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3D Coords & Vectors
[[chap12.pdf]]
Cell: ### ### #### Common Final: [[2022-11-16]] @ 19:00 WebWork for homework
[[Math 243 Chapter 12 Notes.pdf]]
x = independant y = dependant x & y plane = independant z = dependant P 1 = ( 2 , − 1 , 7 ) P 2 = ( 1 , − 3 , 5 ) d ⏟ distance = ∣ P 1 P 2 ∣ = ( x 1 − x 2 ) 2 + ( y 2 − y 1 ) 2 + ( z 2 − z 1 ) 2 \begin{align} P_{1} &= (2, -1, 7) \\ P_{2} &= (1, -3, 5) \\ \underbrace{ d }_{ \text{distance} } &= |P_{1}P_{2}|= \sqrt{ (x_{1}-x_{2})^{2} + (y_{2}-y_{1})^{2} + (z_{2}-z_{1})^{2}} \end{align}
Set of all pts eq from a given pt called centre, labeled by ( h , k , l ) (h, k, l)
r 2 = ( x − h ) 2 + ( y − k ) 2 + ( z − l ) \begin{align} r^{2} = (x - h)^{2} + (y - k)^{2} + (z - l) \\ \end{align}
Give the centre and radius of the sphere x 2 + y 2 + z 2 + 4 x − 6 y + 2 z + 6 = 0 x^{2} + y^{2} + z^{2} + 4x - 6y + 2z + 6 = 0 r = 8 r = \sqrt{ 8 }
![[Pasted image 20220909083000.png]]
projection - set a variable to zero to put it on a single plane
traces - let z = different values
Vectors
Vector: Anythign that is fully described by magnitude and direction may be modelled by a vector
2 vecs are equivalent if they have the same mag. & dir.
Geometric: Arrow from a pt (Initial Point) to another pt (Terminal Point) and/or an angle θ \theta Cartesian: F ⃗ = F = < F x ⏟ x component , F y , F z > \vec{F} = \mathbf{F} = \left<\underbrace{ F_{x} }_{ \text{x component} }, F_{y}, F_{z}\right> F x = x 2 − x 1 F y = y 2 − y 1 F z = z 2 − z 1 \begin{align} F_{x} &= x_{2} - x_{1} \\ F_{y} &= y_{2} - y_{1} \\ F_{z} &= z_{2} - z_{1} \end{align}
Denote vector formed from ( 1 , 2 ) (1, 2) ( 3 , 7 ) (3, 7)
Geometric : Arrow from ( 1 , 2 ) (1, 2) ( 3 , 7 ) (3, 7)
Cartesian : a ⃗ = < 3 − 1 , 7 − 2 > = < 2 , 5 > \vec{a} = \left<3 - 1, 7 - 2\right> = \left<2, 5\right>
Polar F ⃗ = < F x , F y > \vec{F} = \left<F_{x}, F_{y}\right> F x = ∣ F ⃗ ∣ cos θ F y = ∣ F ⃗ ∣ sin θ \begin{align} F_{x} = |\vec{F}|\cos \theta \\ F_{y} = |\vec{F}|\sin \theta \end{align}
Unit Axial Basis
i ^ = < 1 , 0 , 0 > j ^ = < 0 , 1 , 0 > k ^ = < 0 , 0 , 1 > \begin{align} \hat{i} &= \left<1, 0, 0\right> \\ \hat{j} &= \left<0, 1, 0\right> \\ \hat{k} &= \left<0, 0, 1\right> \end{align}
Add/Sub like components Result = resultant vector
c a ⃗ = c < a x , a y , a z > = < c a x , c a y , c a z > \begin{align} c\vec{a} &= c\left<a_{x}, a_{y}, a_{z}\right> \\ &= \left<ca_{x}, ca_{y}, ca_{z}\right> \end{align}
A special vector: unit vector of a ⃗ \vec{a}
mag of 1 1
u ⃗ = a ⃗ ∣ a ⃗ ∣ \vec{u} = \frac{\vec{a}}{|\vec{a}|}
