| 789 |
position vector of a point.
Given here by $\vect{r}$
of point $\left(x,y,z\right)$
|
\[
\vect{r}=\left\langle x,y,z\right\rangle
=x\vect{i}+y\vect{j}+z\vect{k}
\]
|
| 790 |
distance between points
$P_1\left(x_1,y_1,z_1\right)$
and
$P_2\left(x_2,y_2,z_2\right)$
|
\[
\abs{\overrightarrow{P_1 P_2}}
=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}
\]
|
| 792 |
standard equation for the sphere
of radius $a$ and center
$P_0\left(x_0,y_0,z_0\right)$
|
\[
\left(x-x_0\right)^2+\left(y-y_0\right)^2+\left(z-z_0\right)^2=a^2
\]
|
| 793 |
midpoint between points
$P_1\left(x_1,y_1,z_1\right)$
and
$P_2\left(x_2,y_2,z_2\right)$
|
\[
\left(
\frac{x_1+x_2}{2},
\frac{y_1+y_2}{2},
\frac{z_1+z_2}{2}
\right)
\]
|
| 802 |
cross product (vector product)
|
\[
\vect{u}\times\vect{v}=
\begin{vmatrix}
\vect{i}&\vect{j}&\vect{k}\\
u_1&u_2&u_3\\
v_1&v_2&v_3
\end{vmatrix}
\]
|
| 801 |
area of parallelogram.
The area $A$ of the parallelogram with base
$\abs{\vect{u}}$ and height
$\abs{\vect{v}}\sin\theta$.
|
\[
A=\abs{\vect{u}\times\vect{v}}
\]
|
| 802 |
area of triangle.
The area $A$ of the triangle with base
$\abs{\vect{u}}$ and height $\abs{\vect{v}}\sin\theta.$
Note. Given three points, you can use either formula to find
the area of the parallelogram or triangle they form without
having to find the angle $\theta.$
|
\[
A=\frac{1}{2}\abs{\vect{u}\times\vect{v}}
\]
|
| 803 |
triple scalar product (box product)
|
\[
\left(\vect{u}\times\vect{v}\right)\cdot\vect{w}=
\begin{vmatrix}
u_1 &u_2 &u_3\\
v_1 &v_2 &v_3\\
w_1 &w_2 &w_3
\end{vmatrix}
\]
|
| 804 |
volume of parallelpiped
|
\[
V=\abslr{
\left(\vect{u}\times\vect{v}\right)\cdot\vect{w}
}
\]
|
| 803 |
torque
|
\[
\]
|
|
Vector Equation for a Line in Space.
If $\vect{r}_0$ is the position vector of
$P_0\left(x_0,y_0,z_0\right),$ then
a vector equation for the line $L$
through $P_0\left(x_0,y_0,z_0\right)$
parallel to $\vect{v}$ is given. This defines
$\vect{r}$ as the position vector for each
point $P\left(x,y,z\right)$ on $L$ at $t.$
It is often more convenient to use the second form.
|
\[
\vect{r}\left(t\right)
=\vect{r}_0+t\vect{v}\quad
-\infty\lt t\lt\infty
\\
\vect{r}\left(t\right)
=\left(x_0+tv_1\right)\vect{i}
+\left(y_0+tv_2\right)\vect{j}
+\left(z_0+tv_3\right)\vect{k}
\]
|
|
Parametric Equations for a Line in Space.
The standard parametrization of the line
passing through $P_0\left(x_0,y_0,z_0\right)$
parallel to
$\vect{v}=v_1\vect{i}+v_2\vect{j}+v_3\vect{k}.$
|
\[
x=x_0+tv_1\\
y=y_0+tv_2\\
z=z_0+tv_3
\]
|
|
Vector Parallel to a Line in Space.
If the parametrization
\[
x=x_0+tv_1\\
y=y_0+tv_2\\
z=z_0+tv_3
\]
of a line is given, then the vector
is parallel to the line.
|
\[
\vect{v}=v_1\vect{i}+v_2\vect{j}+v_3\vect{k}
\]
|
| 811 |
Vector Normal to a Plane, Equation for a Plane.
