| 12 |
vector between two points
$P\left(x_1, y_1\right)$
and
$Q\left(x_2, y_2\right)$
|
\[
\vect{v}
= \overrightarrow{PQ}
= \left(
x_2 - x_1,\
y_2 - y_1
\right)
\]
|
| 15 |
equation of the line (point-direction form)
through
$\vect{r}_\vect{0}=P_0\left(x_0,y_0,z_0\right)$
parallel to
$\vect{v}=\left(v_1,v_2,v_3\right), -\infty\lt t\lt\infty.$
(The equations given are equivalent forms illustrating
different notations commonly used.)
The last line of equations is called the
standard parametrization of the line.
|
\begin{align*}
\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}\\
&= \vect{r_0}+t\vect{v}\\
&= \vect{r_0}+\overrightarrow{P_0 P}\\
&= 0\\
\vect{c}\left(t\right)
&= \vect{x}_0+t\vect{v}\\
\vect{l}\left(t\right)
&= \vect{x}_0+t\vect{v}\\
\overrightarrow{P_0 P}
&= \vect{r}-\vect{r}_0\\
&= t\vect{v}\\
x &= x_0+tv_1\\
y &= y_0+tv_2\\
z &= z_0+tv_3
\end{align*}
|
| 17 |
equation of the line
through two points
(endpoints of vectors) $a$ and $b,$
(
point-point form.
)
The last line of equations is the
standard parametrization of the line.
|
\begin{align*}
\vect{l}\left(t\right)
&= \left(1-t\right)\vect{a}+t\vect{b}\\
\vect{l}\left(t\right)
&= \vect{a}+t\left(\vect{b}-\vect{a}\right)\\
x &= x_1+\left(x_2-x_1\right)t\\
y &= y_1+\left(y_2-y_1\right)t\\
z &= z_1+\left(z_2-z_1\right)t
\end{align*}
|
| 32 |
Triangle Inequality
|
|
|
distance between points
$P_1\left(x_1,y_1,z_1\right)$
and
$P_2\left(x_2,y_2,z_2\right)$
|
\[
\left| \overrightarrow{P_1 P_2} \right|
= \sqrt{
\left( x_2 - x_1 \right)^2
+ \left( y_2 - y_1 \right)^2
+ \left( z_2 - z_1 \right)^2
}
\]
|
|
midpoint between points
$P_1\left(x_1,y_1,z_1\right)$
and
$P_2\left(x_2,y_2,z_2\right)$
|
\[
P_\mathrm{midpoint} =
\left(
\frac{x_1 + x_2}{2},
\frac{y_1 + y_2}{2},
\frac{z_1 + z_2}{2}
\right)
\]
|
|
polar form of a vector.
$\theta$ is the angle between $\vect{v}$ and
the positive $x$ (polar) axis.
|
\[
\vect{v}
= \left(\left|\vect{v}\right|\cos{\theta},
\left|\vect{v}\right|\sin{\theta}\right)
\]
|
|
dot product of vectors
$\vect{u}$ and $\vect{v}$
|
\[
\vect{u}\cdot\vect{v}
= \left|\vect{u}\right|\left|\vect{v}\right|\cos{\theta}
\]
|
|
angle between vectors
$\vect{u}$ and $\vect{v}$
|
\[
\theta =
\cos^{-1}
\frac{\vect{u}\cdot\vect{v}}
{\left|\vect{u}\right|\left|\vect{v}\right|}
\]
|
|
(orthogonal) projection
of $\vect{u}$ onto $\vect{v}.$
|
\[
\mathrm{proj}_\vect{v}\vect{u}
= \left(
\frac{\vect{u}\cdot\vect{v}}
{\left|\vect{v}\right|^2}
\right)
\vect{v}
= \frac{\vect{u}\cdot\vect{v}}
{\vect{v}\cdot\vect{v}}
\frac{\vect{v}}
{\left|\vect{v}\right|}
\]
|
|
scalar component
of $\vect{u}$
in direction of $\vect{v}.$
$\theta$ is the angle between $\vect{u}$ and $\vect{v}$
|
\[
\left|\vect{u}\right|\cos{\theta}
= \frac{\vect{u}\cdot\vect{v}}
{\left|\vect{v}\right|}
\]
|
| not in book |
equation of line
normal to
$\vect{n},$
parallel to
$\vect{v},$
and passing through point
$P_0\left(x_0,y_0\right),$
$\overrightarrow{P_0P}=\vect{r}-\vect{r}_0.$
The first two equations are called
the
point-normal form,
the next two the
general form.
Note that different
values of $C$ with $A$ and $B$ fixed produces different lines,
but that $\vect{n}$ is normal to each of them since
$\vect{n}$ does not depend on $C$ but only on $A$ and $B.$
Thus, all such lines are parallel and $C$
determines the shift of the line.