a ⃗ = 3 i ^ + 2 j ^ − 5 k ^ = 3 < 1 , 0 , 0 > + 2 < 0 , 1 , 0 > − 5 < 0 , 0 , 1 > = < 3 , 2 , − 5 > \begin{align} \vec{a} &= 3\hat{i} + 2\hat{j} - 5\hat{k} \\ &= 3\left<1, 0, 0\right> + 2\left<0, 1, 0\right> - 5\left<0, 0, 1\right> \\ &= \left<3, 2, -5\right> \end{align}
Dot Product = Inner Product = Scalar Product
a ⃗ = < a x , a y , a z > b ⃗ = < b x , b y , b z > a ⃗ ⋅ b ⃗ = a x b x + a y b y + a z b z \begin{align} \vec{a} &= <a_{x}, a_{y}, a_{z}> \\ \vec{b} &= <b_{x}, b_{y}, b_{z}> \\ \vec{a} \cdot \vec{b} &= a_{x}b_{x}+a_{y}b_{y}+a_{z}b_{z} \end{align} < 2 , 3 > ⋅ < − 3 , 1 > = 2 ( − 3 ) + 3 ( 1 ) = − 6 + 3 = 3 \begin{align} \left<2, 3\right> \cdot \left<-3, 1\right> &= 2(-3) + 3(1) \\ &= -6 + 3 \\ &= 3 \end{align}
This results in a number
ϕ \phi
a ⃗ ⋅ b ⃗ = ∣ a ⃗ ∣ ∣ b ⃗ ∣ cos ϕ \vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}|\cos \phi
#dotproduct-aro if a ⃗ ⋅ b ⃗ = 0 \vec{a} \cdot \vec{b} = 0 a ⃗ ⊥ b ⃗ \vec{a} \perp \vec{b} > 0 >0 < 0 <0
ϕ = cos − 1 [ a ⃗ ⋅ b ⃗ ∣ a ⃗ ∣ ∣ b ⃗ ∣ ] \phi=\cos^{-1}\left[ \frac{\vec{a}\cdot \vec{b}}{|\vec{a}| |\vec{b}|} \right]
work = force * displacement
w = F ⃗ ⋅ D ⃗ = ∣ F ⃗ ∣ ∣ D ⃗ ∣ cos ϕ w = \vec{F} \cdot \vec{D} = |\vec{F}||\vec{D}|\cos \phi
Reference 12.3#1
Component projection of b ⃗ \vec{b} a ⃗ \vec{a}
b x = ∣ b ∣ cos ϕ a ⋅ b = ∣ a ∣ ∣ b ∣ cos ϕ a ⋅ b ∣ a ∣ = ∣ b ∣ cos ϕ = b x \begin{align} b_{x} &= |b|\cos \phi \\ a \cdot b &= |a| |b| \cos \phi \\ \frac{a \cdot b}{|a|} &= |b|\cos \phi=b_{x} \end{align}
c o m p a ⃗ b ⃗ = a ⃗ ⋅ b ⃗ ∣ a ⃗ ∣ comp_{\vec{a}}\vec{b} = \frac{\vec{a}\cdot \vec{b}}{|\vec{a}|}
Vector projection of b ⃗ \vec{b} a ⃗ \vec{a}
\begin{align}
\frac{\vec{a}\cdot \vec{b}}{|\vec{a}|} \cdot \frac{\vec{a}}{|\vec{a}|} \
proj_{\vec{a}}\vec{b} = \frac{\vec{a}\cdot \vec{b}}{|\vec{a}|^{2}}\vec{a} \end{align} $$
Cross Product = Outer Product → \to ⊥ \perp
a × b = c a ⊥ c and b ⊥ c < c 1 , c 2 , c 3 > = < a 2 b 3 − a 3 b 2 , a 3 b 1 − a 1 b 2 − a 2 b 1 > \begin{align} a \times b = c \\ a \perp c \text{ and } b \perp c \\ \left<c_{1},c_{2},c_{3}\right> = \left<a_{2}b_{3}-a_{3}b_{2},a_{3}b_{1}-a_{1}b_{2}-a_{2}b_{1}\right> \end{align}
%%![[Pasted image 20220912085241.png]]%%
∣ a × b ∣ = ∣ a ∣ ∣ b ∣ sin ϕ |a \times b| = |a| |b| \sin \phi
if ∣ a × b ∣ = 0 |a \times b| = 0 a ∥ b a \parallel b
Given 3 vectors (Reference 12.4#1)
∣ a ⃗ ⋅ ( b ⃗ × c ⃗ ) ∣ = volume |\vec{a}\cdot(\vec{b}\times\vec{c})| = \text{volume}
![[Pasted image 20220914081430.png]] (08:14) Plane = two vectors between the points
