Wize University Calculus 2 Textbook > Vector Functions

Vector Geometry

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3D Coordinate System
2D space R2\colorOne{R^2}R2: 
We have the x and y axes
Our points have coordinates(x,y)\left(x,y\right)(x,y)
The equation involving xxx and yyy describes a curve in R2R^2R2

3D space R3\colorOne{R^3}R3:
We have the x, y, and z axes
Our points have coordinates (x,y,z)\left(x,y,z\right)(x,y,z)
The equation involving x, yx,\ yx, y and zzz describes a surface in R3R^3R3
Ex. The equation of a sphere in R3R^3R3 is (x−h)2+(y−k)2+(z−l)2=r2\left(x-h\right)^2+\left(y-k\right)^2+\left(z-l\right)^2=r^2(x−h)2+(y−k)2+(z−l)2=r2 -- the graph of this surface will be all the points that are on the surface of this sphere
The distance between two points P1(x1, y1, z1)P_1\left(x_1,\ y_1,\ z_1\right)P1​(x1​, y1​, z1​) and P2(x2,y2,z2)P_2\left(x_2,y_2,z_2\right)P2​(x2​,y2​,z2​) is ∣P1P2∣=(x2−x1)2+(y2−y1)2+(z2−z1)2\left|P_1P_2\right|=\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2+\left(z_2-z_1\right)^2}∣P1​P2​∣=(x2​−x1​)2+(y2​−y1​)2+(z2​−z1​)2​

Example
Describe the surface represented by the equation y=x2y=x^2y=x2 in R3R^3R3

In the R2R^2R2 space, y=x2y=x^2y=x2 describes a parabola with vertex at the origin (0, 0)
So, in R3R^3R3 space, this describes the stack of identical parabolas with vertices on the z-axis

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Vectors
[Everything we discuss in this lesson applies to vectors in any space RnR^nRn]

A vector in R3R^3R3 is represented by v⃗=⟨x,y,z⟩\vec{v}=\left\langle x,y,z\right\ranglev=⟨x,y,z⟩
*x,y,zx, y, zx,y,z are called the components of this vector

Geometrically
A vector v⃗=⟨x,y,z⟩\vec{v}=\left\langle x,y,z\right\ranglev=⟨x,y,z⟩ is a directed line segment that goes from the origin (0,0,0)(0,0,0)(0,0,0) and points to the point (x,y,z)(x,y,z)(x,y,z), but this line segment (vector) can be moved around in space

Given any two points P1(x1,y1,z1)P_1\left(x_1,y_1,z_1\right)P1​(x1​,y1​,z1​) and P2(x2,y2,z2)P_2\left(x_2,y_2,z_2\right)P2​(x2​,y2​,z2​), the position vector P1P2→=⟨x2−x1, y2−y1, z2−z1⟩\overrightarrow{P_1P_2}=\left\langle x_2-x_1,\ y_2-y_1,\ z_2-z_1\right\rangleP1​P2​​=⟨x2​−x1​, y2​−y1​, z2​−z1​⟩ is the vector that starts at P1P_1P1​ and points to P2P_2P2​

