Vector geometry · Géométrie vectorielle
| English | Français |
|---|---|
| collinear/ˈkɒlɪnɪə/ | aligné |
| position vector/pəˈzɪʃn ˈvektə/ | vecteur position |
| parallel/ˈpærəlel/ | parallèle |
| midpoint/ˈmɪdpɔɪnt/ | point milieu |
Proving points line up
- How do you prove that three points are collinear 共线 (lie on the same line)?
- Show that the vector from the first to the second is a scalar multiple of the vector from the second to the third.
Position vectors 位置向量 and · et $\overrightarrow{AB}$
- The · Le position vector of a point is the vector from the origin $O$ to it.
- The vector from $A$ to · à $B$:
If $\mathbf{a} = \begin{pmatrix} 1 \\ 3 \end{pmatrix}$ and · et $\mathbf{b} = \begin{pmatrix} 4 \\ 7 \end{pmatrix}$, then $\overrightarrow{AB} = \begin{pmatrix} 4-1 \\ 7-3 \end{pmatrix} = \begin{pmatrix} 3 \\ 4 \end{pmatrix}$.
Vector geometry · Géométrie vectorielle
resultant = a + b · résultante = a + b
Combine vectors to reach a point — the resultant is the direct route. · Combinez des vecteurs pour atteindre un point — la résultante est le trajet direct.
If a and b are the position vectors of A and B, then the vector AB is: · Si a et b sont les vecteurs de position de A et B, alors le vecteur AB est :
AB = (position of B) − (position of A) = b − a. · AB = (position de B) − (position de A) = b − a.
A has position vector (1, 3) and B has position vector (4, 7). The top component of AB is? · A a un vecteur de position (1, 3) et B a un vecteur de position (4, 7). La composante haute de AB est ?
AB = b − a: top = 4 − 1 = 3. · AB = b − a : haut = 4 − 1 = 3.
Parallel · Parallèle 平行 vectors and collinear points
- Two vectors are parallel · parallèle if one is a scalar multiple of the other: $\mathbf{p} = k\mathbf{q}$.
- Three points · Trois points $A, B, C$ are collinear if · si $\overrightarrow{AB}$ and · et $\overrightarrow{BC}$ are parallel and share the point $B$.
Parallel alone isn't enough. Two vectors can be parallel without the points being collinear — they could be on separate parallel lines. You need a shared point too.
Two vectors are parallel if one is a scalar multiple of the other. · Deux vecteurs sont parallèles si l'un est un multiple scalaire de l'autre.
Parallel vectors point the same (or opposite) way, so one is k times the other. · Les vecteurs parallèles pointent dans la même (ou direction opposée), donc l'un est k fois l'autre.
To prove three points are collinear, show the vectors between them are ______ and share a point. · Pour prouver que trois points sont alignés, montrez que les vecteurs entre eux sont ______ et partagent un point.
Collinear points have parallel vectors between them AND a shared point. · Les points alignés ont des vecteurs parallèles entre eux ET un point partagé.
Midpoint 中点 of a segment
- $M$ is the midpoint of $AB$, where $\overrightarrow{OA} = \mathbf{a}$ and · et $\overrightarrow{OB} = \mathbf{b}$.
- The midpoint's position vector is the average of the two endpoints.
The magnitude of a column vector is found with Pythagoras
M is the midpoint of AB. The position vector OM is: · M est le milieu de AB. Le vecteur de position OM est :
OM = a + ½(b − a) = ½(a + b).
A = (2, 4) and B = (8, 10). The midpoint M has x-coordinate? · A = (2, 4) et B = (8, 10). Quelle est la coordonnée x du milieu M ?
½(2 + 8) = 5.
Why vectors matter
- Video games, physics engines, and GPS all use vector mathematics to track positions and movements.
Position vectors locate points on the grid — the foundation for proving collinearity and finding midpoints.
A complete vector proof
- If $\overrightarrow{OA}=\mathbf a$, $\overrightarrow{OB}=\mathbf b$, and M is the midpoint of AB, then $\overrightarrow{OM}=(\mathbf a+\mathbf b)/2$.
- Define C by $\overrightarrow{OC}=2\mathbf b-\mathbf a$. Then $\overrightarrow{BC}=\overrightarrow{OC}-\overrightarrow{OB}=\mathbf b-\mathbf a=\overrightarrow{AB}$. Equal vectors share the direction and meet at B, so A, B, C are collinear and B is the midpoint of AC (for distinct A and B).
A = (1,3), B = (4,7) and OC = 2OB − OA. Find the x-coordinate of C. · A = (1,3), B = (4,7) et OC = 2OB − OA. Trouver l'abscisse de C.
C = (8,14) − (1,3) = (7,11).
You've got it
- $\overrightarrow{AB} = \mathbf{b} - \mathbf{a}$ (end minus start)
- parallel · parallèle vectors are scalar multiples; that's how you show points are collinear
- midpoint of $AB$: $\;\overrightarrow{OM} = \dfrac{1}{2}(\mathbf{a} + \mathbf{b})$