Pure mathematics

Vectors: 2D, 3D, Magnitude, Direction and Position Vectors

Year 12 · Year 13

  • By the end of this lesson students will be able to understand and use vector notation in 2D and 3D.
  • By the end of this lesson students will be able to perform vector addition, subtraction, and scalar multiplication.
  • By the end of this lesson students will be able to calculate the magnitude of a vector in 2D and 3D.
  • By the end of this lesson students will be able to find a unit vector in the direction of a given vector.
  • By the end of this lesson students will be able to use position vectors to solve geometrical problems, including collinearity and parallel vectors.

Key concepts

What is a Vector?

A vector is a quantity that has both magnitude (size) and direction. This contrasts with a scalar, which has only magnitude. Vectors are often represented geometrically by an arrow, where the length of the arrow represents the magnitude and the arrowhead indicates the direction. In coordinate form, a 2D vector can be written as a column vector (xy)\begin{pmatrix} x \\ y \end{pmatrix} or using unit vectors i\mathbf{i} and j\mathbf{j} as xi+yjx\mathbf{i} + y\mathbf{j}. A 3D vector is written as (xyz)\begin{pmatrix} x \\ y \\ z \end{pmatrix} or xi+yj+zkx\mathbf{i} + y\mathbf{j} + z\mathbf{k}, where i\mathbf{i}, j\mathbf{j}, and k\mathbf{k} are unit vectors in the direction of the positive x, y, and z axes respectively.

Vector Operations

Vectors can be added, subtracted, and multiplied by a scalar. To add or subtract vectors, you add or subtract their corresponding components. For example, if a=(a1a2a3)\mathbf{a} = \begin{pmatrix} a_1 \\ a_2 \\ a_3 \end{pmatrix} and b=(b1b2b3)\mathbf{b} = \begin{pmatrix} b_1 \\ b_2 \\ b_3 \end{pmatrix}, then a+b=(a1+b1a2+b2a3+b3)\mathbf{a} + \mathbf{b} = \begin{pmatrix} a_1+b_1 \\ a_2+b_2 \\ a_3+b_3 \end{pmatrix} and ab=(a1b1a2b2a3b3)\mathbf{a} - \mathbf{b} = \begin{pmatrix} a_1-b_1 \\ a_2-b_2 \\ a_3-b_3 \end{pmatrix}. To multiply a vector by a scalar kk, you multiply each component by kk. So, ka=(ka1ka2ka3)k\mathbf{a} = \begin{pmatrix} ka_1 \\ ka_2 \\ ka_3 \end{pmatrix}.

Magnitude of a Vector

The magnitude of a vector, also known as its modulus or length, is a scalar quantity representing its size. For a 2D vector a=(xy)\mathbf{a} = \begin{pmatrix} x \\ y \end{pmatrix}, its magnitude is found using Pythagoras' theorem. For a 3D vector a=(xyz)\mathbf{a} = \begin{pmatrix} x \\ y \\ z \end{pmatrix}, the magnitude extends this concept.

a=x2+y2|\mathbf{a}| = \sqrt{x^2 + y^2} (for 2D) or a=x2+y2+z2|\mathbf{a}| = \sqrt{x^2 + y^2 + z^2} (for 3D)
Unit Vectors and Direction

A unit vector is a vector with a magnitude of 1. A unit vector in the same direction as a given vector a\mathbf{a} is denoted by a^\hat{\mathbf{a}} (read as 'a-hat'). It is found by dividing the vector by its magnitude. This process normalises the vector. The unit vector gives the direction of the original vector without regard to its magnitude.

a^=1aa\hat{\mathbf{a}} = \frac{1}{|\mathbf{a}|}\mathbf{a}
Position Vectors and Geometry

A position vector describes the position of a point in space relative to a fixed origin, O. If A is a point, its position vector is OA\vec{OA}, often denoted by a lowercase bold letter, a\mathbf{a}. If we have two points A and B with position vectors a\mathbf{a} and b\mathbf{b} respectively, the displacement vector from A to B is given by AB=ba\vec{AB} = \mathbf{b} - \mathbf{a}. This concept is fundamental for solving geometrical problems using vectors, such as determining if points are collinear (lie on the same straight line) or if lines are parallel.

