Subtracting two vectors means adding the first vector to the opposite of the
second.
u−v=u+(−v)
Collinear vectors
Let u be a non-zero vector and v an arbitrary vector.
If u and v are collinear, then there is a unique real number
λ such that v=λu.
If there exists λ∈R such that v=λu, then u and v are
collinear.
Let a and b be two non-collinear vectors. For any vector v∈ in V
there exist α,β∈R such that v=αa+βb. The scalars α and β with this
property are unique.
Theorem:
For any u and v∈V, u+v=v+u (commutativity)
For any u, v, w∈V, (u+v)+w=u+(v+w) (associativity)
For any u∈V, u+0=0+u (0 is the identity element)
For any u∈V, u+(−u)=(−u)+u=0 (every vector has an opposite).
Theorem: For any α,β∈R and any u, v∈V we have:
(α+β)v=αv+βv;
α(u+v)=αu+αv;
α(βv)=(αβ)v;
1⋅v=v.
Definition: Let Oxy be a rectangular coordinate system in the plane and
the points A(1,0) and B(0,1). We write the vectors OA=i, OB=j; i,j
are called the unit vectors of the coordinate axes of the systemOxy. The
pair (i,j) is called the basis of the systemOxy.
Definition: Let Oxy be a rectangular coordinate system in the plane. For
any point M in the plane, the vector OM is called the position vector of
the point M.
Definition: In space we fix three axes Ox, Oy, Oz with the same
origin O, pairwise perpendicular. The resulting system is written Oxyz
and is called a three-dimensional rectangular (or Cartesian) coordinate
system. The orientation of the axes is usually chosen as the arrows in the
adjacent figure show.
Definition: In the system Oxyz, each point M(x,z,y) in space is determined
by the position vector OM=x⋅i+y⋅j+z⋅k; x,y,z are called the coordinates of the vector
OM in the basis (i,j,k); x is called the abscissa, y the ordinate, and
z the applicate of the point M(x,y,z).
Definition: The norm of a vector is the distance between its endpoints.
Let the points M1(x1,y1,z1) and M2(x2,y2,z2). Then M1M2=d(M1,M2)=(x1−x2)2+(y1−y2)2+(z1−z2)2.
The scalar product. For any two vectors v1,v2∈V3, the real number v1⋅v2=v1⋅v2⋅cos(α), where
α=m(∢(v1,v2)) is called the scalar product of the vectors v1 i v2. If v1 sau v2 is zero,
then by definition the scalar product v1⋅v2 is zero.
i⋅i=1j⋅j=1
k⋅k=1
i⋅j=0i⋅k=0
j⋅k=0
Properties of the scalar product.
Let u,v,w∈V3.
u⋅v=v⋅u (commutativity)
u⋅v=∣u∣⋅Pruv (Pruv is the scalar projection of v onto u)
u⋅(v+w)=u⋅v+u⋅w (distributivity)
(α⋅u)⋅v=α(u⋅v)=u⋅(αv) (moving the scalar)
If u=x⋅i+y⋅j+z⋅k and v=r⋅i+s⋅j+t⋅k, then u⋅v=x⋅r+y⋅s+z⋅t
The scalar product is zero if and only if u⊥v or u=0 or v=0.
Theorem. Let u=x⋅i+y⋅j+z⋅k and v=x1⋅i+y1⋅j+z1⋅k be vectors, the points M(x,y,z), M1(x1,y1,z1), and
let α be the angle between the vectors u and v. Then cos(α)=∣v1∣⋅∣v2∣v1⋅v2=x2+y2+z2⋅x12+y12+z12x⋅x1+y⋅y1+z⋅z1
Definition. Let v=OM be the position vector of the point M. The angles
which the direction (line, vector) OM makes with the positive directions of
the coordinate axes Ox, Oy, Oz are called direction angles. The
cosines of these angles are called direction cosines.
Theorem. Let v be the position vector of the point M(x,y,z). The vector
v makes with the axes the direction angles α,β,γ with cos(α)=x2+y2+z2x, cos(β)=x2+y2+z2y,
cos(γ)=x2+y2+z2z.
cos2(α)+cos2(β)+cos2(γ)=1
If cos(α),cos(β),cos(γ) are the direction cosines of the line d and p≥0, then x⋅cos(α)+y⋅cos(β)+z⋅cos(γ)−p=0
is the normal equation of the plane perpendicular to the line d, at distance
p from the origin.
The equation of the line passing through the point M(xM,yM) with direction vector
u(α,β) is: y−yM=αβ(x−xM).
The equation of the line passing through the points A(xA,yA), B(xB,yB) is:
y−yA=xB−xAyB−yA(x−xA).
The distinct lines y=mx+n and y=m′x+n′ are parallel if and only if m=m′, and
perpendicular if and only if mm′=−1.