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13. Vectors in two dimensions

Syllabus
0606–2028–2029
Topic
13
Level
—

Learning objectives

13.1Use vector notation• Understand and use vector notation in forms including column vectors, directed line segments, p, and ai - bj.• Use correct vector notation.13.2Use position and unit vectors• Know and use position vectors and unit vectors.• Find the unit vector in the same direction as a given vector.13.3Calculate and combine vectors• Find the magnitude of a vector.• Add and subtract vectors and multiply vectors by scalars.• Includes equating like vectors and solving vector-geometry problems, with diagrams given in more complex cases.13.4Compose and resolve velocities• Compose and resolve velocities.• Determine a resultant vector by adding two or more vectors.• Use a velocity vector to determine position and solve contextual problems such as particles colliding.

Read and write vector notation precisely

A vector records both magnitude and direction. Different notations can describe the same vector, but the notation must show that the quantity is a vector rather than an ordinary number.

Form What it identifies
a\mathbf{a} or p\mathbf{p} a named vector
AB→\overrightarrow{AB} the directed displacement from AA to BB
(xy)\begin{pmatrix}x\\y\end{pmatrix} horizontal and vertical components
xi+yjx\mathbf{i}+y\mathbf{j} the same components in unit-vector form

If A=(2,−1)A=(2,-1) and B=(7,3)B=(7,3), then AB→=(7−23−(−1))=(54)=5i+4j\overrightarrow{AB}=\begin{pmatrix}7-2\\3-(-1)\end{pmatrix}=\begin{pmatrix}5\\4\end{pmatrix}=5\mathbf{i}+4\mathbf{j}. These are three descriptions of the same directed displacement.

AB→\overrightarrow{AB} and BA→\overrightarrow{BA} point in opposite directions, so BA→=−AB→\overrightarrow{BA}=-\overrightarrow{AB}. By contrast, ABAB without an arrow normally denotes the scalar distance between the points. Do not drop bold type or the arrow when a vector is required.

Use position vectors and normalize direction

A position vector locates a point from a fixed origin. If R=(x,y)R=(x,y) relative to origin OO, then OR→=(xy)\overrightarrow{OR}=\begin{pmatrix}x\\y\end{pmatrix}. A unit vector keeps a direction but has magnitude 11.

|\mathbf{a}|=\sqrt{x^2+y^2},\qquad \widehat{\mathbf{a}}=\frac{\mathbf{a}}{|\mathbf{a}|}\quad(\mathbf{a}\ne\mathbf{0})

Find the magnitude first, then divide every component by that same positive magnitude. This scales the vector without changing its direction.

For a=(5−12)\mathbf{a}=\begin{pmatrix}5\\-12\end{pmatrix}, ∣a∣=52+(−12)2=13|\mathbf{a}|=\sqrt{5^2+(-12)^2}=13. The unit vector in the same direction is 113(5−12)=(5/13−12/13)\frac1{13}\begin{pmatrix}5\\-12\end{pmatrix}=\begin{pmatrix}5/13\\-12/13\end{pmatrix}. Its magnitude checks as 11.

The zero vector has no direction, so it cannot be normalized. Dividing by a negative number would reverse the direction; the denominator in the unit-vector formula is the non-negative magnitude ∣a∣|\mathbf a|.

Calculate with vectors and prove geometry

Vector calculations are component-wise: combine horizontal components together and vertical components together. Equality of two vectors therefore gives one equation from each component.

Operation Component rule
addition (ab)+(cd)=(a+cb+d)\begin{pmatrix}a\\b\end{pmatrix}+\begin{pmatrix}c\\d\end{pmatrix}=\begin{pmatrix}a+c\\b+d\end{pmatrix}
subtraction (ab)−(cd)=(a−cb−d)\begin{pmatrix}a\\b\end{pmatrix}-\begin{pmatrix}c\\d\end{pmatrix}=\begin{pmatrix}a-c\\b-d\end{pmatrix}
scalar multiplication k(ab)=(kakb)k\begin{pmatrix}a\\b\end{pmatrix}=\begin{pmatrix}ka\\kb\end{pmatrix}
magnitude ∣(ab)∣=a2+b2\left|\begin{pmatrix}a\\b\end{pmatrix}\right|=\sqrt{a^2+b^2}

For example, (41)+k(−23)=r(−105)\begin{pmatrix}4\\1\end{pmatrix}+k\begin{pmatrix}-2\\3\end{pmatrix}=r\begin{pmatrix}-10\\5\end{pmatrix} gives 4−2k=−10r4-2k=-10r and 1+3k=5r1+3k=5r. Solving the two scalar equations gives k=−3/2k=-3/2 and r=−7/10r=-7/10.

In vector geometry, build a route using directed segments. If OA→=a\overrightarrow{OA}=\mathbf a, OB→=b\overrightarrow{OB}=\mathbf b and PP divides ABAB with AP:PB=1:3AP:PB=1:3, then AB→=b−a\overrightarrow{AB}=\mathbf b-\mathbf a and OP→=a+14(b−a)=34a+14b\overrightarrow{OP}=\mathbf a+\frac14(\mathbf b-\mathbf a)=\frac34\mathbf a+\frac14\mathbf b.

To prove points are collinear, show that two directed vectors with a common point are scalar multiples. A positive scalar gives the same direction and a negative scalar gives the opposite direction, but both lie on the same straight line. Equal magnitudes alone do not prove parallelism or collinearity.

Resolve velocity and model moving positions

A velocity vector combines speed with direction. Resolve every velocity into components on the same axes; resultant velocities are then found by component-wise addition, and position changes linearly when velocity is constant.

\mathbf p(t)=\mathbf p_0+t\mathbf v

With i\mathbf i east and j\mathbf j north, a speed ss on bearing θ\theta has velocity ssin⁡θ i+scos⁡θ js\sin\theta\,\mathbf i+s\cos\theta\,\mathbf j, because a bearing is measured clockwise from north. If an angle α\alpha is measured above the positive horizontal axis instead, use scos⁡α i+ssin⁡α js\cos\alpha\,\mathbf i+s\sin\alpha\,\mathbf j.

A cyclist travelling at 4 m s−14\,\mathrm{m\,s^{-1}} on bearing 015∘015^\circ has velocity 4sin⁡15∘ i+4cos⁡15∘ j4\sin15^\circ\,\mathbf i+4\cos15^\circ\,\mathbf j. If another velocity acts at the same time, add its i\mathbf i and j\mathbf j components separately to obtain the resultant.

For two particles, write both position vectors using the same time variable. They collide only if both component equations give the same admissible time. For instance, if the xx-components meet at one time but the yy-components meet at another, the paths may cross geometrically but the particles do not collide.

Keep displacement, velocity and speed distinct: displacement and velocity are vectors, while speed is the scalar magnitude of velocity. Include the initial position p0\mathbf p_0 unless the particle starts at the origin.