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Vectors CPT Mindmap, what is a vector? - Coggle Diagram

- - - - Vectors with the same magnitude but opposite directions
    - - Same magnitude and same direction
    - - AB→ represents a vector running from A to B with its tail at A and head at B
      - |AB| represents only the magnitude
  - - - To determine the sum of two vectors when placed tail to tail, their sum is the diagonal of the constructed parallelogram
    - - Triangle law of addition is used when the vectors are head to tail
      - Triangle law can also be used to subtract vectors
    - - when two opposite vectors are added the resultant is a zero vector
  - - - When two vectors are parallel or lie on the same straight line, these vectors cannot be translated
      - 1/x x is a unit vector
  - - - a + b= b + a
    - - (a+b) + c = a + (b+c)
    - - k(a+b) = ka + kb
    - - m(na)=(mn)a=mna
    - - (m+n)a=ma+na
  - - - The location of every point is unique and every
      - In R2, P(a, b) is the point that is units from O(0,0) along the x asis and b units parallel to the y axis
      - In R3, P(a, b, c) is a point that is units from O(0, 0, 0) along the x axis and is b units parallel to the y axis and c units parallel to the z axis
    - - In R3, three mutually perpendicular axes form a right handed system
  - - - position vector OP can be represented as either OP=(a, b) or OP = ai + bj
      - Every vector in R2 can be written in terms of its components as i and j
    - - to determine the sum of two algebraic vectors add their corresponding x and y components
        
        OA + OD = ai + bj + ci + dj
    - - To multiply an algebraic vector by a scalar multiply both x and y components by the scalar
        
        mOP = m(a, b)
  - - - OP = (a, b, c) or OP = ai + bj +ck where O is (0, 0, 0)
    - - OB = pi + qj + sk, OC = mi + nj + rk, OD = fi + gj + hk
        
        OP = OD + (OB + OC)
      - mOP = m(a, b, c) = (ma, mb, mc)
  - - - for non collinear vectors u and v a linear combination of those vectors is au + bv where a and b are scalars
    - - the set of vectors {i,j} form a spanning set for R2
- - - - the resultant of several forces is a single force that can be used to represent the combined effort of all forces, the individual forces that make it up are referred to as components of the resultant
    - - The equilibrate of several forces is a single force that opposes the resultant of forces acting on an object
    - - when three noncollinear vectors are in a state of equilibrium these vectors will always lie in the same plane and form a linear combination
      - when three vectors are arranged head to tail the result is a triangle because the resultant of two forces is opposed by the third
        
        -c = a + b or a + b + c = (-c) + (c) = 0
    - - Resolving a force is taking a single force and decomposing it into two components
      - Vectors are resolves into two components that are right angles from each other
        
        referred to as horizontal and vertical components
        
        |fx| = |f|cos θ and |fy| = |f|sin θ
  - - - velocity of an object is stated relative to a frame of reference
      - resultant velocity is v v1 + v2
  - - - AC · AB = |AC| |AB| cos θ
        
        0 < θ < 90, a · b > 0
        
        θ = 90, a · b = 0
        
        90 < θ < 180, a· b < 0
      - result of dot product is always scalar
    - - Communtative Property
        
        p · q = q · p
      - Distributive Property
        
        p · (q + r) = p · q + p · r
      - Magnitudes Property
        
        p · p = |p|²
      - Associative Property with a scalar K
        
        (kp) · q = p · (kq) = k(p · q)
  - - - R2
        
        x · y = |x||y|cosθ = x1y1 + x2y2
      - R3
        
        x · y = |x||y|cosθ = x1y1 + x2y2 + x3y3
    - - cosθ = a·b/|a⃗||b⃗|
  - - - θ = 90°
      - 0 ≤ θ < 90°
      - 90° < θ ≤180°
    - - a·b/|b|
      - a·b/|a|
    - - cosα = a/√a² + b² + c²
      - cosβ = b/√a² + b² + c²
      - cosγ = c/√a² + b² + c²
    - - a · b/(b· b)b
  - - - b x a and the opposite is a x b
    - - k = 1
        
        a x b = (a2b3 - a3b2, a3b1 - a1b3,a1b2 - a2b1)
      - k = -1
        
        a x b = (a3b2 - a2b3, a1b3 - a3b1,a2b1 - a1b2)
      - add image for cross product
  - - - W = f · s
    - - the cross product can be used in forces that involve rotation on a point or axis
        
        the torque or turning effect of the force f about the point M is defined by r x f, this vector is perpendicular to the plane formed by the vectors r and f. the magnitude of the torque f about M is (|r|sinθ)(|f|) = |r x f|
- - - - r = r0 + tm
    - - x = x0 + ta and y = y0 + tb
  - - - Ax + By + C = 0 where n = (A, B) is normal to the line
    - - these planes are parallel if n1 = kn2 where k is nonzero
      - these planes are perpendicular if n1·n2 = 0
  - - - r = r0 + tm
    - - x = x0 + ta and y = y0 + tb and z = z0 + tc
    - - x-x0/a
      - y-y0/b
      - z-z0/c
  - - - r0 = (x0, y0, z0)
      - a = (a1, a2, a3)
      - b = (b1, b2, b3)
    - - x = x0 + sa1 + tb1
      - y = y0 + sa2 + tb2
      - z = z0 + sa3 + tb3
  - - - Case 1(one variable)
        
        Two of the coefficients(two of A, B or C) equal 0 and D = 0
        
        these are three coordinate planes
        
        two of the coefficients equal 0 and D does not equal 0
        
        parallel to the three coordinate planes
      - Case 2(two variable)
        
        One of the coefficients(one of A, B or C) equal 0 and D doesn't equal 0
        
        Find two. intercepts and draw a plane parallel to the missing variable axis
        
        One of the coefficients(one of A, B or C) equal 0 and D = 0
        
        Find a point with the missing variable set equal to zero and draw the plane connecting it
      - Case 3(three variables)
        
        None of the coefficients equal zero and D doesn't equals 0
        
        Find three intercepts and draw a plane through these points
        
        None of the coefficients equal zero and D equals 0
        
        Determine two points in addition to (0, 0, 0) and draw the plane through these points
  - - - Ax + By + Cz + D = 0
    - - n1 · n2 /|n1||n2|