nyuuzyou/wikis huggingface dataset:
Sec1Sess15

equations of lines
We have seen: line =
intersection of 2 planes
another way to think of line:
trajectory of moving point
= "parametric equation"
example line through:
Q_0 = (-1,2,2)
Q_1 = (1,3,-1)
Q(t) = moving point
Q(0) = Q_0
moves at constant speed
position at time t is \vec{Q_0}\ Q(t)
= t * (\vec{Q_0,Q_1}) due to traveling at constant time
Q_0 = (-1,2,2)
Q_1 = (1,3,-1)
t * (\vec{Q_0,Q_1}) = t * \langle (1-\ -1),(3-2),(-1-2) \rangle = t * \langle 2, 1, -3 \rangle 
Q_0 = (-1,2,2)
Q(t) = (x(t),y(t),z(t))
x(t) = -1 + 2t
y(t) = 2 + t
z(t) = 2 - 3t
Q(t) = Q_0 + t*Q_0*Q_1
= (-1 + 2t, 2 + t, 2 - 3t)
= first parametric equation example in the class
in a (x,y, and z) parametric equation the independent variable is t and dependent variables are x,y,z
Example 1: Line through P_0 = (1,2,3) and parallel to vector <1,3,5>
 P_0 = (1,2,3)
P = <x,y,z>
->P_0P = <x-1,y-2,z-3>
->P_0P being parallel to <1,3,5> makes it a scalar multiple of <1,3,5>
therefore <x-1,y-2,z-3> = t*<1,3,5>
t = x - 1, 3t = y - 2, 5t = z - 3
x = 1 + t, y = 2 + 3t, z = 3 + 5t
<1 + t, 2 + 3t, 3 + 5t>
In general, the line through P0 = (x0, y0, z0) in the direction of (i.e., parallel
to) v = <v1, v2, v3> has parametrization
<x, y, z> = <x0 + tv1, y0 + tv2, z0 + tv3>
= x = x0 + tv1, y = y0 + tv2 z = z0 + tv3.
Example 4: Find the line through the point p_0 = (1,2,3)\ and\ p_1 = (2,5,8)
(\vec{p_0,p_1}) = <2-1,5-2,8-3> = <1,3,5> = v
(\vec{O,p_0}) = <1,2,3> where O is origin
line = (\vec{O,p_0})+t*(\vec{p_0,p_1}) = <1 + t, 2 + 3t, 3 + 5t>
Recitation 2:
L_1 <2-t, 1+t>
L_2 <2+t, 4 + 2t>
Set x in L_2 to be equal to x in L_2 and the same for y
create new variable "u"
L_2 x = 2 + u = 2 - t
 y = 4 + 2u = 1 + t
-u = t
insert into t in y equations
4 + 2u = 1 - u
3u = -3
u = -1
t = 1
L_1:
x = 1
y = 2
L_2:
x = 2+u
y = 4+2u
x = 1
y = 2
Point of intersection is <1,2>
Sec1Sess16

