6AStudyGuide_Final.pdf

# 2 summary a vector is an arrow between two points in

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SUMMARY A vector is an arrow between two points in the cartesian plane or space. Given two points A = ( x 1 , y 1 , z 1 ) and B = ( x 2 , y 2 , z 2 ) then the vector between them is ~ AB = h x 2 - x 1 , y 2 - y 1 , z 2 - z 1 i . The norm (or magnitude) of a vector ~v = h a, b, c i is k ~v k = a 2 + b 2 + c 2 . Vectors can be added and scaled as follows: h a, b, c i + h a 0 , b 0 , c 0 i = h a + a 0 , b + b 0 , c + c 0 i , λ · h a, b, c i = h λa, λb, λc i . The dot product between two vectors ~v = h a, b, c i and ~w = h a 0 , b 0 , c 0 i is defined by ~v · ~w = aa 0 + bb 0 + cc 0 . It can be shown that if θ is the angle between the vectors ~v and ~w then ~v · ~w = k ~v kk ~w k cos θ. It then follows that ~v and ~w are orthogonal if and only if ~v · ~w = 0. There is another product of vectors whose output is a vector. This is called the cross product: ~v × ~w = det i j k a b c a 0 b 0 c 0 = ( bc 0 - b 0 c ) i - ( ac 0 - a 0 c ) j + ( ab 0 - a 0 b ) k , where i = h 1 , 0 , 0 i , j = h 0 , 1 , 0 i and k = h 0 , 0 , 1 i . From the properties of determinants it follows that ~v × ~w is orthogonal to both ~v and ~w , i.e., ~v · ( ~v × ~w ) = 0 and ~w · ( ~v × ~w ) = 0 . It is also easy to check that ~v × ~w = - ~w × ~v . The norm of the cross product is k ~v × ~w k = k ~v kk ~w k sin θ, where θ is the angle between ~v and ~w . 3
Therefore ~v and ~w are parallel if and only if ~v × ~w = ~ 0. The area of the parallelogram generated by the vectors ~v and ~w is A = k ~v × ~w k . Likewise, the volume of the parallelepiped generated by the vectors ~u , ~v , and ~w is V = ~u · ( ~v × ~w ) . The parametric equations of the line with direction vector ~v = h a, b, c i and passing through the point P 0 = ( x 0 , y 0 , z 0 ) are: x = at + x 0 y = bt + y 0 z = ct + z 0 . The equation of the plane with normal vector ~n and passing through the point P 0 = ( x 0 , y 0 , z 0 ) is: ~n · h x - x 0 , y - y 0 , z - z 0 i = 0 . To compute the equation of a plane containing three points A , B , and C can be obtained by taking ~n = ~ AB × ~ AC . We can project one vector over another by using the formula Proj ~v ~u = ~u · ~v k ~v k 2 ~v. The norm of the vector projection is called the scalar projection and is computed by k Proj ~v ~u k = | ~u · ~v | k ~v k . To compute the distance from a point P to a plane with normal vector ~n passing through the point P 0 we can proceed as follows: 1. Compute the vector ~ P 0 P . 2. The distance from P to the plane is computed by the scalar projection d = Proj ~n ~ P 0 P = | ~n · ~ P 0 P | k ~n k . 4

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3. If ~n = h A, B, C i and P 0 = ( x 0 , y 0 , z 0 ) then the equation of the plane is A ( x - x 0 ) + B ( y - y 0 ) + C ( z - z 0 ) = 0 . Therefore the distance from the point P = ( x 1 , y 1 , z 1 ) to the plane is given by d = A ( x 1 - x 0 ) + B ( y 1 - y 0 ) + C ( z 1 - z 0 ) A 2 + B 2 + C 2 . The distance from a point P to a line with direction vector ~v passing through the point P 0 can be computed as follows: 1. Compute the vector ~ P 0 P . 2. Compute the projection Proj ~v ~ P 0 P = ~ P 0 P · ~v k ~v k 2 ~v. 3. The vector ~ P 0 P - Proj ~v ~ P 0 P has the same magnitude as the distance from P 0 to the line.
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