![[Pasted image 20220914081951.png]] (8:19) Use the scalar triple product to show that the vectors ... are coplanar Volume of parallelipiped = 0? ∣ a ⃗ ⋅ ( b ⃗ × c ⃗ ) ∣ = 0 |\vec{a}\cdot(\vec{b}\times \vec{c})|=0
τ = r ⃗ × F ⃗ ∣ τ ∣ = ∣ r ⃗ × F ⃗ ∣ = ∣ r ⃗ ∣ ∣ F ⃗ ∣ sin ϕ \begin{align} \tau &= \vec{r}\times \vec{F} \\ |\tau| &= |\vec{r}\times \vec{F}| = |\vec{r}| |\vec{F}| \sin \phi \end{align}
2 Special Vector Functions
Line
in 2d: y = m x + b y=mx+b
m → m \to b → b \to y 0 y_{0}
in vectors: L ⃗ = v ⃗ t + r 0 ⃗ \vec{L} = \vec{v}t+\vec{r_{0}}
$$ \begin{align}
\vec{v} &\to \text{direction} \ \vec{r_{0}} &\to \text{terminal point on the line of a given victor}\ t &\to \text{time/independent variable} \end{align} $$
![[Pasted image 20220914090554.png]] (9:03 ish)
![[Pasted image 20220914090325.png]] (9:05)
Eq of a line segment between 2 pts r 0 ⃗ \vec{r_{0}} r 1 ⃗ \vec{r_{1}}
M ⃗ = ( 1 − t ) r 0 ⃗ + t r 1 ⃗ 0 ≤ t ≤ 1 \begin{align} \vec{M} = (1-t)\vec{r_{0}}+t\vec{r_{1}} \\ 0\leq t\leq 1 \end{align}
![[Pasted image 20220916081046.png]]
Check that the coefficient of each component is a scaled value
not ∥ \parallel v 1 ⃗ ≠ c v 2 ⃗ \vec{v_{1}}\neq c\vec{v_{2}}
$$ \begin{align} 1 + 2t &= -11 + 3s \ 17 + 7t &= -22 + 9s \ \
-3 ( 2t - 3s &= -12 ) \ 7t - 9s &= -39 \ \
t &= -3 \ s &= 2 \ \end{align} $$
Plug in the values for the z z
Given some pt r 0 ⃗ = < x 0 , y 0 , z 0 > \vec{r_{0}}= \left<x_{0},y_{0},z_{0}\right> r ⃗ = < x , y , z > \vec{r}= <x, y, z> n ⃗ ⋅ ( r ⃗ − r 0 ⃗ ) = 0 < a , b , c > ⏟ A normal to a plane ⋅ < x − x 0 , y − y 0 , z − z 0 > ⏟ Every vector in plane whose initial pt is r 0 ⃗ = 0 a ( x − x 0 ) + b ( y − y 0 ) + c ( z − z 0 ) = 0 \begin{align} \vec{n} \cdot (\vec{r}-\vec{r_{0}}) &= 0 \\ \underbrace{ <a,b,c> }_{ \text{A normal to a plane} } \cdot \underbrace{ <x-x_{0},y-y_{0},z-z_{0}> }_{ \text{Every vector in plane whose initial pt is } \vec{r_{0}} } &= 0 \\ a(x-x_{0}) + b(y-y_{0}) + c(z-z_{0}) &= 0 \end{align}
![[Pasted image 20220916082652.png]]
Cross two vectors to get the normal Plug into above ![[Pasted image 20220916083308.png]]
Plug equations into the plane equation to get t 4 ( 2 + 3 t ) + 5 ( − 4 t ) − 2 ( 5 + t ) = 18 4(2+3t)+5(-4t)-2(5+t)=18 Put the resultant t t ![[Pasted image 20220916083656.png]]
Find the angle between the normals ϕ = cos − 1 [ n 1 ⋅ n 2 ∣ n 1 ∣ ∣ n 2 ∣ ] \phi=\cos^{-1}\left[\frac{n_{1}\cdot n_{2}}{|n_{1}| |n_{2}|}\right] acute, obtuse, or right? #dotproduct-aro Cylinder : Surface that consists of all lines ∥ \parallel
12.6#1 y = x 2 \begin{align} y = x^2 \end{align}
$$ \begin{align} Ax^{2}+by^{2}+Cz^{2}+Dxy+Eyz+Fxz+Gx+Hg+Iz+J=0 \
\end{align} $$
Need to know: ![[Pasted image 20220914084706.png]]