An important approach to finding parallel vectors and directions is to look for the dot product with a
line's (cross product for plane) normal and another line (plane, etc.) to equal 0.
How to find the line intersection with a plane
Consider the plane x+2y+4y=7
Where does the line through Q_0, Q_1 intersect the plane?
plane equation: x+2y+4z = P
Q_0 = (-1,2,2)
Q_1 = (1,3,-1)
P\ .\ Q_0 = (-1)+2*2+4*2=11
11 > 7
P . Q_1 = 1 + 2*3 + 4*-1=3
3 < 7
because no Q points resulted in 7 none were found to be in the plane
Due to Q_0 passing through 7 as a result of the plane equation to get to Q_1 
the 2 Q points are on opposite sides of the plane
Point of Intercept of a line with a plane:
use Q_0 (point 1 of line) and Q_1 (point 2 of line)
to find direction of line (x2-x1,y2-y1,z2-z1) to find parametric equation
x = (x2-x1)t, y = (y2-y1)t, z = (z2-z1)t
substitute the parametric equation into the equation of the plane:
plane = a + b + c = d
a(x)+b(y)+c(z)=d
solve t for d
apply t found to parametric equation to find x,y,z of point of intercept
Finding at what point the line intersects the plane:
what about Q(t)?
x(t)+2y(t)+4z(t)
 Q_1 \ -\ Q_0 = (1-\ -1,3-2,-1-2) = (2,1,-3) = Q_{1,0} = change from time 0 to time 1
p = x+2y+4z
Q(t)\ =\ p * Q_0 + t*(Q_1-Q_0)
=(-1+2t)+2(2+t)+4(2-3t)
=-8t+11
is the result greater or equal to 7?
Q(t) is in the plane when -8t+11=7
when t = 1/2 the result is 7. time (t) is 1/2 when intersection occurs
Q(1/2) = (0,5/2,1/2) = intersection of the plane
if no solution for the equation of the plane's result (7 in this case) occurs than the line could be parallel to the plane or in the plane.
What is an equation for a line that lies on a plane?
Line should be perpendicular from normal of plane, therefore parallel to the plane.
Test for perpendicularity with dot product and normal. Normal of plane is given by
the equation of the plane.
E.g. plane: 3x+\ y+\ -5z=9
N = <3i, j, -5k>
Perpendicularity is tested by <x,y,z> . N = 0 where <x,y,z> is the line.
E.g. x = at, y=bt, z=ct where 3x + y + -5z=0, a,b,c, not all 0
What is an equation for a line that lies on a plane and intersects a point?
plane: 3x + y + -5z=9
point: <4,8,10>
x = 4+at, y = 8+bt, z = 10+ct where 3x + y + -5z=0, a,b,c, not all 0
practice problem 1
plane = p = 2x + y -4z = 4
line = x = t, y = 2 + 3t, z = t
Q_0\ =\ (0,2,0)
Q_1\ =\ (1,5,1)
Q_{1,0}\ =\ (1-0,5-2,1-0) = (1,3,1)
Q(t)\ =\ p\ *\ Q_0\ +\ t*(Q_1-Q_0)
t*Q_0*Q_1\ =\ (t,3t,t)
Q_0\ +\ t*(Q_1-Q_0)\ =\ (0+t,2+3t,0+t)\ =\ (t,2+3t,t)
Q(t) = 2(t)+(2+3t)-4(t)=4
=5t-4t+2=t+2=4
when t+2=4 there is the intersection, therefore
t=2
since line = x = t, y = 2 + 3t, z = t
intersection point is at (2, 8, 2)
"parameterization of the plane" =\ Q_0\ +\ t*(Q_1-Q_0)
Sec1Sess17

Parametric curves and equations
More generally, parametric equations for arbitrary motion in the plane or in space.
Example: t (time) can be the parameter and functions for x, y, and z are parametric equations
Position vector: r(t) = x(t)i + y(t)j + z(t)k = <x(t),y(t),z(t)>. The vector from the origin to the moving point.
Distance vector: OP = r(t) = x(t)i + y(t)j = <x,y>
Circle example:
Parametric form: x(t) = a*cos(t) , y(t) = a*sin(t)
Symmetric form: x^2 + y^2 = a^2
Eclipse example:
Parametric form: x(t) = a*cos(t) , y(t) = a*sin(t)
Symmetric form: x^2/a^2 + y^2/a^2 = 1
Line Example:
Parameterize the line through (x_0,y_0)\mathrm{\ parallel\ to\ }\langle b_1,b_2 \rangle by:
\mathrm{x(t) = x_0 +t(b_1), y(t) = y_0 +t*b_2\ equivalent\ to\ r(t) = (x, y) = (x_0 +t(b_1), y_0 +t(b_2)) = (x_0, y_0)+t(b_1, b_2).}
\mathrm{Advanced\ line:\ parametric:\ x\ =\ 3(cos^2*t),\ y\ =\ 3(sin^2*t)\ .\ symmetric:\ x\ +\ y\ =\ 3,\ with\ x,\ y\ non-negative.} (symmetric form looses info because the line reverses back in direction on itself and that is not shown in that form.)
Helix: r(t) = cos(t)*i + sin(t)*j + a*t*k
Cycloid: wheel of radius a rolling on floor = x-axis
P = a point on the rim of a wheel starts at 0. What happens?
Curve with no simple symmetric form, but a practical parametric form is created
Question: position (x(t), y(t)) of the point P?
instead of t, theta is used. The (x(theta), y(theta)) is a function of the angle, theta, by which the wheel (circle) has rotated.
A = point where circle touches x-axis
a = radius of circle
B = circle center
O = origin (can be <0,0>)
T = \mathrm{\theta} , theta, angle of rotation 
P = a point on the rim of a wheel starts at 0.
\vec{OP}\ =\ \vec{OA}\ +\ \vec{AB}\ +\ \vec{BP}
\vec{OA} = \langle aT,0 \rangle 
this is the distance the circle has traveled on the x-axis
= arclength from A to P
the arclength along the circle once flattened equals the distance traveled by the circle on the x-axis
\vec{AB} = \langle 0,a \rangle 
this is just a line that points strait up from the point the circle touches the x-axis
\vec{BP}, things known include |\vec{BP}| = a, angle T with vertical
= <-a*sin(T),-a*cos(T)>
\vec{OP} = <aT - a*sin(T), a - a*cos(T)>
* Finding the parametric equations for a point P on the circumfrance of a disk rotating over time t (cycloid):
1. start at origin
2. add <0,i> to go to center of disk where i is y at y-intercept
3. add <t,0> to go to center of disk at time t
4. add -<sin(angle),cos(angle)> to shift over by one unit at the given angle if movement is clockwise, adjust sin, cos to fit 
if movement is counterclockwise
Finding angle: after time t the center of the disk has moved t units to the right, distance travelled 
is r*angle where r is radius and angle is the central angle in radians. Therefore t = r*angle and angle = t/r.
OP(t) = <0, i> + <t, 0> + <− sin(t/r), − cos(t/r)>
since <0,i> applies to y axis than it is added on the y axis
since <t,0> applies to x axis than it is added on the x axis
r = r(t) = <t - sin(t/r), i - cos(t/r)>
Sec1Sess18