Vector Operations
Length of a vector: ∣⟨a,b,c⟩∣=a2+b2+c2\left|\left\langle a,b,c\right\rangle\right|=\sqrt{a^2+b^2+c^2}∣⟨a,b,c⟩∣=a2+b2+c2​
Scalar multiplication: k⟨a,b,c⟩=⟨ka, kb, kc⟩k\left\langle a,b,c\right\rangle=\left\langle ka,\ kb,\ kc\right\ranglek⟨a,b,c⟩=⟨ka, kb, kc⟩
Vector addition/subtraction: ⟨a,b,c⟩±⟨p,q,r⟩=⟨a±p, b±q, c±r⟩\left\langle a,b,c\right\rangle\pm\left\langle p,q,r\right\rangle=\left\langle a\pm p,\ b\pm q,\ c\pm r\right\rangle⟨a,b,c⟩±⟨p,q,r⟩=⟨a±p, b±q, c±r⟩
A unit vector in the same direction as v⃗\vec{v}v is defined as v⃗∣v⃗∣\frac{\vec{v}}{\left|\vec{v}\right|}∣v∣v​
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Properties
u⃗, v⃗, w⃗\vec{u},\ \vec{v},\ \vec{w}u, v, w are vectors in R3R^3R3 and c, dc,\ dc, d are scalars
Addition‾u⃗+v⃗=v⃗+u⃗u⃗+(v⃗+w⃗)=(u⃗+v⃗)+w⃗Scalar Multiplication‾(cd)u⃗=c(du⃗)c(u⃗+v⃗)=cu⃗+cv⃗(c+d)u⃗=cu⃗+du⃗The 0 and 1 in Vectors‾1u⃗=u⃗u⃗+(−u⃗)=0⃗u⃗+0⃗=u⃗ where 0⃗=⟨0,0,0⟩\begin{array}{|lll|}
\hline
\underline{\text{Addition}}\\
\vec{u}+\vec{v}=\vec{v}+\vec{u}&
\vec{u}+\left(\vec{v}+\vec{w}\right)=\left(\vec{u}+\vec{v}\right)+\vec{w}\\\\
\hline
\underline{\text{Scalar Multiplication}}\\
\left(cd\right)\vec{u}=c\left(d\vec{u}\right)&
c\left(\vec{u}+\vec{v}\right)=c\vec{u}+c\vec{v}&
\left(c+d\right)\vec{u}=c\vec{u}+d\vec{u}\\\\
\hline
\underline{\text{The 0 and 1 in Vectors}}\\
1\vec u=\vec u&\vec u+(-\vec u)=\vec 0&\vec{u}+\vec{0}=\vec{u}\text{ where }\vec 0=\langle0,0,0\rangle\\\\\hline
\end{array}Addition​u+v=v+uScalar Multiplication​(cd)u=c(du)The 0 and 1 in Vectors​1u=u​u+(v+w)=(u+v)+wc(u+v)=cu+cvu+(−u)=0​(c+d)u=cu+duu+0=u where 0=⟨0,0,0⟩​​

Standard basis vectors
These are vectors with length 1 (unit vector) that point in the direction along the positive x, y, z axes
i=⟨1,0,0⟩i=\left\langle1,0,0\right\ranglei=⟨1,0,0⟩
j=⟨0,1,0⟩j=\left\langle0,1,0\right\ranglej=⟨0,1,0⟩
k=⟨0,0,1⟩k=\left\langle0,0,1\right\ranglek=⟨0,0,1⟩

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Dot Product
The dot product between two vectors in R3R^3R3 is defined as ⟨a,b,c⟩⋅⟨p,q,r⟩=ap+bq+cr\left\langle a,b,c\right\rangle\cdot\left\langle p,q,r\right\rangle=ap+bq+cr⟨a,b,c⟩⋅⟨p,q,r⟩=ap+bq+cr
*The result is a number!

Properties
u⃗⋅v⃗=v⃗⋅u⃗\vec{u}\cdot\vec{v}=\vec{v}\cdot\vec{u}u⋅v=v⋅u
u⃗⋅(v⃗+w⃗)=u⃗⋅v⃗+u⃗⋅w⃗\vec{u}\cdot\left(\vec{v}+\vec{w}\right)=\vec{u}\cdot\vec{v}+\vec{u}\cdot\vec{w}u⋅(v+w)=u⋅v+u⋅w
(cu⃗)⋅v⃗=c(u⃗⋅v⃗)=u⃗⋅(cv⃗)\left(c\vec{u}\right)\cdot\vec{v}=c\left(\vec{u}\cdot\vec{v}\right)=\vec{u}\cdot\left(c\vec{v}\right)(cu)⋅v=c(u⋅v)=u⋅(cv)
u⃗⋅u⃗=∣u⃗∣2\vec{u}\cdot\vec{u}=\left|\vec{u}\right|^2u⋅u=∣u∣2
0⃗⋅u⃗=0\vec{0}\cdot\vec{u}=00⋅u=0
cos⁡θ=u⃗⋅v⃗∣u⃗∣∣v⃗∣\cos\theta=\frac{\vec{u}\cdot\vec{v}}{\left|\vec{u}\right|\left|\vec{v}\right|}cosθ=∣u∣∣v∣u⋅v​ where θ\thetaθ is the acute angle between the vectors
Example
If u⃗=⟨1,3,6⟩\vec{u}=\left\langle1,3,\sqrt 6\right\rangleu=⟨1,3,6​⟩ and v⃗=⟨3,0,−4⟩\vec{v}=\left\langle3,0,-4\right\ranglev=⟨3,0,−4⟩, find the cosine of the angle between these two vectors.