In particular,
$D=Ax_0+By_0+Cz_0$
if and only if
$P_0\left(x_0,y_0,z_0\right)$
is on the line. Note that since
$\vect{v}$
is parallel to the line,
then the slope of the line can be read off of
$ \vect{v}:m=\left.\frac{dy}{dx}\right|=\frac{\mp A}{\pm B}.$
|
\begin{align*}
&\vect{n}\cdot\overrightarrow{P_0P}
= \vect{n}\cdot\left(\vect{r}-\vect{r}_0\right)\\
&A\left(x-x_0\right)+B\left(y-y_0\right)+C
= 0\\
\end{align*}
\begin{alignat*}{2}
Ax+By &= C
&\quad\textrm{if}\quad
&C=Ax_0+By_0\\
Ax+By+C &= 0
&\quad\textrm{if}\quad
&C = -\left(Ax_0 + By_0 \right)\\
\end{alignat*}
\begin{align*}
\vect{n}
&= \left(A,B\right)\\
&= A\vect{i} + B\vect{j}\\
\vect{v}
&= \pm B\vect{i}\mp A\vect{j}
\end{align*}
|
| not in book |
orthogonal vectors.
If
$\vect{v}$
and
$\vect{n}$
are as given,
then
$\vect{v}$
is
tangent to a curve at a point
if and only if
$\vect{n}$
is
normal to the curve at that point.
|
\begin{align*}
\vect{v}&=\left(a,b\right)\\
\vect{n}&=\left(\pm b,\mp a\right)
\end{align*}
|
| not in book |
slope of line tangent to curve
$y=f\left(x\right)$
that has a tangent vector
$\vect{v}$
at
$x = c.$
|
\begin{align*}
m &= \left.\frac{dy}{dx}\right|_{x=c}=\frac{b}{a}\\
\vect{v} &= \left(a,b\right)\
\end{align*}
|
| 51 |
equation of plane
normal to
$\vect{n}$
and passing through point
$P_0\left(x_0,y_0,z_0\right).$
The first two equations are
called the
point-normal form
and the next two the
general form.
Note that different
values of $D$ with $A,$ $B,$ $C$
fixed produces different
planes, but that
$\vect{n}$
is normal
to each of them since
$\vect{n}$
does
not depend on $D$ but only on
$A,$ $B,$ $C.$
Thus, all such planes
are parallel and $D$ determines
a shift of a plane. In
particular, $D=Ax_0+By_0+Cz_0$
if and only if
$P_0\left(x_0,y_0,z_0\right)$
is on the plane.
|
\[
\begin{array}{l}
\vect{n}\cdot\overrightarrow{P_0P} = 0\\
\vect{n}\cdot\left(x-x_0,y-y_0,z-z_0\right) = 0\\
A\left(x-x_0\right)+B\left(y-y_0\right)+C\left(z-z_0\right)=0\\
Ax+By+Cz=D\quad\textrm{if}\quad D=Ax_0+By_0+Cz_0\\
Ax+By+Cz+D=0\quad\textrm{if}\quad D=-\left(Ax_0+By_0+Cz_0\right)\\
\vect{n}=\left(A,B,C\right)=A\vect{i}+B\vect{j}+C\vect{k}
\end{array}
\]
|
| 52 |
two conditions for
parallel vectors
$\vect{n}_1$
and
$\vect{n}_2.$
$\sigma$ is a real constant.
|
\[
\begin{array}{l}
\vect{n}_1 = \sigma \vect{n}_2\\
\vect{n}_1 \times \vect{n}_2 = \vect{0}
\end{array}
\]
|
| 53 |
Distance from a Point to a Plane.
The distance $d$ from a point
$S\left(x_0,y_0,z_0\right)$
to the plane $Ax+By+Cz+D=0$
that passes through point $P$
and is normal to
$\vect{n}=A\vect{i}+B\vect{j}+C\vect{k}.$
|
\begin{align*}
d &= \left|\overrightarrow{PS}\cdot
\frac{\vect{n}}
{\left|\vect{n}\right|}\right|\\
&= \frac{\left|Ax_0+By_0+Cz_0+D\right|}
{\sqrt{A^2+B^2+C^2}}
\end{align*}
|
| 43 |
cross product (vector product)
|
\[
\begin{array}{ll}
\vect{u}\times\vect{v}
&= \begin{vmatrix}
\vect{i} &\vect{j} &\vect{k}\\
u_1 &u_2 &u_3\\
v_1 &v_2 &v_3\\
\end{vmatrix}\\
&= (u_2v_3-u_3v_2)\vect{i}
- (u_1v_3-u_3v_1)\vect{j}
+ (u_1v_2-u_2v_1)\vect{k}\\
&= (u_2v_3-u_3v_2)\vect{i}
+ (u_3v_1-u_1v_3)\vect{j}
+ (u_1v_2-u_2v_1)\vect{k}\\
&= (u_2v_3-u_3v_2, u_3v_1-u_1v_3, u_1v_2-u_2v_1)
\end{array}
\]
|
| 46 |
magnitude of cross product
|
\[
\vect{u}\times\vect{v}
= \left|\vect{u}\right|
\left|\vect{v}\right|
\left|\sin\theta\right|
\]
|
| 48 |
area of parallelogram
with base
$|\vect{u}|$
and height $|\vect{v}|\sin\theta.$
|
\[
A=\|\vect{u}\times\vect{v}\|
\]
|
| 48 |
area of triangle
with base
$|\vect{u}|$
and height
$|v|\sin\theta.$
Note. Given 3 points, you can use
either formula to find the area of
the parallelogram or triangle they
form without having to find
the angle $\theta.$
|
\[
A={\frac 1 2}\|\vect{u}\times\vect{v}\|
\]
|
| 49 |
triple scalar product (box product).