AB=ba\vec{AB} = \mathbf{b} - \mathbf{a}

Key facts to remember

  • 1A vector has both magnitude and direction, whereas a scalar has only magnitude.
  • 2Vectors can be represented as column vectors (xy)\begin{pmatrix} x \\ y \end{pmatrix} or (xyz)\begin{pmatrix} x \\ y \\ z \end{pmatrix}, or using i,j,k\mathbf{i}, \mathbf{j}, \mathbf{k} notation.
  • 3The magnitude of a vector a=xi+yj+zk\mathbf{a} = x\mathbf{i} + y\mathbf{j} + z\mathbf{k} is a=x2+y2+z2|\mathbf{a}| = \sqrt{x^2 + y^2 + z^2}.
  • 4A unit vector in the direction of a\mathbf{a} is a^=1aa\hat{\mathbf{a}} = \frac{1}{|\mathbf{a}|}\mathbf{a}.
  • 5The position vector of a point A, denoted a\mathbf{a} or OA\vec{OA}, is the vector from the origin O to A.
  • 6The displacement vector from point A to point B is given by AB=ba\vec{AB} = \mathbf{b} - \mathbf{a}, where a\mathbf{a} and b\mathbf{b} are the position vectors of A and B respectively.
  • 7Two vectors u\mathbf{u} and v\mathbf{v} are parallel if u=kv\mathbf{u} = k\mathbf{v} for some scalar kk.
  • 8Three points A, B, C are collinear if AB=kBC\vec{AB} = k\vec{BC} (or AC=kAB\vec{AC} = k\vec{AB}) for some scalar kk, and they share a common point.

Worked examples

Example 1

Given vectors a=(32)\mathbf{a} = \begin{pmatrix} 3 \\ -2 \end{pmatrix} and b=(15)\mathbf{b} = \begin{pmatrix} -1 \\ 5 \end{pmatrix}, find: (a) a+2b\mathbf{a} + 2\mathbf{b} (b) ab|\mathbf{a} - \mathbf{b}| (c) A unit vector in the direction of a\mathbf{a}.

I(a) First, calculate 2b2\mathbf{b}: 2b=2(15)=(2×12×5)=(210)2\mathbf{b} = 2\begin{pmatrix} -1 \\ 5 \end{pmatrix} = \begin{pmatrix} 2 \times -1 \\ 2 \times 5 \end{pmatrix} = \begin{pmatrix} -2 \\ 10 \end{pmatrix}.
IINow, add a\mathbf{a} to 2b2\mathbf{b}: a+2b=(32)+(210)=(3+(2)2+10)=(18)\mathbf{a} + 2\mathbf{b} = \begin{pmatrix} 3 \\ -2 \end{pmatrix} + \begin{pmatrix} -2 \\ 10 \end{pmatrix} = \begin{pmatrix} 3 + (-2) \\ -2 + 10 \end{pmatrix} = \begin{pmatrix} 1 \\ 8 \end{pmatrix}.
III(b) First, calculate ab\mathbf{a} - \mathbf{b}: ab=(32)(15)=(3(1)25)=(47)\mathbf{a} - \mathbf{b} = \begin{pmatrix} 3 \\ -2 \end{pmatrix} - \begin{pmatrix} -1 \\ 5 \end{pmatrix} = \begin{pmatrix} 3 - (-1) \\ -2 - 5 \end{pmatrix} = \begin{pmatrix} 4 \\ -7 \end{pmatrix}.
IVNow, find the magnitude of ab\mathbf{a} - \mathbf{b}: ab=42+(7)2=16+49=65|\mathbf{a} - \mathbf{b}| = \sqrt{4^2 + (-7)^2} = \sqrt{16 + 49} = \sqrt{65}.
V(c) First, find the magnitude of a\mathbf{a}: a=32+(2)2=9+4=13|\mathbf{a}| = \sqrt{3^2 + (-2)^2} = \sqrt{9 + 4} = \sqrt{13}.
VINow, divide a\mathbf{a} by its magnitude to find the unit vector: a^=113(32)=(313213)\hat{\mathbf{a}} = \frac{1}{\sqrt{13}}\begin{pmatrix} 3 \\ -2 \end{pmatrix} = \begin{pmatrix} \frac{3}{\sqrt{13}} \\ \frac{-2}{\sqrt{13}} \end{pmatrix}.

Answer

(a) (18)\begin{pmatrix} 1 \\ 8 \end{pmatrix} (b) 65\sqrt{65} (c) (313213)\begin{pmatrix} \frac{3}{\sqrt{13}} \\ \frac{-2}{\sqrt{13}} \end{pmatrix}

Remember to rationalise the denominator if specifically asked, but for vectors, leaving it as 113\frac{1}{\sqrt{13}} or in the component form shown is usually acceptable.

Example 2

A vector v\mathbf{v} is given by 2i3j+6k2\mathbf{i} - 3\mathbf{j} + 6\mathbf{k}. (a) Find the magnitude of v\mathbf{v}. (b) Find a unit vector in the direction of v\mathbf{v}.