cos⁡θ=⟨1,3,6⟩⋅⟨3,0,−4⟩∣⟨1,3,6⟩∣∣⟨3,0,−4⟩∣=3−46(4)(5)=3−4620\displaystyle \cos\theta=\frac{\left\langle1,3,\sqrt{6}\right\rangle\cdot\left\langle3,0,-4\right\rangle}{\left|\left\langle1,3,\sqrt{6}\right\rangle\right|\left|\left\langle3,0,-4\right\rangle\right|}=\frac{3-4\sqrt 6}{(4)(5)}=\frac{3-4\sqrt 6}{20}cosθ=​⟨1,3,6​⟩​∣⟨3,0,−4⟩∣⟨1,3,6​⟩⋅⟨3,0,−4⟩​=(4)(5)3−46​​=203−46​​

Orthogonal Vectors
Two vectors are orthogonal (perpendicular) if and only if their dot product is 0


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Directional Cosines & Angles
The angle that a non-zero vector v⃗=⟨a,b,c⟩\vec{v}=\left\langle a,b,c\right\ranglev=⟨a,b,c⟩ forms with the x, y, and z axes are given below
cos⁡α=a∣v⃗∣\cos\alpha=\frac{a}{\left|\vec{v}\right|}cosα=∣v∣a​ → α=cos⁡−1(a∣v⃗∣)\to\ \alpha=\cos^{-1}\left(\frac{a}{\left|\vec{v}\right|}\right)→ α=cos−1(∣v∣a​)
cos⁡β=b∣v⃗∣\cos\beta=\frac{b}{\left|\vec{v}\right|}cosβ=∣v∣b​ → β=cos⁡−1(b∣v⃗∣)\to\ \beta=\cos^{-1}\left(\frac{b}{\left|\vec{v}\right|}\right)→ β=cos−1(∣v∣b​)
cos⁡γ=c∣v⃗∣\cos\gamma=\frac{c}{\left|\vec{v}\right|}cosγ=∣v∣c​ → γ=cos⁡−1(c∣v⃗∣)\to\ \gamma=\cos^{-1}\left(\frac{c}{\left|\vec{v}\right|}\right)→ γ=cos−1(∣v∣c​)

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Projections
The projection of u⃗\vec{u}u onto v⃗\vec{v}v is denoted by projv⃗u⃗\text{proj}_{\vec{v}}\vec{u}projv​u, and it describes the "shadow" u⃗\vec{u}u casts on v⃗\vec{v}v
Vector projection: proj⁡v⃗u⃗=(v⃗⋅u⃗∣v⃗∣2)v⃗\displaystyle \operatorname{proj}_{\vec{v}}\vec{u}=\left(\frac{\vec{v}\cdot\vec{u}}{\left|\vec{v}\right|^2}\right)\vec{v}projv​u=(∣v∣2v⋅u​)v
Scalar projection (length of the projection): compv⃗u⃗=v⃗⋅u⃗∣v⃗∣\displaystyle \text{comp}_{\vec{v}}\vec{u}=\frac{\vec{v}\cdot\vec{u}}{\left|\vec{v}\right|}compv​u=∣v∣v⋅u​
Example
Find the projection of ⟨1,2,3⟩\left\langle1,2,3\right\rangle⟨1,2,3⟩ onto ⟨4,0,−1⟩\left\langle4,0,-1\right\rangle⟨4,0,−1⟩

