8 Vectors and transformation geometry
- Syllabus
- 2016
- Topic
- 8
- Level
- —
A scalar has magnitude only. A vector has both magnitude and direction; in this syllabus vectors are two-dimensional.
| Scalar | Vector |
|---|---|
| mass, time, temperature | displacement, velocity, force |
| distance | displacement |
| speed | velocity |
Ask whether changing direction while keeping the numerical size would change the quantity. If yes, the quantity is vector-valued.
A negative scalar is still a scalar. A vector is not identified merely by having two numbers: the numbers must encode directed components.
\overrightarrow{OA}=\mathbf a=\begin{pmatrix}a_x\a_y\end{pmatrix}
The top component is horizontal: positive right, negative left. The bottom component is vertical: positive up, negative down.
\overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}=\mathbf b-\mathbf a
Order matters: AB=−BA. Do not swap the two components or omit the direction arrow when naming a vector between points.
A directed line segment shows a vector by its length and arrow direction. Its starting position is irrelevant: equal vectors may be drawn in different places.
A(x_1,y_1),\ B(x_2,y_2)\Rightarrow\overrightarrow{AB}=\begin{pmatrix}x_2-x_1\y_2-y_1\end{pmatrix}
From any chosen start, move by the horizontal component, then the vertical component, and place the arrowhead at the endpoint.
A segment without an arrow has no specified direction. Parallel segments of different lengths are not equal vectors, and reversing an arrow changes the sign.
| Type | Test |
|---|---|
| parallel vectors | one is a scalar multiple of the other |
| unit vector | magnitude is 1 |
| position vector of P | vector OP from the origin |
\mathbf b=k\mathbf a\Rightarrow \mathbf a\parallel\mathbf b\quad(k<0\text{ gives opposite directions})
\widehat{\mathbf a}=\frac{\mathbf a}{|\mathbf a|}\quad(\mathbf a\ne\mathbf0)
Parallel does not mean equal: the multiplier may change length or reverse direction. The zero vector has no unit direction and cannot be normalised.
\begin{pmatrix}a\b\end{pmatrix}+\begin{pmatrix}c\d\end{pmatrix}=\begin{pmatrix}a+c\b+d\end{pmatrix}
For addition, place vectors head-to-tail; the resultant joins the first tail to the last head. Subtracting b means adding the reversed vector −b.
\overrightarrow{AB}+\overrightarrow{BC}=\overrightarrow{AC}
Only combine matching horizontal and vertical components. Vector subtraction is not commutative, and a valid path must respect each arrow's direction.
\left|\begin{pmatrix}x\y\end{pmatrix}\right|=\sqrt{x^2+y^2}
Magnitude is the non-negative length of the vector. It follows from Pythagoras applied to the horizontal and vertical components.
\left|\begin{pmatrix}-6\8\end{pmatrix}\right|=\sqrt{36+64}=10
Square each signed component before adding and give a non-negative result. ∣a∣ is a scalar, not a column vector.
k\begin{pmatrix}a\b\end{pmatrix}=\begin{pmatrix}ka\kb\end{pmatrix}
| Multiplier k | Effect |
|---|---|
| k>1 | same direction, longer |
| 0<k<1 | same direction, shorter |
| k<0 | reverse direction, length scaled by ∣k∣ |
| k=0 | zero vector |
|k\mathbf a|=|k|,|\mathbf a|
Multiply every component. A negative multiplier reverses direction; it does not create a negative magnitude.
The resultant is the single vector with the same overall effect as all the given vectors applied in sequence.
\mathbf r=\mathbf v_1+\mathbf v_2+\cdots+\mathbf v_n
Choose a consistent positive direction, convert every vector to components, add components, then state the result as a directed vector. A closed head-to-tail path has resultant 0.
Do not add magnitudes unless all vectors lie on the same line in the same direction. Opposing components must carry opposite signs.
| Goal | Vector evidence |
|---|---|
| collinear points | connecting vectors are scalar multiples |
| parallel lines | direction vectors are scalar multiples |
| same midpoint | position-vector averages are equal |
| concurrency | independently derived position vectors give the same point |
\overrightarrow{OP}=\overrightarrow{OA}+\lambda\overrightarrow{AB}
Express every route from a common origin, simplify to the same vector basis, and compare coefficients or scalar multiples. State the geometric conclusion after the algebra.
Equal-looking coefficients are meaningful only in a consistent vector basis. A scalar-multiple result proves parallelism; extra point-sharing or position information is needed to conclude collinearity.
| Transformation | Complete description |
|---|---|
| reflection | mirror line |
| rotation | centre, angle, direction |
| translation | column vector |
| enlargement | centre and scale factor |
Transform every vertex, preserve vertex order, and join the images. Use perpendicular equal distances for reflections, constant centre-distance for rotations, and centre-to-point rays for enlargements.
\begin{pmatrix}x\y\end{pmatrix}\mapsto\begin{pmatrix}x+a\y+b\end{pmatrix}
A transformation is not fully described by its type alone. For negative enlargement factors the image lies on the opposite ray from the centre; do not infer a centre or mirror line by appearance alone.
Apply transformations in the stated order and use the first image as the input to the second. The order can change the final result.
P\xmapsto{T_1}P'\xmapsto{T_2}P''
Track one labelled vertex at a time, record intermediate coordinates, then transform the remaining vertices consistently. To describe a combined result, compare the original and final shapes only after the chain is complete.
Do not collapse two transformations by guessing from the picture. Rotations, reflections and translations generally do not commute, so reversing the order may produce another image.
\begin{pmatrix}a&b\c&d\end{pmatrix}\begin{pmatrix}x\y\end{pmatrix}=\begin{pmatrix}ax+by\cx+dy\end{pmatrix}
The columns are the images of the basis vectors: transform (1,0)T to get column 1 and (0,1)T to get column 2.
M\begin{pmatrix}1\0\end{pmatrix}=\begin{pmatrix}a\c\end{pmatrix},\qquad M\begin{pmatrix}0\1\end{pmatrix}=\begin{pmatrix}b\d\end{pmatrix}
A 2imes2 matrix here represents only transformations that leave the origin fixed. Ordinary translations cannot be encoded by such a matrix; keep point vectors as (x,y)T.