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
intersection of lines in 3D
vector equation of a plane
position vector + λ * direction vector + μ* direction vector
finding the Cartesian equation of a plane
finding the normal equation of a plane
how can a plane and a line intersect
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
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’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