The vector projection is: proj⁡⟨4,0,−1⟩⟨1,2,3⟩=(117)⟨4,0,−1⟩ or ⟨417,0,−117⟩\operatorname{proj}_{\left\langle4,0,-1\right\rangle}\left\langle1,2,3\right\rangle=\left(\frac{1}{17}\right)\left\langle4,0,-1\right\rangle\ or\ \left\langle\frac{4}{17},0,-\frac{1}{17}\right\rangleproj⟨4,0,−1⟩​⟨1,2,3⟩=(171​)⟨4,0,−1⟩ or ⟨174​,0,−171​⟩

The scalar projection is comp⟨4,0,1⟩⟨1,2,3⟩=117\text{comp}_{\left\langle4,0,1\right\rangle}\left\langle1,2,3\right\rangle=\frac{1}{17}comp⟨4,0,1⟩​⟨1,2,3⟩=171​

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Cross Product
The cross product between two vectors in   R3  ‾\underline{\ \ R^3\ \ }  R3  ​ gives us a thrid vector in R3R^3R3 that is orthogonal to the original two vectors. It is defined by
⟨u1, u2, u3⟩×⟨v1, v2, v3⟩=⟨u2v3−u3v2, u3v1−u1v3, u1v2−u2v1⟩\left\langle u_1,\ u_2,\ u_3\right\rangle\times\left\langle v_1,\ v_2,\ v_3\right\rangle=\left\langle u_2v_3-u_3v_2,\ u_3v_1-u_1v_3,\ u_1v_2-u_2v_1\right\rangle⟨u1​, u2​, u3​⟩×⟨v1​, v2​, v3​⟩=⟨u2​v3​−u3​v2​, u3​v1​−u1​v3​, u1​v2​−u2​v1​⟩

Properties
u⃗×v⃗=−(v⃗×u⃗)\vec{u}\times\vec{v}=-\left(\vec{v}\times\vec{u}\right)u×v=−(v×u)
(cu⃗)×v⃗=c(u⃗×v⃗)=u⃗×(cv⃗)\left(c\vec{u}\right)\times\vec{v}=c\left(\vec{u}\times\vec{v}\right)=\vec{u}\times\left(c\vec{v}\right)(cu)×v=c(u×v)=u×(cv)
u⃗×(v⃗+w⃗)=u⃗×v⃗+u⃗×w⃗\vec{u}\times\left(\vec{v}+\vec{w}\right)=\vec{u}\times\vec{v}+\vec{u}\times\vec{w}u×(v+w)=u×v+u×w
(u⃗+v⃗)×w⃗=u⃗×w⃗+v⃗×w⃗\left(\vec{u}+\vec{v}\right)\times\vec{w}=\vec{u}\times\vec{w}+\vec{v}\times\vec{w}(u+v)×w=u×w+v×w
∣u⃗×v⃗∣=∣u⃗∣∣v⃗∣sin⁡θ\left|\vec{u}\times\vec{v}\right|=\left|\vec{u}\right|\left|\vec{v}\right|\sin\theta∣u×v∣=∣u∣∣v∣sinθ where θ\thetaθ is the acute angle between the two vectors
u⃗×v⃗=0\vec{u}\times\vec{v}=0u×v=0 if and only if u⃗\vec{u}u and v⃗\vec{v}v are parallel

Application
The volume of a parallelepiped defined by the vectors u⃗, v⃗, and w⃗\vec{u},\ \vec{v},\ \text{and}\ \vec{w}u, v, and w is calculated by 
V=∣u⃗⋅(v⃗×w⃗)∣V=\left|\vec{u}\cdot\left(\vec{v}\times\vec{w}\right)\right|V=∣u⋅(v×w)∣
*Choose the simpler looking vectors to perform the cross product