T = theta = angle
For cycloid:
\vec{OP} = \langle aT - a*sin(T), a - a*cos(T) \rangle 
What happens when the point measured on the cycloid reaches near the bottom?
take length unit = radius: a = 1
x(T) = T - sin(T)
y(T) = 1 - cos(T)
For small T
sin(T) \approx T
cos(T) \approx 1
Better Approximation?
Use Taylor approximation (similar to Taylor expansion seen before):
for t small,
\mathrm{f(t) \approx f'(0) + t*f'(0)+(t^2)/2*f(0)+(t^3)/6*f'(0)}
add higher level derivatives of f(x) for more accuracy.
Better approximation for small angles:
sin(T) \approx T - T^3/6
cos(T) \approx 1 - T^2/2
Even better approximation
sin(T) \approx T - (T - T^3/6) \approx T^3/6
cos(T) \approx 1 - (1 - T/2) \approx T^2/2
|X| << |y|
\mathrm{y/x\ \approx \ T^2/T^3\ =\ 2/T\ ->inf\ (as\ it\ approaches\ infinity,\ an\ asymptotic\ convergence)}
slope at origin is therefore infinite (inflection point as the left side of the plotted cycloid point falls and the right side raises)
Sec1Sess19

Parametric Equation
Velocity and acceleration
Position vector: r(t) = <x(t),y(t),z(t)>
\vec{r(t)} = \langle t - sin(t), 1 - cos(t) \rangle 
the position vector shows y = t*x, so t = y/x and can be substituted into equations 
to find xy-equations of a curve
velocity = derivative of position vector with respect to time = d(r)/d(t)
\mathrm{ =\ \langle d(x)/d(t),d(y)/d(t),d(z)/d(t) \rangle \ =\ velocity\ =\ dr/dt\ =\ dx/dt*i\ +\ dy/dt*j\ =\ x'*i\ +\ y'*j\ =\ <x',\ y'>.}
Cycloid example:
\vec{v}\ =\ \langle\ 1\ -\ cos(t),\ sin(t)\ \rangle 
at t = 0: v = 0
Speed (scalar) = |\vec{v}| = magnitude of v
= sqrt((1-cos(t))^2+(sin(t))^2)
= sqrt((1-cos(t))^2+sin^2(t))
= sqrt(1-2*cos(t)+cos^2(t)+sin^2(t))
= sqrt(2-2*cos(t))
Acceleration:
the derivative of the velocity vector. therefore the second derivative of the position vector.
\vec{a} = d(\vec{v})/d(t)
\mathrm{e.g.\ cycloid:\ }\vec{a}\mathrm{\ =\ \langle sin(t),cos(t) \rangle }
\mathrm{At\ t\ =\ 0:}\vec{a}\mathrm{\ = \langle 0,1 \rangle }
|d(r)/d(t)| <> d(|r|)/d(t)
differentiating the length of a vector is tricky, there is no simple formula for it, it does not need to be calculated often
Tangent vector: geometric name for d(r)/d(t) (the velocity vector). As change(t) shrinks to 0 the vector change(r)/change(t) becomes tangent to the curve of the change in position.
When computing the vectors with a rate of time involving rotations, radians of pi should be used as a unit to represent the value of the angle change created by the rotations. This can represent the angle variable in theta, e.g. 3 revolutions per second = 6*pi radians / sec = T.
Sec1Sess2