The plane normal to any vector
\(
\vect{n}=\left(A,B,C\right)=A\vect{i}+B\vect{j}+C\vect{k}
\)
passing through the point
$P_0\left(x_0,y_0,z_0\right).$
|
\[
\begin{align*}
&\textbf{Vector Form:}\\
&\vect{n}\cdot\overrightarrow{P_0 P}=0\\
&\textbf{Component Form:}\\
&A(x-x_0)+B(y-y_0)+C(z-z_0)=0\\
&\textbf{Component Form (Simplified):}\\
&Ax+By+Cz=D\\
&(D=Ax_0+By_0+Cz_0)
\end{align*}
\]
|
|
Conversely, the
vector normal to a plane
$Ax+By+Cz=D$.
|
\[
\vect{n}=A\vect{i}+B\vect{j}+C\vect{k}
\]
|
|
The equation of the line $L$ normal to
$\vect{n}=\left(A,B,C\right)=A\vect{i}+B\vect{j}+C\vect{k}$
and passing through the point
$P_0\left(x_0,y_0,z_0\right)$
|
\[
Ax+By=C\\
\left(C=Ax_0+By_0\right)
\]
|
|
Vector Normal to a Line.
Conversely, the vector normal to any line $Ax+By=C.$
Note that the slope of any line normal to $\vect{n}$ is
$-\frac{A}{B}$ and the slope of any line parallel
to $\vect{n}$ is $\frac{B}{A}.$
|
\[
\vect{n} = A\vect{i} + B\vect{j}
\]
|
| 812 |
Vector Parallel to the Line of Intersection of two Planes.
If $\vect{n}_1$ and $\vect{n}_2$ are vectors normal to planes
$P_1$ and $P_2$ respectively, then
$\vect{n}_1\times \vect{n}_2$ is a vector parallel to the line of
intersection of the two planes. If
$\vect{n}_1\times\vect{n}_2=\zeros$ then $P_1$ and $P_2$ are parallel.
|
\[
\]
|
| 812 |
Equation of the Line of Intersection of Two Planes.
If $\vect{n}=A\vect{i}+B\vect{j}+C\vect{k}$ is a
vector parallel to the line of intersection of two
planes and if the point $P_0\left(x_0,y_0,z_0\right)$
lies on the line then the equation of the line is as given.
|
\[
x=x_0+At\\
y=y_0+Bt\\
z=z_0+Ct\\
\]
|
| 814 |
Angle between Intersecting Planes.
If $\vect{n}_1$ and $\vect{n}_2$ are vectors normal
to planes $P_1$ and $P_2$ respectively,
then the angle between the planes is as given.
|
\[
\theta = \cos^{-1} \frac{\vect{n}_1\cdot\vect{n}_2}{\abs{\vect{n}_1}\abs{\vect{n}_2}}
\]
|
| 814 |
Distance from a Point to a Line.
The distance $d$ from a point $S$
to the line that passes through
point $P$ parallel to $\vect{v}.$
|
\[
d=\frac{\abslr{\overrightarrow{PS}\times\vect{v}}}
{\abs{\vect{v}}}
\]
|
| 814 |
Distance from a Point to a Plane.
The distance $d$ from a point $S$ to
the plane that passes through
point $P$ and is normal to
$\vect{n}=A\vect{i}+B\vect{j}+C\vect{k}.$
|
\[
d=\abslr{
\overrightarrow{PS}
\cdot
\frac{\vect{n}}{\abs{\vect{n}}}
}
\]
|
|
Equation of the Line of Intersection of two Planes.
If $\vect{n}_1$ and $\vect{n}_2$ are vectors normal
to planes $P_1$ and $P_2$ respectively,
then
$\vect{n}=\vect{n}_1\times\vect{n}_2=$ vector
parallel to the line of intersection of the
two planes.
If $\vect{n}_1\times\vect{n}_2=\zeros$ then $P_1$ and $P_2$
are parallel. If the point
$P_0\left(x_0,y_0,z_0\right)$
lies on the line then the equation of the
line is as given.
|
\[
\vect{r}\left(t\right)=\vect{r}_0+t\vect{v}\\
\text{where}\\
\vect{v}=\vect{n}_1\times\vect{n}_2=v_1\vect{i}+v_2\vect{j}+v_3\vect{k}\\
x=x_0+tv_1\\
y=y_0+tv_2\\
z=z_0+tv_3
\]
|
| 826 |
Equations for Curves in Space
where
\[
x=f\left(t\right)\\
y=g\left(t\right)\\
z=h\left(t\right)
\]
|
\[
\vect{r}\left(t\right)=f\left(t\right)\vect{i}+g\left(t\right)\vect{j}+h\left(t\right)\vect{k}
\]
|
| 829 |
Line Tangent to Curve
traced by $\vect{r}.$
If
$
\vect{r}\left(t\right)
=f\left(t\right)\vect{i}
+g\left(t\right)\vect{j}
+h\left(t\right)\vect{k}
$
and
$P_0\left(x_0,y_0,z_0\right)$
is the terminal point of
$\vect{r}\left(t_0\right)$
then the line tangent to the curve traced by $\vect{r}$ at
$P_0$ (i.e. at
$t=t_0$)
is given.