Example
Find the volume of the parallelpiped defined by the vecotrs ⟨1,2,0⟩, ⟨−1,4,5⟩, ⟨0,1,1⟩\left\langle1,2,0\right\rangle,\ \left\langle-1,4,5\right\rangle,\ \left\langle0,1,1\right\rangle⟨1,2,0⟩, ⟨−1,4,5⟩, ⟨0,1,1⟩

V=∣⟨−1,4,5⟩⋅(⟨1,2,0⟩×⟨0,1,1⟩)∣V=\left|\left\langle-1,4,5\right\rangle\cdot(\langle1,2,0\rangle\times\langle0,1,1\rangle)\right|V=∣⟨−1,4,5⟩⋅(⟨1,2,0⟩×⟨0,1,1⟩)∣
V=∣⟨−1,4,5⟩⋅⟨2,−1,1⟩∣V=\left|\left\langle-1,4,5\right\rangle\cdot\left\langle2,-1,1\right\rangle\right|V=∣⟨−1,4,5⟩⋅⟨2,−1,1⟩∣
V= ∣−1∣V=\ \left|-1\right|V= ∣−1∣
V=1V=1V=1

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Equations of Lines & Planes
The equation of a line or plane gives us all the points on that line or plane

Lines in R3
If a line contains the point P(p1, p2, p3)P\left(p_1,\ p_2,\ p_3\right)P(p1​, p2​, p3​) and has a direction vector (vector parallel to the line) v⃗=(v1, v2, v3)\vec{v}=\left(v_1,\ v_2,\ v_3\right)v=(v1​, v2​, v3​), then the following are equations of the line:

Vector equation⟨x,y,z⟩=⟨p1,p2,p3⟩+t⟨v1,v2,v3⟩Parametric equationsx=p1+t v1y=p2+t v2z=p3+t v3Symmetric equationsx−p1v1=y−p2v2=z−p3v3\begin{array}{|c|c|}
\hline
\text{Vector equation}&\langle x,y,z\rangle=\langle p_1,p_2,p_3\rangle+t\langle v_1,v_2,v_3\rangle\\
\hline
\text{Parametric equations}&\begin{array}{c}x=p_1+t \ v_1\\y=p_2+t \ v_2\\z=p_3+t \ v_3\end{array}\\
\hline
\text{Symmetric equations}&\displaystyle \frac{x-p_1}{v_1}=\frac{y-p_2}{v_2}=\frac{z-p_3}{v_3}\\
\hline
\end{array}Vector equationParametric equationsSymmetric equations​⟨x,y,z⟩=⟨p1​,p2​,p3​⟩+t⟨v1​,v2​,v3​⟩x=p1​+t v1​y=p2​+t v2​z=p3​+t v3​​v1​x−p1​​=v2​y−p2​​=v3​z−p3​​​​
*ttt is a parameter and can take on any value

Line Segment
A line segment from the point P(p1, p2, p3)P\left(p_1,\ p_2,\ p_3\right)P(p1​, p2​, p3​) to Q(q1, q2, q3)Q\left(q_1,\ q_2,\ q_3\right)Q(q1​, q2​, q3​) is given by 
⟨x,y,z⟩=(1−t)⟨p1, p2, p3⟩+t⟨q1, q2, q3⟩\left\langle x,y,z\right\rangle=\left(1-t\right)\left\langle p_{1,}\ p_2,\ p_3\right\rangle+t\left\langle q_1,\ q_2,\ q_3\right\rangle⟨x,y,z⟩=(1−t)⟨p1,​ p2​, p3​⟩+t⟨q1​, q2​, q3​⟩ 
where 0≤t≤10\le t\le10≤t≤1

Example
a.) Find the equation of the line that passes through the point (1,2,3)\left(1,2,3\right)(1,2,3) and is parallel to the line ⟨0,−1,3⟩+s⟨2,6,−4⟩\left\langle0,-1,3\right\rangle+s\left\langle2,6,-4\right\rangle⟨0,−1,3⟩+s⟨2,6,−4⟩

