\text{v}=〈x,y,z〉=x\text{i}+y\text{j}+z\text{k}.\]

Example

Try it.

Let \(\overset{\to}{PQ}\) be the vector with initial point \(P=(3,12,6)\) and terminal point \(Q=(-4,-3,2)\) as shown in . Express \(\overset{\to}{PQ}\) in both component form and using standard unit vectors.

Solution

In component form,

\[\begin{array}{ll}\overset{\to}{PQ} & =〈{x}_{2}-{x}_{1},{y}_{2}-{y}_{1},{z}_{2}-{z}_{1}〉 \\ & =〈-4-3,-3-12,2-6〉=〈-7,-15,-4〉.\end{array}\]

In standard unit form,

\[\overset{\to}{PQ}=-7\text{i}-15\text{j}-4\text{k}.\]

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 ().

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.

Condensed — the full section is in OpenStax Calculus Volume 3.

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
\[d=\sqrt{{({x}_{2}-{x}_{1})}^{2}+{({y}_{2}-{y}_{1})}^{2}+{({z}_{2}-{z}_{1})}^{2}}.\]
In three dimensions, the equations \(x=a,y=b,\text{and}\ z=c\) describe planes that are parallel to the coordinate planes.
The standard equation of a sphere with center \((a,b,c)\) and radius \(r\) is
\[{(x-a)}^{2}+{(y-b)}^{2}+{(z-c)}^{2}={r}^{2}.\]
In three dimensions, as in two, vectors are commonly expressed in component form, \(\text{v}=〈x,y,z〉,\) or in terms of the standard unit vectors, \(x\text{i}+y\text{j}+z\text{k}.\)
Properties of vectors in space are a natural extension of the properties for vectors in a plane. Let \(\text{v}=〈{x}_{1},{y}_{1},{z}_{1}〉\) and \(\text{w}=〈{x}_{2},{y}_{2},{z}_{2}〉\) be vectors, and let \(k\) be a scalar.
- Scalar multiplication: \(k\text{v}=〈k{x}_{1},k{y}_{1},k{z}_{1}〉\)
- Vector addition: \(\text{v}+\text{w}=〈{x}_{1},{y}_{1},{z}_{1}〉+〈{x}_{2},{y}_{2},{z}_{2}〉=〈{x}_{1}+{x}_{2},{y}_{1}+{y}_{2},{z}_{1}+{z}_{2}〉\)
- Vector subtraction: \(\text{v}-\text{w}=〈{x}_{1},{y}_{1},{z}_{1}〉-〈{x}_{2},{y}_{2},{z}_{2}〉=〈{x}_{1}-{x}_{2},{y}_{1}-{y}_{2},{z}_{1}-{z}_{2}〉\)
- Vector magnitude: \(‖\text{v}‖=\sqrt{{x}_{1}{}^{2}+{y}_{1}{}^{2}+{z}_{1}{}^{2}}\)
- Unit vector in the direction of v: \(\frac{\text{v}}{‖\text{v}‖}=\frac{1}{‖\text{v}‖}〈{x}_{1},{y}_{1},{z}_{1}〉=〈\frac{{x}_{1}}{‖\text{v}‖},\frac{{y}_{1}}{‖\text{v}‖},\frac{{z}_{1}}{‖\text{v}‖}〉,\) \(\text{v}\ne 0\)

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:	\({(x-a)}^{2}+{(y-b)}^{2}+{(z-c)}^{2}={r}^{2}\)

Vectors in Three Dimensions

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

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

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

For the following exercises, express vector \(\overset{\to}{PQ}\) with the initial point at \(P\) and the terminal point at \(Q\)

in component form and
by using standard unit vectors.

For the following exercises, use the given vectors \(\text{a}\) and \(\text{b}\) to find and express the vectors \(\text{a}+\text{b},\) \(4\text{a},\) and \(-5\text{a}+3\text{b}\) in component form.

For the following exercises, vectors u and v are given. Find the magnitudes of vectors \(\text{u}-\text{v}\) and \(-2\text{u}.\)

For the following exercises, find the unit vector in the direction of the given vector \(\text{a}\) and express it using standard unit vectors.

Condensed — the full section is in OpenStax Calculus Volume 3.

Practice (40)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

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

Одкриј го одговорот

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

Одкриј го одговорот
Find the distance between points \({P}_{1}=(3,\text{-}1,5)\) and \({P}_{2}=(2,1,\text{-}1).\)

Одкриј го одговорот

Substitute values directly into the distance formula:
\[\begin{array}{ll}d({P}_{1},{P}_{2}) & =\sqrt{{({x}_{2}-{x}_{1})}^{2}+{({y}_{2}-{y}_{1})}^{2}+{({z}_{2}-{z}_{1})}^{2}} \\ & =\sqrt{{(2-3)}^{2}+{(1-(-1))}^{2}+{(-1-5)}^{2}} \\ & =\sqrt{(-1{)}^{2}+{2}^{2}+{(-6)}^{2}} \\ & =\sqrt{41}.\end{array}\]
Find the distance between points \({P}_{1}=(1,-5,4)\) and \({P}_{2}=(4,-1,-1).\)

Одкриј го одговорот

\(5\sqrt{2}\)
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 \((1,-2,-3).\)
Одкриј го одговорот
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 \(y=-2.\)
Write an equation of the plane passing through point \((1,-6,-4)\) that is parallel to the xy-plane.