Section 1 Session 2
Magnitude = |A| = sqrt(a_1^2+a_2^2) which is the euclidean distance from the origin (0,0) 
Law of Cosines:
c^2 = a^2 + b^2 - 2*a*b*cos(C)
|\vec{C}|^2 = |\vec{A}|^2 + |\vec{B}|^2 - 2|\vec{A}|*|\vec{B}|*cos(angle_{ab})
|\vec{C}|^2 = |\vec{C}| . |\vec{C}|
|\vec{C}|^2 = (|\vec{A}| - |\vec{B}|) . (|\vec{A}| - |\vec{B}|)
effect of cos(): (|\vec{A}| . |\vec{B}|) = |\vec{A}|*|\vec{B}|*cos(angle_{ab})
the angle between a vector and itself is 0
cos(0) = 1
|\vec{A}| . |\vec{A}| = |\vec{A}|^2*Cos(angle_{aa})
Combining points into a vector:
NOTE: Order is important, with AB vector from points A and B the calculations are:
\vec{AB} = <b1-a1,b2-a2,c3-a3>
DOT PRODUCT 
A . B = a_1*b_1 + a_2*b_2
E.g. A = i + 2j, B = 3i + 4j
A . B = 1 * 3 + 2 * 4 = 11
Dot product works the same for 3-dimensional vectors
A . B = a1*b1 + a2*b2 + a3*b3
Test for orthagonality (vectors 90 degrees from each other):
cos(90) = 0, therefore A . B = 0 = A * B * cos(90)
Find length of 3rd triangle side given 2 sides:
Given side A and B, the 3rd side is A-B .
E.x. A = 2i + 3j + 5k and B = i - 2j + 4k, side C = i + 5j + k
Plotted geometric examples of finding a 3rd side are here:
http://tutorial.math.lamar.edu/Classes/CalcII/DotProduct.aspx
Sec1Sess20

Arc Length and Velocity
Distance traveled along a curve
s = arc length = distance traveled along a trajectory (curve)
r = position = net displacement (total difference in position, start vs. end)
v = velocity = tangent vector
lots more notation here (not to repeat the list here):
http://ocw.mit.edu/courses/mathematics/18-02sc-multivariable-calculus-fall-2010/1.-vectors-and-matrices/part-c-parametric-equations-for-curves/session-20-velocity-and-arc-length/MIT18_02SC_notes_12.pdf
NOTE: <x,y,z> notation for vectors, (x,y,z) notation for points
can be s vs. t
d(s) a.k.a. d(s)/d(t)
= speed = |\vec{v}| (length of velocity vector) = sqrt(v^2)
arc length = int_{t1,t2}(d(s)/d(t)) where t1, t2 are time1, time2
note: for solution on continuous integral subtract value at t2 from t1
Example: length of an arch of cycloid is:
int_{0,2pi}(sqrt(2-2*cot(t)))
want to integrate t = 0 to 2*pi
(this will be solved later not by directly taking integral but by another method)
unit tangent vector T
the vector tangent to the curve at a point, converted to a unit length (length of 1) by dividing by the length of \vec{v}.
T_{hat}\mathrm{\ =\ }\vec{v}\mathrm{\ /\ |}\vec{v}\mathrm{|}
\vec{v}\ =\ d(\vec{r})/d(t)\ =\ d(\vec{r})/d(s)\ *\ d(s)/d(t)
\mathrm{d(}\vec{r}\mathrm{)/d(s)\ =\ T_{hat}}
\mathrm{d(s)/d(t)\ =\ |}\vec{v}\mathrm{|}
\vec{v}\mathrm{\ =\ T_{hat}\ *\ |}\vec{v}\mathrm{|}
velocity has:
1. direction: tangent to trajectory (curve), T_{hat}, d(r)/d(s)
2. length: speed, |\vec{v}|, d(s)/d(t)
r(t+t_{change}) is an approximation of the distance along the curve because a straight line is used to measure between point 1,t, and point 2,t_{change}.
\mathrm{change(}\vec{r}\mathrm{)\ \approx \ T_{hat}*s_{change}}
\mathrm{=change(}\vec{r}\mathrm{)\ /\ change(d)\ \approx \ T_{hat}*(s_{change\ /\ change(t)})}
Limit as change_{t} -> 0
gives d(r)/d(t) = T_{hat}*d(s)/d(t)
d(s)/d(t) = d(r)/d(t) = sqrt((d(x)/d(t))^2+(d(y)/d(t))^2)
when given the x and y of the position vector, take their derivatives to find the
x and y of the velocity vector.
e.g. x(t) = 20t, y(t) = 5t + 6
v = <20, 5>
Examples of finding v, s, r, etc.: http://ocw.mit.edu/courses/mathematics/18-02sc-multivariable-calculus-fall-2010/1.-vectors-and-matrices/part-c-parametric-equations-for-curves/session-20-velocity-and-arc-length/MIT18_02SC_we_11_comb.pdf
Sec1Sess21