We need a point on the line and the direction vector of the line:
Point: (1,2,3)(1,2,3)(1,2,3)
Direction vector: any vector parallel to ⟨2,6,−4⟩\left\langle2,6,-4\right\rangle⟨2,6,−4⟩ --> ⟨1,3,−2⟩\left\langle1,3,-2\right\rangle⟨1,3,−2⟩
The vector equation of the line is ⟨1,2,3⟩+t⟨1,3,−2⟩\left\langle1,2,3\right\rangle+t\left\langle1,3,-2\right\rangle⟨1,2,3⟩+t⟨1,3,−2⟩
The parametric equations are x=1+t, y=2+3t, z=3−2tx=1+t,\ y=2+3t,\ z=3-2tx=1+t, y=2+3t, z=3−2t
The symmetric equations are x−11=y−23=z−3−2\frac{x-1}{1}=\frac{y-2}{3}=\frac{z-3}{-2}1x−1​=3y−2​=−2z−3​

b.) Determine if the point (3,8,1)\left(3,8,1\right)(3,8,1) is on the line described in part a.)

Solving the parametric equations, we get
3=1+t→t=23=1+t\to t=23=1+t→t=2
8=2+3t→t=28=2+3t\to t=28=2+3t→t=2
1=3−2t→t=11=3-2t\to t=11=3−2t→t=1
Since not all of the equations result in the same t value, the point is not on the line.

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Planes in R3
If a line contains the point P(p1, p2, p3)P\left(p_1,\ p_2,\ p_3\right)P(p1​, p2​, p3​) and has a normal vector (vector orthogonal to the plane) n⃗=⟨a, b, c⟩\vec{n}=\left\langle a,\ b,\ c\right\ranglen=⟨a, b, c⟩, then the following are equations of the plane:
Vector Equation/Point-Normal Form⟨a, b, c⟩⋅(⟨x, y, z⟩−⟨p1, p2, p3⟩)=0Scalar Equation/Standard forma(x−p1)+b(y−p2)+c(z−p3)=0orax+by+cz+d=0, where d=−⟨a,b,c⟩⋅⟨p1,p2,p3⟩\begin{array}{|c|c|}
\hline
\begin{array}{c}\text{Vector Equation/}\\\text{Point-Normal Form}\end{array}&\langle a,\ b,\ c\rangle\cdot (\langle x,\ y,\ z\rangle-\langle p_1,\ p_2,\ p_3\rangle)=0\\
\hline

\begin{array}{c}\text{Scalar Equation/}\\\text{Standard form}\end{array}&\begin{array}{c}a(x-p_1)+b(y-p_2)+c(z-p_3)=0\\\text{or}\\ax+by+cz+d=0\text{, where }d=-\langle a,b,c\rangle\cdot\langle p_1,p_2,p_3\rangle\end{array}\\\hline
\end{array}Vector Equation/Point-Normal Form​Scalar Equation/Standard form​​⟨a, b, c⟩⋅(⟨x, y, z⟩−⟨p1​, p2​, p3​⟩)=0a(x−p1​)+b(y−p2​)+c(z−p3​)=0orax+by+cz+d=0, where d=−⟨a,b,c⟩⋅⟨p1​,p2​,p3​⟩​​​

Distance
The distance from a point (p1, p2, p3)(p_1,\ p_2,\ p_3)(p1​, p2​, p3​) to the plane ax+by+cz+d=0ax+by+cz+d=0ax+by+cz+d=0 is
D=∣a(p1)+b(p2)+c(p3)+d∣a2+b2+c2\displaystyle D=\frac{\left|a\left(p_1\right)+b\left(p_2\right)+c\left(p_3\right)+d\right|}{\sqrt{a^2+b^2+c^2}}D=a2+b2+c2​∣a(p1​)+b(p2​)+c(p3​)+d∣​

Example
a. Find the equation of the plane that contains the points A(0, 2, 1)A\left(0,\ 2,\ 1\right)A(0, 2, 1), B(1, 1, 2)B\left(1,\ 1,\ 2\right)B(1, 1, 2) and C(1, 0, 1)C\left(1,\ 0,\ 1\right)C(1, 0, 1)
b. Find the distance between the point (1, 1, 1)\left(1,\ 1,\ 1\right)(1, 1, 1) and the plane found in part a.)