Note that this time the determinant
is not a cross product so it is not
a vector but a scalar.
The box product is a real number.
|
\begin{align*}
\left(\vect{u}\times\vect{v}\right)\cdot\vect{w}
&= \vect{w}\cdot\left(\vect{u}\times\vect{v}\right)\\
&= \begin{vmatrix}
u_1 &u_2 &u_3\\
v_1 &v_2 &v_3\\
w_1 &w_2 &w_3
\end{vmatrix}
\end{align*}
|
| 49 |
identities involving box product
|
\begin{align*}
\left(\vect{u}\times\vect{v}\right)\cdot\vect{w}
&= \vect{w}\cdot\left(\vect{u}\times\vect{v}\right)\\
&= \vect{u}\cdot\left(\vect{v}\times\vect{w}\right)\\
&= \left(\vect{v}\times\vect{w}\right)\cdot\vect{u}\\
&= \left(\vect{w}\times\vect{u}\right)\cdot\vect{v}
\end{align*}
|
| 49 |
volume of parallelpiped
|
\begin{align*}
V &= \left|
\left(\vect{u}\times\vect{v}\right)
\cdot\vect{w}
\right|\\
&= \mathrm{Abs}
\left(\begin{vmatrix}
u_1 &u_2 &u_3\\
v_1 &v_2 &v_3\\
w_1 &w_2 &w_3
\end{vmatrix}\right)
\end{align*}
|
| not in book |
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
$\vect{n} = \vect{n}_1\times\vect{n}_2$
is the vector parallel to
the line of intersection of the two planes.
If $\vect{n}_1 \times \vect{n}_2 = \vect{0}$
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.
(See Thomas' Calculus, p. 812 for this.)
|
\begin{align*}
\vect{r}\left(t\right)
&= \vect{r}_0+t\vect{v}\\
& \textrm{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
\end{align*}
|
| not in book |
equation of the line tangent to the curve
formed by the intersection of two surfaces
$f\left(x,y,z\right)=0$
and
$g\left(x,y,z\right)=0$
at the point
$\vect{r}\left(t\right)=P_0\left(x_0,y_0,z_0\right).$
(See Thomas' Calculus, p. 922 for this.)
|
\begin{align*}
\vect{r}\left(t\right)
&= \vect{r}_0+t\vect{v}\\
& \textrm{where}\\
\vect{v}
&= \nabla f\times\nabla g\\
x &= x_0+tv_1\\
y &= y_0+tv_2\\
z &= z_0+tv_3\\
\end{align*}
|
| not in book |
distance from a point to a line.
The distance $d$ from a point
$S=\left(x_0,y_0,z_0\right)$
to the line
$Ax+By+C=0$
that passes through point $P$
parallel to $v.$
|
\[
d = \frac{\left|\overrightarrow{PS}\times\vect{v}\right|}
{\left|\vect{v}\right|}
= \frac{\left|Ax_0+By_0+C\right|}
{\sqrt{A^2+B^2}}
\]
|
| 66 |
polar coordinates
|
\[
\begin{array}{l}
x=r\cos{\theta}\\
y=r\sin{\theta}
\end{array}
\]
|
| 66 |
cylindrical coordinates
|
\[
\begin{array}{l}
x=r\cos{\theta}\\
y=r\sin{\theta}\\
z=z
\end{array}
\]
|
| 69 |
spherical coordinates
|
\begin{align*}
x&=\rho\sin{\varphi}\cos{\theta}\\
y&=\rho\sin{\varphi}\sin{\theta}\\
z&=\rho\cos{\varphi}\\
&\textrm{where}\\
\rho &\geq 0\\
0 &\le \theta\le 2\pi\\
0 &\le \varphi\le\pi
\end{align*}
|
|
distance formula in $n\textrm{-space}$
|
|
|
Angle between Intersecting Planes.
If $\vect{n}_1$ and $\vect{n}_2$
are vectors normal
to planes $P1$ and $P2$ respectively,
then the angle between the planes is given.
|
\[
\theta
= {cos}^{-1}
\frac{\vect{n}_1\cdot\vect{n}_2}
{\left|\vect{n}_1\right|\left|\vect{n}_2\right|}
\]
|
Alex Notes