Kepler's Second Law
Motion of planets is in a plane,
and the area is swept out (encompased inside) by the line from
sun to planet at a constant rate.
once orbit is known than it can be found how fast
a planet moves on that orbit
Newton
Later explained the 2cnd law by the formula for
gravitational attraction
Kepler's law in terms of vectors?
express area in terms of vectors
eclipse vectors are 1. center to starting point and 2. center to point traveled to
area approximately equals 1/2 area created by parallelogram of vectors in elipse
\mathrm{\approx \ 1/2|}\vec{r}\mathrm{\ x\ change(}\vec{r}\mathrm{)|\ where\ r} is the eclipse position vectors notation
= area swept after time occurs change(time) (small)
\mathrm{change}(\vec{r})\ \approx \ \vec{v}\ *\ \mathrm{change(t)\ therefore}
\mathrm{1/2*|}\vec{r}\ \mathrm{x\ change}(\vec{r}\mathrm{)|\ \approx \ 1/2\ *\ |}\vec{r}\mathrm{\ x\ }\vec{v}\mathrm{|\ *\ change(t)}
change(t) is able to be taken out of length section because it is a scalar
law says: |\vec{r}\ \mathrm{x}\ \vec{v}| = constant = rate at which area is swept by position vector
Part of law 2:
plane of motion
both eclipse vectors are in the plane
vector1: center to starting point
vector2: velocity of point moving on eclipse
Direction of \vec{r} x -> v is normal to the plane of motion
(motion will stay in the plane)
kepler's 2cnd law <=> \vec{r} x -> v = constant vector
\mathrm{<=>\ d(}\vec{r}\mathrm{\ x\ }\vec{v}\mathrm{) = 0}
Product rule is ok for taking derivative of dot product and cross product
Order of vectors matters at least for cross product.
similar to product rule: d(r . s) = d(r) . s + r . d(s)
d(r x s) = d(r) x s + r x d(s)
d(r)/d(t) x v + r x d(v)/d(t) = 0
v = d(r)/d(t)
a = d(v)/d(t)
v x v + r x a = 0
a vector cross itself is always 0
r x a = 0 because the acceleration is parallel to the position
this is stated in keplar's 2cnd law
<=> gravitational force is parallel to the position vector
For motion under any type of central force, the path of motion will lie in a plane and area will be swept out by the radius vector at a constant rate
Sec1Sess3

1st application: Finding Angles:
in 3d triangle find theta (angle, @): 
point_1 e.g. point_1 = <-1,1,0>
A = point_1->point_2
B = point_1->point_3
|A| = \sqrt{point1^2}
|A|.|B| = |A|*|B|*cos(angle)
cos(angle) = A . B/|A|*|B|
notation can also be written as cos(angle) = A-> . B-> / |A->| * |B->|
use cos^{-1}(angle) to find @ in degrees
Sign (positive or negative) of |A|.|B|: 
greater than 0 if @ < 90*
equals 0 if @ = 90*
less than 0 if @ > 90*
2cnd Application: 
Finding orthogonality (perpendicular directions, e.g. 90* angle)
If the dot product is 0 than the vectors are perpendicular
orthogonality only occurs when cos(@) = 0 = 90* because the angle must be 90*
orthogonality symbol is _|_ 
Note:
diagonal of two vectors can be found by A+B, i.e. traveling \vec{A} then \vec{B} and arriving at the end point of the diagonal from the origin of the diagonal. A and B have the same origin, IDK if they could have different origins for this to work.
This is an example of adding vectors also
Sec1Sess4