a. AB→=⟨1, −1, 1⟩\overrightarrow{AB}=\left\langle1,\ -1,\ 1\right\rangleAB=⟨1, −1, 1⟩ and  AC→=⟨1,−2,0⟩\overrightarrow{AC}=\left\langle1,-2,0\right\rangleAC=⟨1,−2,0⟩ are direction vectors parallel to the plane.
To find the normal vector to the plane, we find AB→×AC→=⟨2,1,−1⟩\overrightarrow{AB}\times\overrightarrow{AC}=\left\langle2,1,-1\right\rangleAB×AC=⟨2,1,−1⟩
We can pick any one of the 3 given points to use in our equation of the plane: 2x+y−z−⟨2,1,−1⟩⋅⟨0,2,1⟩=02x+y-z-\left\langle2,1,-1\right\rangle\cdot\left\langle0,2,1\right\rangle=02x+y−z−⟨2,1,−1⟩⋅⟨0,2,1⟩=0
Therefore, the equation of the plane is 2x+y−z=12x+y-z=12x+y−z=1

b. Using the distance formula: D=∣2(1)+(1)−(1)−1∣22+12+12=16D=\frac{\left|2\left(1\right)+\left(1\right)-\left(1\right)-1\right|}{\sqrt{2^2+1^2+1^2}}=\frac{1}{\sqrt{6}}D=22+12+12​∣2(1)+(1)−(1)−1∣​=6​1​

Cylinder & Quadric Surfaces
Cylinders
Cylinders are 3D surfaces that have uniform cross sections.

Examples


Although not always true, typically, the equations will only involve 2 variables. For example
y=x2y=x^2y=x2
z=x2+5z=x^2+5z=x2+5
y2+z2=4y^2+z^2=4y2+z2=4
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Quadric Surfaces
In 3D space, quadric surfaces are a set of points with general (x,y,z)(x,y,z)(x,y,z)  coordinates satisfying a polynomial equation. We can sketch these surfaces by using traces which are curves generated by intersection of the surface with the planes parallel to the coordinate planes.
This will allow us to classify the surfaces using the following table from your textbook:

Traces are also known as level surface. To find level surfaces you should takes one variable constant. For example, if we take z=kz=kz=k, then we will have a curve in the plane z=kz=kz=k.

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Determine and sketch the domain for

f(x,y)=20−x2+4x−y2+2yf\left(x,y\right)=\sqrt{20-x^2+4x-y^2+2y}f(x,y)=20−x2+4x−y2+2y​

z=20−x2+4x−y2+2yz=\sqrt{20-x^2+4x-y^2+2y}z=20−x2+4x−y2+2y​
⇒ z2=20−x2+4x−y2+2y\Rightarrow\ z^2=20-x^2+4x-y^2+2y⇒ z2=20−x2+4x−y2+2y

from complete square method: 
x2+ax=(x+a2)2−a24x^2+ax=\left(x+\frac{a}{2}\right)^{^2}-\frac{a^2}{4}x2+ax=(x+2a​)2−4a2​

⇒ z2=20−(x−2)2+4−(y−1)2+1\Rightarrow\ z^2=20-(x-2)^2+4-(y-1)^2+1⇒ z2=20−(x−2)2+4−(y−1)2+1
⇒ z2+(x−2)2+(y−1)2=25\Rightarrow\ z^2+(x-2)^2+(y-1)^2=25⇒ z2+(x−2)2+(y−1)2=25

Sphere with center C=(2,1,0)C=(2,1,0)C=(2,1,0), with radius R=5R=5R=5

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Find the equation of a sphere with centre C(3,−5,−2)C\left(3,-5,-2\right)C(3,−5,−2) and radius 7. What is the intersection of this sphere with the xyxyxy-plane?

