Definition
A vector AB is determined by three things:
- A direction (that of line (AB))
- A sense (from A to B)
- A magnitude (the length AB)
Two vectors with same direction, sense and magnitude are equal, even if they start from different points.
Coordinates in a frame
In a frame (O,i,jβ), if A(xAβ,yAβ) and B(xBβ,yBβ):
AB(xBββxAβyBββyAββ)
Watch out: βendpoint minus startβ, not the other way.
Operations
Addition (Chasles)
AB+BC=AC
In coordinates: add component-wise.
Scalar multiplication
If kβR and u=(x,y):
ku=(kx,ky)
- k>0: same sense, norm multiplied by β£kβ£.
- k<0: opposite sense.
- k=0: zero vector.
Norm of a vector
β₯uβ₯=x2+y2β
The vectorβs length.
Example: u(3,4) β β₯uβ₯=5.
Collinearity
Vectors u(x1β,y1β) and v(x2β,y2β) are collinear iff:
det(u,v)=x1βy2ββx2βy1β=0
Uses:
- 3 points A,B,C collinear β AB,AC collinear.
- Parallel lines β collinear direction vectors.
Dot product
uβ
v=xuβxvβ+yuβyvβ
Also:
uβ
v=β₯uβ₯β₯vβ₯cos(ΞΈ)
Key property (orthogonality)
uβ₯vβΊuβ
v=0
The perpendicularity test.
Midpoint
If I is the midpoint of [AB]:
I(2xAβ+xBββ,2yAβ+yBββ)
Vectorially: AI=IB.
Sample exercise
A(1,2), B(4,6), C(7,10).
Q1: Are they collinear?
AB=(3,4), AC=(6,8).
det=3Γ8β6Γ4=0. Yes.
Q2: Distance AB.
β₯ABβ₯=25β=5.
Q3: Midpoint M of [AC].
M(4,6). Notice: M=B! (points are collinear).
Common mistakes
- Confusing point and vector coordinates: AB=(xBββxAβ,yBββyAβ).
- Using slope to test collinearity: determinant is safer (no division by zero).
- Adding norms: β₯u+vβ₯ξ =β₯uβ₯+β₯vβ₯ generally.
Takeaways
- Vector coordinates: endpoint β start.
- β₯uβ₯=x2+y2β.
- Collinear β det=0.
- Orthogonal β dot product = 0.
- Chasles: AB+BC=AC.
- Midpoint: average of coordinates.
Going further