Одкриј го одговорот

\(z=-4\)
Find the standard equation of the sphere with center \((10,7,4)\) and point \((-1,3,-2),\) as shown in .

Одкриј го одговорот

Use the distance formula to find the radius \(r\) of the sphere:
\[\begin{array}{ll}r & =\sqrt{{(-1-10)}^{2}+{(3-7)}^{2}+{(-2-4)}^{2}} \\ & =\sqrt{{(-11)}^{2}+{(-4)}^{2}+{(-6)}^{2}} \\ & =\sqrt{173}.\end{array}\]
The standard equation of the sphere is
\[{(x-10)}^{2}+{(y-7)}^{2}+{(z-4)}^{2}=173.\]
Find the standard equation of the sphere with center \((-2,4,-5)\) containing point \((4,4,-1).\)

Одкриј го одговорот

\({(x+2)}^{2}+{(y-4)}^{2}+{(z+5)}^{2}=52\)
Let \(P=(-5,2,3)\) and \(Q=(3,4,-1),\) and suppose line segment \(PQ\) forms the diameter of a sphere (). Find an equation of the sphere.

Одкриј го одговорот

Since \(PQ\) is a diameter of the sphere, we know the center of the sphere is the midpoint of \(PQ.\) Then,
\[\begin{array}{ll}C & =(\frac{-5+3}{2},\frac{2+4}{2},\frac{3+(-1)}{2}) \\ & =(-1,3,1).\end{array}\]
Furthermore, we know the radius of the sphere is half the length of the diameter. This gives
\[\begin{array}{ll}r & =\frac{1}{2}\sqrt{{(-5-3)}^{2}+{(2-4)}^{2}+{(3-(-1))}^{2}} \\ & =\frac{1}{2}\sqrt{64+4+16} \\ & =\sqrt{21}.\end{array}\]
Then, the equation of the sphere is \({(x+1)}^{2}+{(y-3)}^{2}+{(z-1)}^{2}=21.\)
Find an equation of the sphere with diameter \(PQ,\) where \(P=(2,-1,-3)\) and \(Q=(-2,5,-1).\)

Одкриј го одговорот

\({x}^{2}+{(y-2)}^{2}+{(z+2)}^{2}=14\)
Describe the set of points that satisfies \((x-4)(z-2)=0,\) and graph the set.

Одкриј го одговорот

We must have either \(x-4=0\) or \(z-2=0,\) so the set of points forms the two planes \(x=4\) and \(z=2\) ().
Describe the set of points that satisfies \((y+2)(z-3)=0,\) and graph the set.

Одкриј го одговорот

The set of points forms the two planes \(y=-2\) and \(z=3.\)
Describe the set of points in three-dimensional space that satisfies \({(x-2)}^{2}+{(y-1)}^{2}=4,\) and graph the set.

Одкриј го одговорот

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 \(x=2\ \text{and}\ y=1.\) The cylinder extends indefinitely in the z-direction ().
Describe the set of points in three dimensional space that satisfies \({x}^{2}+{(z-2)}^{2}=16,\) and graph the surface.

Одкриј го одговорот

A cylinder of radius 4 centered on the line with \(x=0\ \text{and}\ z=2.\)
Let \(\overset{\to}{PQ}\) be the vector with initial point \(P=(3,12,6)\) and terminal point \(Q=(-4,-3,2)\) as shown in . Express \(\overset{\to}{PQ}\) in both component form and using standard unit vectors.

Одкриј го одговорот

In component form,
\[\begin{array}{ll}\overset{\to}{PQ} & =〈{x}_{2}-{x}_{1},{y}_{2}-{y}_{1},{z}_{2}-{z}_{1}〉 \\ & =〈-4-3,-3-12,2-6〉=〈-7,-15,-4〉.\end{array}\]
In standard unit form,
\[\overset{\to}{PQ}=-7\text{i}-15\text{j}-4\text{k}.\]
Let \(S=(3,8,2)\) and \(T=(2,-1,3).\) Express \(\overset{\to}{ST}\) in component form and in standard unit form.

