Vector geometry
Introduced| English | Meaning |
|---|---|
| collinear/ˈkɒlɪnɪə/ | Lying on the same straight line. |
| position vector/pəˈzɪʃn ˈvektə/ | The vector from the origin to a point, locating the point. |
| parallel/ˈpærəlel/ | Distinct lines in one plane that never meet; non-zero vectors are parallel when one is a scalar multiple of the other. |
| midpoint/ˈmɪdpɔɪnt/ | The point halfway between the endpoints of a line segment. |
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 $\overrightarrow{AB}$
- The position vector of a point is the vector from the origin $O$ to it.
- The vector from $A$ to $B$:
If $\mathbf{a} = \begin{pmatrix} 1 \\ 3 \end{pmatrix}$ and $\mathbf{b} = \begin{pmatrix} 4 \\ 7 \end{pmatrix}$, then $\overrightarrow{AB} = \begin{pmatrix} 4-1 \\ 7-3 \end{pmatrix} = \begin{pmatrix} 3 \\ 4 \end{pmatrix}$.
Vector geometry
resultant = a + b
Combine vectors to reach a point — the resultant is the direct route.
If a and b are the position vectors of A and B, then the vector AB is:
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AB = (position of B) − (position of A) = b − a.
A has position vector (1, 3) and B has position vector (4, 7). The top component of AB is?
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AB = b − a: top = 4 − 1 = 3.
Parallel 平行 vectors and collinear points
- Two vectors are parallel if one is a scalar multiple of the other: $\mathbf{p} = k\mathbf{q}$.
- Three points $A, B, C$ are collinear if $\overrightarrow{AB}$ and $\overrightarrow{BC}$ 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.
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Parallel vectors point the same (or opposite) way, so one is k times the other.
To prove three points are collinear, show the vectors between them are ______ and share a point.
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Collinear points have parallel vectors between them AND a shared point.
Midpoint 中点 of a segment
- $M$ is the midpoint of $AB$, where $\overrightarrow{OA} = \mathbf{a}$ and $\overrightarrow{OB} = \mathbf{b}$.
- 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:
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OM = a + ½(b − a) = ½(a + b).
A = (2, 4) and B = (8, 10). The midpoint M has x-coordinate?
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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.
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C = (8,14) − (1,3) = (7,11).
You've got it
- $\overrightarrow{AB} = \mathbf{b} - \mathbf{a}$ (end minus start)
- parallel vectors are scalar multiples; that's how you show points are collinear
- midpoint of $AB$: $\;\overrightarrow{OM} = \dfrac{1}{2}(\mathbf{a} + \mathbf{b})$
Write from a midpoint x-coordinate of 3
- A is (0, 0) and B is (6, 8).
- Find the x-coordinate of the midpoint.
A is (0, 0) and B is (6, 8). What is the x-coordinate of the midpoint?
Half of 6 is 3.
What is the midpoint x-coordinate?
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The halfway x-value is 3. Other options — 5 (the lesson midpoint x-coordinate); 6 (the x-coordinate of B).