vectors Flashcards
scalar product
multiply corresponding terms then add each of the products
angle between two vectors. cosθ =
cosθ = a.b/(|a||b|)
scalar product of perpendicular lines
a.b = 0
vector product of two vectors
determinant of the two vectors as a matrix
what does the vector product find
a vector perpendicular to the two other vector
general vector equation of a line
position vector + λ * direction vector
finding the Cartesian equation of a line
set each line equal to x, y or z. then rearrange for λ.
finding the intersection of two lines
- equate the two equations
- write down two equations in λ and μ
- solve the equations simultaneously
- substitute back into the original equations to get the positions
intersection of lines in 3D
- can be parallel
- can intersect
- can be skew
vector equation of a plane
position vector + λ * direction vector + μ* direction vector
finding the Cartesian equation of a plane
- vector product the two direction vectors to find the normal
- find the constant by putting in the point is n1x + n2x + n3x = d
finding the normal equation of a plane
- find the normal (vector product the two direction vectors)
- find the constant by scalar producting a point and the normal
- write in the form r.n = d (where r stays as r)
how can a plane and a line intersect
- intersects once
- line parallel to plane (no points)
- line is within the plane
angle between to planes
the angle between the two normal lines
distance between a point (b) and a plane (in normal form r.n = p)
D = |(b.n)-p|/|n|
point (x,y) and a line (ax+by = c) in 2D
|ax + by - c|/√a^2+b^2
cartesian equation of a line in 2D from vector equation
- set the line equal to (x, y)
- rearrange the two equations for λ
- set the equations equal and rearrange into the normal form
distance between skew lines
D = |(b-a).n|/|n|
When finding an unknown line which intersects planes
let x = λ, find y and z in terms of lambda then put line in cartesian form
shortest distance to a point to a line in 3D
|(p-a)xd|/|d|
p is the point, a is the lines position vector, d is the direction vector
angle between plane and line
cosθ = n.d/|n||d|
90-θ
draw picture to help understand
distance between skew lines
|(b-a).n|/|n|
b and a are the position vectors, n is the cross product of the direction vectors
proving the lines are skew
- prove they don’t intersect
- prove they’re not parallel
distance between point and plane
|(b.n)-p|/|n|
b is the position vector of the point, n is the normal to the plane, p is RHS of r.n = p
distance between point and plane cartesian
|ax+by+cz + d|/√(a^2 +b^2 + c^2)
where a, b, c and d is the cartesian equation, and x,y,z is the point
vector product of parallel vectors
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