Одкриј го одговорот

\(\overset{\to}{ST}=〈-1,-9,1〉=\text{-}\text{i}-9\text{j}+\text{k}\)
Let \(\text{v}=〈-2,9,5〉\) and \(\text{w}=〈1,-1,0〉\) (). Find the following vectors.
1. \(3\text{v}-2\text{w}\)
2. \(5‖\text{w}‖\)
3. \(‖5\text{w}‖\)
4. A unit vector in the direction of \(\text{v}\)
Одкриј го одговорот
1. First, use scalar multiplication of each vector, then subtract:
  \[\begin{array}{ll}3\text{v}-2\text{w} & =3〈-2,9,5〉-2〈1,-1,0〉 \\ & =〈-6,27,15〉-〈2,-2,0〉 \\ & =〈-6-2,27-(-2),15-0〉 \\ & =〈-8,29,15〉.\end{array}\]
2. Write the equation for the magnitude of the vector, then use scalar multiplication:
  \[5‖\text{w}‖=5\sqrt{{1}^{2}+{(-1)}^{2}+{0}^{2}}=5\sqrt{2}.\]
3. First, use scalar multiplication, then find the magnitude of the new vector. Note that the result is the same as for part b.:
  \[‖5\text{w}‖=‖〈5,-5,0〉‖=\sqrt{{5}^{2}+{(-5)}^{2}+{0}^{2}}=\sqrt{50}=5\sqrt{2}.\]
4. 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:
  \[\begin{array}{ll}\frac{\text{v}}{‖\text{v}‖} & =\frac{1}{‖\text{v}‖}〈-2,9,5〉 \\ & =\frac{1}{\sqrt{{(-2)}^{2}+{9}^{2}+{5}^{2}}}〈-2,9,5〉 \\ & =\frac{1}{\sqrt{110}}〈-2,9,5〉 \\ & =〈\frac{-2}{\sqrt{110}},\frac{9}{\sqrt{110}},\frac{5}{\sqrt{110}}〉.\end{array}\]
Let \(\text{v}=〈-1,-1,1〉\) and \(\text{w}=〈2,0,1〉.\) Find a unit vector in the direction of \(5\text{v}+3\text{w}.\)

Одкриј го одговорот

\(〈\frac{1}{3\sqrt{10}},-\frac{5}{3\sqrt{10}},\frac{8}{3\sqrt{10}}〉\)
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 \(30\text{^{\circ}}\) (see the following figure). Write the initial velocity vector of the ball, \(\text{v},\) in component form.

Одкриј го одговорот

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 \(\text{w}\) extending from the quarterback’s arm to a point directly above the receiver’s head at an angle of \(30\text{^{\circ}}\) (see the following figure). This vector would have the same direction as \(\text{v},\) 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
\[\text{Dist from QB to receiver}=\sqrt{{15}^{2}+{20}^{2}}=\sqrt{225+400}=\sqrt{625}=25\ \text{yd}.\]
We have \(\frac{25}{‖\text{w}‖}=\text{cos}\ 30\text{^{\circ}}.\) Then the magnitude of \(\text{w}\) is given by
\[‖\text{w}‖=\frac{25}{\text{cos}\ 30\text{^{\circ}}}=\frac{25\cdot 2}{\sqrt{3}}=\frac{50}{\sqrt{3}}\ \text{yd}\]
and the vertical distance from the receiver to the terminal point of \(\text{w}\) is
\[\text{Vert dist from receiver to terminal point of}\ \text{w}=‖\text{w}‖\text{sin}\ 30\text{^{\circ}}=\frac{50}{\sqrt{3}}\cdot \frac{1}{2}=\frac{25}{\sqrt{3}}\ \text{yd}.\]
Then \(\text{w}=〈20,15,\frac{25}{\sqrt{3}}〉,\) and has the same direction as \(\text{v}.\)
Recall, though, that we calculated the magnitude of \(\text{w}\) to be \(‖\text{w}‖=\frac{50}{\sqrt{3}},\) and \(\text{v}\) has magnitude \(60\) mph. So, we need to multiply vector \(\text{w}\) by an appropriate constant, \(k.\) We want to find a value of \(k\) so that \(‖k\text{w}‖=60\) mph. We have
\[‖k\text{w}‖=k‖\text{w}‖=k\frac{50}{\sqrt{3}}\ \text{mph,}\]
so we want
\[\begin{array}{lll} \\ \\ k\frac{50}{\sqrt{3}} & = & 60 \\ k & = & \frac{60\sqrt{3}}{50} \\ k & = & \frac{6\sqrt{3}}{5}.\end{array}\]
Then
\[\text{v}=k\text{w}=k〈20,15,\frac{25}{\sqrt{3}}〉=\frac{6\sqrt{3}}{5}〈20,15,\frac{25}{\sqrt{3}}〉=〈24\sqrt{3},18\sqrt{3},30〉.\]
Let’s double-check that \(‖\text{v}‖=60.\) We have
\[‖\text{v}‖=\sqrt{{(24\sqrt{3})}^{2}+{(18\sqrt{3})}^{2}+{(30)}^{2}}=\sqrt{1728+972+900}=\sqrt{3600}=60\ \text{mph}.\]
So, we have found the correct components for \(\text{v}.\)
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 \(45\text{^{\circ}}.\) Write the initial velocity vector of the ball, \(\text{v},\) in component form.

