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28 Cards in this Set
- Front
- Back
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Translation
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change initial and final points of vector without changing the vector itself. (slope goes up the same
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length of a vector :
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||u|| = SQRT(u1^2+u2^2+u3^2....+un^2)
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Parrallel vectors:
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if one vector is a scalar multiple of another
ex. u = av & v = bu |
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Distance properties
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d(A, B) = D (B,A)
D(A,B) >= 0 D(A,B) = 0 if A=B |
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Norm/length properties
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a) ||v|| >= 0
b) ||v|| = 0 if v = 0 c) ||k*v|| = |k| * ||v|| |
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Unit Vector Formula
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1/||v|| * v
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Standard Unit Vectors
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(2,-3,4,5) = 2i-3j+4k+6L
V= v1e1+v2e2+v3e3....vnen |
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if v=u, then
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u*v=||v||^2
or ||v|| = SQRT(v*v) |
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Angle between 2 vectors:
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cos0 = (u*v)/(||u||*||v||)
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Orthagonal
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1) u*v=0
2) Orthagonal set : all pairs should equal to 0. |
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Cauchy-Shwarz Inequality
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(u*v)^2 =< ||u||^2 * ||v||^2
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Pythagoras Theorem
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||u+v||^2 = ||u||^2 + ||v||^2
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General Equation of a line (R2 and R3)
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R2:
Ax+By=C R3: Ax+By+Cz=D Note** set (ABC) can be considered as Normal vector as well. |
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Vector Equation of a line (R2 and R3)
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R2:
x=x0+tv R3: x= x0+v1t1+v2t2 |
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Parametric equation of a line (R2 and R3)
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x= x0 +at
y= y0 +bt z= z0 +ct |
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Line through two points
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if x0 and x1 are on a line, then we know (x1-x0) is the exact same vector. However, chances are this vector isnt starting at origin, so we use this formula:
x= x0+(x1-x0)t |
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Post normal equation of planes
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n = normal vector
x= arbatrary point on the plane x0= point we are examining n(x-x0)=0 note** n*v=0 means there orthagonal! |
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Homogenous system
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ax+by+cz= 0
* ends with zero ** has at least one solution, x=y=z=0 |
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Matrix trace
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sum of entries on its diagonal, has to be used on a square.
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Matrix inner product
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u^t*v
column vectors have to be the same |
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Matrix outer product
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u*v^t
column vectors don't have to be the same |
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Cancellation law
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if ab=ac, then b=c if a=! 0
NOT TRUE FOR MATRIXES |
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Identity Matrix
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1s on main diagonal and zeros everywhere else, HAS TO BE SQUARE
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Identity property
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AIn= A , InA=A
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Communtative law of multiplication
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ab=ba
*does not neccesaarly work for matrixes |
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Matrix inverses
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if AB=BA, then
AB=BA=In. B is the inverse of A. A can only have one inverse |
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Finding inverse
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if the determinant isnt 0, there is an inverse
det(A) = ac-bd Inverse formula 1/(ac-bd) * ( d -b) (-c a) |
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application of matrixes
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if Ax=B is invertable and square, then we know
A-1b = solution |