Length of a component (vector) along an angle (angle of another vector)
|\vec{A}| * cos (angle) = |\vec{B}|
Where |\vec{A}| is the magnitude (length) of the vector A
To calculate the length at an angle use a unit vector (vector of all ones)
|\vec{A}|.|u|*cos(angle)
Where |u| is |unit vector|
@ (angle) can be angle along individually the x, y, or z axes. lengths along axes can be found that way.
physics forces in different directions can be calculated in this way
A.u = |A|.|u|*cos(angle)
leg parallel to u is the "orthogonal projection" to u
u angle can be found by B/|B| . e.g. B = 3i + 4j. B/|B| = (3/(\sqrt{9+16}))*i + (4/(\sqrt{9+16}))*j
When finding a "component" of a vector in a direction the result is a scalar
"component": one vector projected onto another (plus its +/- sign). it is |vector|*cos(@)
|A|*cos(angle) = (A*B)/|B| = component = (A*B)/(\sqrt{B_{indices}^2})
Sec1Sess5

complex shapes can be cut up into simpler shapes to find areas
triangle area = 1/2 |A|*|B|*sin(angle) = 1/2 * det(A, B) (see det() description below) where points A, B are relative to origin point C
notice sin() instead of cos()
find sin() by using angle finding methods to find cos() and solving for sin^2()+cos^2()=1
due to complimentary angles sin (angle) = cos(angle')
where angle' = pi/2 - angle. (pi/2 = 90*, the complementary angle)
rotate 90 counter clockwise: vector of <x, y> from origin becomes <-y, x> (yes dimensions are flipped)
Finding determinant:
* KEY TERM * Determinant \begin{bmatrix}a_1 a_2 ; b_1 b_2\end{bmatrix} = (a_1*b_2)-(a_2*b_1)
Det() has same notation as magnitude, | | , but has multiple rows.
Determinant measures the area of a parallelogram with sides A and B
when multiplied by 1/2 the area of a triangle can be found
paral. area will be positive but area of a triangle can be negative
Determinant is + or - the area of a parall. Area is the absolute value of det()
Determinant = |A|*|B|*sin(angle) = |A'|*|B|*cos(angle')
Where A' = <-y, x> of A
A is a vector not a point
= A' . B
= <-a_2, a_1>*<b_1, b_2>
= (a_1*b_2) - (a_2*b_1)
= det(A, B) where det() is determinant
* IMPORTANT POINT * determinants only work on parallelograms, splitting into triangles is needed to use formulas to find areas on other types of rectangles.
Calculate 3x3 determinant:
|a b c| = A
|d e f|
|g h i|
det(A) = a(ei − fh) − b(di − fg) + c(dh − eg)
Sec1Sess6

det (A, B, C) = \begin{bmatrix}a1\ a2\ a3 ; b1\ b2\ b3; c1\ c2\ c3\end{bmatrix} = a1 \begin{bmatrix}b2\ b3 ; c2\ c3\end{bmatrix} - a2 \begin{bmatrix}b1\ b3 ; c1\ c3\end{bmatrix} + a3 \begin{bmatrix}b1\ b2 ; c1\ c2\end{bmatrix}
method of calc det is per col and row of an index eliminate that row and col index and take determinant of what is left. minus is in middle calc for rotation of coords reason.
parallelepiped is 3d box with sides of parallelogram, can have volume found from vectors shown above
laplace expansion to find determinant: det(A) = a ( ei - fh ) - b ( di - fg ) + c ( dh - eg ) where A = \langle \langle a b c \rangle \langle d e f \rangle \langle g h i \rangle \rangle 
lots of further notation and methods:
http://ocw.mit.edu/courses/mathematics/18-02sc-multivariable-calculus-fall-2010/1.-vectors-and-matrices/part-a-vectors-determinants-and-planes/session-6-volumes-and-determinants-in-space/MIT18_02SC_MNotes_d1.pdf
I did not want to rewrite here many of the content written in that doc
In laplace transform note a - b + c format of the row or column that the determinants are multiplied by. This is determined by the "checkerboard pattern".
volume can be found using vectors in arbitrary (any) numbers of dimensions with recursive determinants calculations using Laplase transform or other volume finding methods.
Sec1Sess7