Одкриј го одговорот

\(\text{v}=〈16\sqrt{2},12\sqrt{2},20\sqrt{2}〉\)
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
1. the coordinates of the other six vertices of the box and
2. the length of the diagonal of the box determined by the vertices \(O\) and \(A.\)
Одкриј го одговорот

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

Одкриј го одговорот

A union of two planes: \(y=5\) (a plane parallel to the xz-plane) and \(z=6\) (a plane parallel to the xy-plane)
\((z-2)(z-5)=0\)
\({(y-1)}^{2}+{(z-1)}^{2}=1\)

Одкриј го одговорот

A cylinder of radius \(1\) centered on the line \(y=1,z=1\)
\({(x-2)}^{2}+{(z-5)}^{2}=4\)
Write the equation of the plane passing through point \((1,1,1)\) that is parallel to the xy-plane.

Одкриј го одговорот

\(z=1\)
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 \((1,0,-2).\)

Одкриј го одговорот

\(z=-2\)
Find an equation of the plane passing through points \((1,9,2),\) \((1,3,6),\) and \((1,-7,8).\)
Center \(C(-1,7,4)\) and radius \(4\)

Одкриј го одговорот

\({(x+1)}^{2}+{(y-7)}^{2}+{(z-4)}^{2}=16\)
Center \(C(-4,7,2)\) and radius \(6\)
Diameter \(PQ,\) where \(P(-1,5,7)\) and \(Q(-5,2,9)\)

Одкриј го одговорот

\({(x+3)}^{2}+{(y-3.5)}^{2}+{(z-8)}^{2}=\frac{29}{4}\)
Diameter \(PQ,\) where \(P(-16,-3,9)\) and \(Q(-2,3,5)\)
\({x}^{2}+{y}^{2}+{z}^{2}-4z+3=0\)

Одкриј го одговорот

Center \(C(0,0,2)\) and radius \(1\)
\({x}^{2}+{y}^{2}+{z}^{2}-6x+8y-10z+25=0\)
\(P(3,0,2)\) and \(Q(-1,-1,4)\)

Одкриј го одговорот

a. \(\overset{\to}{PQ}=〈-4,-1,2〉;\) b. \(\overset{\to}{PQ}=-4\text{i}-\text{j}+2\text{k}\)
\(P(0,10,5)\) and \(Q(1,1,-3)\)
\(P(-2,5,-8)\) and \(M(1,-7,4),\) where \(M\) is the midpoint of the line segment \(PQ\)

Одкриј го одговорот

a. \(\overset{\to}{PQ}=〈6,-24,24〉;\) b. \(\overset{\to}{PQ}=6\text{i}-24\text{j}+24\text{k}\)
\(Q(0,7,-6)\) and \(M(-1,3,2),\) where \(M\) is the midpoint of the line segment \(PQ\)

Symbols used here

\sqrt{x},\ \sqrt[n]{x}

square root, n-th root
The non-negative number whose square (n-th power) is x.

imaginary unit
i² = −1.

^\circ

degrees
1/360 of a full turn. 180° = π radians.

\neq

not equal
The two sides are different.

\frac{\partial f}{\partial x}

partial derivative
Derivative with respect to x, holding the other variables fixed.

\nabla f

gradient (nabla, del)
Vector of partial derivatives; points uphill.

\int f(x)\,dx,\ \int_a^b

integral
Antiderivative (indefinite) or signed area from a to b (definite).

\iint,\ \oint

double / contour integral
Integral over a region of the plane; integral around a closed curve.

\mathbf{v},\ \vec{v}

vector
A quantity with magnitude and direction; a column of numbers.

\mathbf{u} \cdot \mathbf{v},\ \|\mathbf{v}\|

dot product, norm
Σ u_i v_i; the length of v, √(v·v).

How to: Vectors in Three Dimensions

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
The plane in space that is parallel to the
The plane in space that is parallel to the
The plane in space that is parallel to the

Questions people ask

What is a partial derivative?

The ordinary derivative with respect to one variable while every other variable is frozen — the slope of the surface in one coordinate direction.

What does the gradient point at?

Uphill: the direction of steepest increase, with length equal to that steepest slope. It is perpendicular to the level curves.

Обиди се со себе.

Parts of this page are adapted from OpenStax Calculus Volume 3 (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.

Повеќе во Multivariable Calculus

Functions of several variables Partial derivatives and the gradient Optimisation in several variables Double and triple integrals Vector fields, line integrals and the big theorems