|
\[
x=x_0+f'\left(t_0\right)\left(t-t_0\right)\\
y=y_0+g'\left(t_0\right)\left(t-t_0\right)\\
z=z_0+h'\left(t_0\right)\left(t-t_0\right)
\]
|
| 848 |
helix equation
|
\[
\vect{r}\left(t\right)
=\left(a\cos t\right)\vect{i}
+\left(a\sin t\right)\vect{j}
+bt\vect{k}
\]
|
|
Arc Length Parameter with Base Point $P(t_0).$
If
\(
\vect{r}\left(t\right)
=f\left(t\right)\vect{i}
+g\left(t\right)\vect{j}
+h\left(t\right)\vect{k}
\)
is a smooth curve traced exactly once on the interval
$[t_0, t],$
and if
$\vect{v}=\frac{d\vect{r}}{dt},$
then the curve’s length on that interval is given.
$s$ is called an
arc length parameter.
|
\[
\begin{align*}
s(t)&=s(t_0,t)\\
&=\int\limits_{t_0}^t \abs{\vect{v}(\tau)}\,d\tau\\
&=\int\limits_{t_0}^t
\sqrt{
\left(\frac{df}{d\tau}\right)^2
+\left(\frac{dg}{d\tau}\right)^2
+\left(\frac{dh}{d\tau}\right)^2
}\,d\tau\\
&=\int\limits_{t_0}^t
\sqrt{
\left(\frac{dx}{d\tau}\right)^2
+\left(\frac{dy}{d\tau}\right)^2
+\left(\frac{dz}{d\tau}\right)^2
}\,d\tau\\
&=\int\limits_{t_0}^t
\sqrt{
\left[x'(\tau)\right]^2
\left[y'(\tau)\right]^2
\left[z'(\tau)\right]^2
}\,d\tau
\end{align*}
\]
|
|
Speed on a Smooth Curve
|
\[
\frac{ds}{dt}
=\left|\vect{v}\left(t\right)\right|
=\left|\vect{v}\right|
\]
|
|
Unit Tangent Vector
|
\[
\vect{T}=\frac{d\vect{r}}{\ds}
=\frac{d\vect{r}/\dt}{\ds/\dt}
=\frac{\vect{v}}{\left|\vect{v}\right|}
\]
|
|
Curvature
If $\vect{r}(t)$ is a smooth curve then the
curvature (a scalar) of $\vect{r}$ is given.
|
\[
\begin{align*}
\kappa=\abslr{\frac{d\vect{T}}{ds}}
&=\abslr{\frac{d}{ds}\left(\frac{d\vect{r}}{ds}\right)}\\
&=\abslr{\frac{d^2\vect{r}}{ds^2}}\\
&=\frac{1}{\abslr{\vect{v}}}\abslr{\frac{d\vect{T}}{dt}}\\
&=\frac{1}{\abs{\vect{v}}}
\abslr{
\frac{d}{dt}
\left(
\frac{\vect{v}}{\abslr{\vect{v}}}
\right)
}
\end{align*}
\]
|
|
Principal Unit Normal
|
\[
\vect{N}
=\frac{1}{\kappa}\frac{d\vect{T}}{ds}
=\frac{d\vect{T}/dt}{\abs{d\vect{T}/dt}}
\]
|
|
Radius of Curvature
|
\[
\rho=\frac{1}{\kappa}
\]
|
|
Binormal Vector
|
\[
\vect{B}=\vect{T}\times\vect{N}
\]
|
|
Curvature
|
\[
\kappa=\abslr{\frac{d\vect{T}}{ds}}
=\frac{\abslr{\vect{v}\times\vect{a}}}{\abslr{\vect{v}}^3}
\]
|
| 850 |
Torsion
|
\[
\tau=-\frac{d\vect{B}}{ds}\cdot\vect{N}
=\frac{
\begin{vmatrix}
\dot x&\dot y&\dot z\\
\ddot x&\ddot y&\ddot z\\
\dddot x&\dddot y&\dddot z\\
\end{vmatrix}
}{
\abslr{\vect{v}\times\vect{a}}^2
}
\]
|
| 851 |
Acceleration Tangential and Normal Components
|
\[
a=a_T\vect{T}+a_N\vect{N}
\]
|
| 851 |
acceleration, tangential component
|
\[
a_T=\frac{d}{dt}\left|\vect{v}\right|
\]
|
| 851 |
acceleration, normal component
|
\[
a_N=\kappa\left|\vect{v}\right|^2=\sqrt{\left|\vect{a}\right|^2-\vect{a}_T^2}
\]
|
Alex Notes