dot product of A and B creates matrix with first row as vector with variables \hat{i}, \hat{j}, and\ \hat{k}
\hat{i}, \hat{j}, and\ \hat{k} are used to denote the first column values that are created from the determinants of the values not in the same row or col as the \langle \widehat{variable} \rangle i.e. \hat{i} .
length: distance in space of vector A and vector B with euclidean distance formula? 
Cross product, A x B, is a vector perpendicular to the plane of A and B
example of way to calculate it: 
For A = <a1, a2, a3> and B = <b1, b2, b3>:
A x B = (a2*b3 − a3*b2)*i - (a3*b1 − a1*b3)*j + (a1*b2 − a2*b1)*k
+ and - between determinants in crossproduct follow checkerboard pattern
direction: dir(A*B): dir() is cross product formula described earlier that creates a vector perpendicular to vectors A and B
IMPORTANT NOTE: Cross product is done with vectors for finding normals for use in finding equations of planes (described later). If only points are given the points need to be combined to create vectors for use in the cross product to get solutions.
Find angles with cross products:
|A × B| = |A||B|sin(@)
with right hand rule the dir() perpendicular line that is wanted always points "up" not "down"
dir() can help find volume by finding the n_^ (perpendicular vector) of the x and y vectors in 3d space
volume = |B*C|*(A*n_^) = |Y*Z|*(X*n_^) where n_^ is vector which is right angle formed from dir(X*Y)
n_^ can be found by cross product. 
n_^ = B*C/|B*C| ?
n_^ = A.(B*C) ?
n_^ = det(A,B,C)
Sec1Sess8

A way to find if a point is in a plane region is finding if det( \langle P_1,P_{new} \rangle \ \langle P_1,P_2 \rangle \ \langle P_1,P_3 \rangle) = 0 where P_{new} is the point being tested
This tests if a volume of 0 is made if the new point is connected to the plane's points
Different way to check if point is in plane:
Normal vector is one that is perpendicular to a plane (points out at a 90* angle from a plane)
if vector \langle P_1, P_{new} \rangle is _|_ (perpendicular) to the vector \vec{N} (the normal vector) than the point in question is on the plane
This tests if the new point is perpendicular to the plane.
To find if vector are perpendicular it can be looked for if their dot products are 0:
(\vec{P_1,P_{new}})\ .\ \vec{N} = 0 where "." is dot product
How can \vec{N} be found? The cross product of points can be taken: \vec{N} = (\vec{P_1,P_2})\ \ x \ \ (\vec{P_1,P_3}) . 
It seems the cross product can be taken (\vec{P_1,P_3}) x (\vec{P_1,P_2}) and this also creates N
"Triple Product":
Since N can be replaced than as seen above:
0 = (\vec{P_1,P_{new}})\ .\ ((\vec{P_1,P_3}) x (\vec{P_1,P_2}))
shows P_{new} in the plane
This is the "triple product" and is equivalent to the determinant
"equation of the plane"
use P_{new} = <x,y,z> symbolically
"equation of the plane" = N . (\vec{P_1,P_{new}}) = 0 and create symbolic solution
Example:
if N = <-2,1,1> and p_1 = <1,3,1>
N . (P_1,P_{new}) = <-2,1,1> . <x-1,y-3,z-1> = 0
-2(x-1) + (y - 3) + (z-1) = 0
-2(x) + y + z = 2
equation of the plane = -2(x) + y + z = 2
http://ocw.mit.edu/courses/mathematics/18-02sc-multivariable-calculus-fall-2010/1.-vectors-and-matrices/part-a-vectors-determinants-and-planes/session-8-equations-of-planes/MIT18_02SC_we_7_comb.pdf
Sec1Sess9

Change of coordinate systems is a linear relationship
Matrix Product:
for <x, y, z>
<2x, 3y, 3z> = u1
<2x, 4y, 5z> = u2
<x, y, 2z> = u3
can be found by
A = <<2, 3, 4>,<2,4,5><1,1,2>>
X = <x,y,z>
<<2, 3, 4>,<2,4,5><1,1,2>> . <x,y,z>=<u1,u2,u3>
A . X =<u1,u2,u3>
Matrix Product:
Dot products between rows of A and columns of X
Multiplying matrix A with matrix B
width A must equal height B