Generally (x−a)2+(y−b)2+(z−c)2=R2(x-a)^2+(y-b)^2+(z-c)^2=R^2(x−a)2+(y−b)2+(z−c)2=R2, C(a,b,c)C(a,b,c)C(a,b,c)
⇒ (x−3)2+(y+5)2+(z+2)2=49\Rightarrow\ (x-3)^2+(y+5)^2+(z+2)^2=49⇒ (x−3)2+(y+5)2+(z+2)2=49
xyxyxy-plain: z=0z=0z=0
⇒(x−3)2+(y+5)2+22=49⇒(x−3)2+(y+5)2=45\Rightarrow (x-3)^2+(y+5)^2+2^2=49 \Rightarrow (x-3)^2+(y+5)^2=45⇒(x−3)2+(y+5)2+22=49⇒(x−3)2+(y+5)2=45

this is a circle at C(3,−5)C(3,-5)C(3,−5)with radius R=45R=\sqrt{45}R=45​.

Cylindrical & Spherical Coordinates
In 3-space, a point can be represented using the Cartesian coordinates (x,y,z)(x,y,z)(x,y,z).
 
Cylindrical Coordinates
If we rewrite the coordinates of this point (x,y,z)(x,y,z)(x,y,z) on the xy-plane into polar coordinates and keep the z coordinate as our height, we get cylindfrical coordinates. 
θ\thetaθ is the angle measured counter clockwise from the positive x-axis to the point (x,y)(x,y)(x,y) on the xy-plane
rrr is the length of the line segment from the origin to the point (x,y)(x,y)(x,y) on the xy-plane
zzz is the "height" of the point (the original z-coordinate)

Cylindrical ➡ Cartesian coordinates: x=rcos⁡θ       y=rsin⁡θ       z=z\boxed{x=r\cos\theta~~~~~~~y=r\sin\theta~~~~~~~z=z}x=rcosθ       y=rsinθ       z=z​

Cartesian ➡ Cylindrical coordinates: r2=x2+y2       tan⁡θ=yx       z=z\boxed{r^2=x^2+y^2~~~~~~~\tan\theta=\dfrac{y}{x}~~~~~~~z=z}r2=x2+y2       tanθ=xy​       z=z​

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Spherical Coordinates
If we draw a line segment connecting our point (x,y,z)(x,y,z)(x,y,z) to the origin, and measure the angle this line makes with the positive z-axis, we get spherical coordinates.

θ\thetaθ is the angle measured counter clockwise from the positive x-axis to the point (x,y)(x,y)(x,y) on the xy-plane
ρ\rhoρ is the length of the line segment connecting the origin to the point (x,y,z)(x,y,z)(x,y,z) in R3
ϕ\phiϕ is the angle that is between 0 and π\piπ measured from the positive z-axis to the line segment connecting the origin to the point (x,y,z)(x,y,z)(x,y,z)
Spherical ➡ Cartesian coordinates: x=ρsin⁡ϕcos⁡θ       y=ρsin⁡ϕsin⁡θ       z=ρcos⁡θ\boxed{x=\rho\sin\phi\cos\theta~~~~~~~y=\rho\sin\phi\sin\theta~~~~~~~z=\rho\cos\theta}x=ρsinϕcosθ       y=ρsinϕsinθ       z=ρcosθ​

We also know that 

Extra Practice

Find a parameterization of the intersection of x + z = 4 and z2  = y. 

Wize University Calculus 2 Textbook > Vector Functions

Vector Geometry

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3D Coordinate System

2D space $\colorOne{R^2}$ :

3D space $\colorOne{R^3}$ :

Vectors

Geometrically

Vector Operations

Properties

Standard basis vectors

Dot Product

Properties

Orthogonal Vectors

Directional Cosines & Angles

Projections

Cross Product

Properties

Application

Equations of Lines & Planes

Lines in R3

Line Segment

Planes in R3

Distance

Cylinder & Quadric Surfaces

Cylinders

Quadric Surfaces

Cylindrical & Spherical Coordinates

Cylindrical Coordinates

Spherical Coordinates

Extra Practice