I(a) The vector v\mathbf{v} can be written as (236)\begin{pmatrix} 2 \\ -3 \\ 6 \end{pmatrix}.
IIThe magnitude of v\mathbf{v} is v=22+(3)2+62|\mathbf{v}| = \sqrt{2^2 + (-3)^2 + 6^2}.
IIIv=4+9+36=49=7|\mathbf{v}| = \sqrt{4 + 9 + 36} = \sqrt{49} = 7.
IV(b) A unit vector in the direction of v\mathbf{v} is v^=1vv\hat{\mathbf{v}} = \frac{1}{|\mathbf{v}|}\mathbf{v}.
VSubstitute the magnitude found in part (a): v^=17(2i3j+6k)\hat{\mathbf{v}} = \frac{1}{7}(2\mathbf{i} - 3\mathbf{j} + 6\mathbf{k}).
VIThis can also be written as 27i37j+67k\frac{2}{7}\mathbf{i} - \frac{3}{7}\mathbf{j} + \frac{6}{7}\mathbf{k} or (273767)\begin{pmatrix} \frac{2}{7} \\ -\frac{3}{7} \\ \frac{6}{7} \end{pmatrix}.

Answer

(a) 77 (b) 17(2i3j+6k)\frac{1}{7}(2\mathbf{i} - 3\mathbf{j} + 6\mathbf{k}) or 27i37j+67k\frac{2}{7}\mathbf{i} - \frac{3}{7}\mathbf{j} + \frac{6}{7}\mathbf{k}

Always ensure the final unit vector has a magnitude of 1 as a quick check.

Example 3

The position vectors of points A, B and C are a=2i+jk\mathbf{a} = 2\mathbf{i} + \mathbf{j} - \mathbf{k}, b=3i2j+2k\mathbf{b} = 3\mathbf{i} - 2\mathbf{j} + 2\mathbf{k} and c=5i8j+8k\mathbf{c} = 5\mathbf{i} - 8\mathbf{j} + 8\mathbf{k} respectively. Show that A, B and C are collinear.

ITo show that A, B and C are collinear, we need to show that AB\vec{AB} is parallel to BC\vec{BC} (or AC\vec{AC}) and that they share a common point (which B does).
IIFirst, find the displacement vector AB\vec{AB}: AB=ba\vec{AB} = \mathbf{b} - \mathbf{a}.
IIIAB=(3i2j+2k)(2i+jk)\vec{AB} = (3\mathbf{i} - 2\mathbf{j} + 2\mathbf{k}) - (2\mathbf{i} + \mathbf{j} - \mathbf{k})
IVAB=(32)i+(21)j+(2(1))k=i3j+3k\vec{AB} = (3-2)\mathbf{i} + (-2-1)\mathbf{j} + (2-(-1))\mathbf{k} = \mathbf{i} - 3\mathbf{j} + 3\mathbf{k}.
VNext, find the displacement vector BC\vec{BC}: BC=cb\vec{BC} = \mathbf{c} - \mathbf{b}.
VIBC=(5i8j+8k)(3i2j+2k)\vec{BC} = (5\mathbf{i} - 8\mathbf{j} + 8\mathbf{k}) - (3\mathbf{i} - 2\mathbf{j} + 2\mathbf{k})
VIIBC=(53)i+(8(2))j+(82)k=2i6j+6k\vec{BC} = (5-3)\mathbf{i} + (-8-(-2))\mathbf{j} + (8-2)\mathbf{k} = 2\mathbf{i} - 6\mathbf{j} + 6\mathbf{k}.
VIIINow, compare AB\vec{AB} and BC\vec{BC}. We can see that BC=2(i3j+3k)\vec{BC} = 2(\mathbf{i} - 3\mathbf{j} + 3\mathbf{k}).
9Therefore, BC=2AB\vec{BC} = 2\vec{AB}.
10Since BC\vec{BC} is a scalar multiple of AB\vec{AB}, they are parallel. As they also share a common point B, the points A, B and C must be collinear.

Answer

Since BC=2AB\vec{BC} = 2\vec{AB} and B is a common point, A, B and C are collinear.

You could also calculate AC\vec{AC} and show it's a scalar multiple of AB\vec{AB} or BC\vec{BC}.

Common mistakes

  • Confusing the position vector of a point with its coordinates. While numerically similar, they represent different mathematical concepts.
  • Incorrectly calculating the magnitude, often by forgetting to square root the sum of squares, or by squaring the sum of components.
  • Calculating the displacement vector AB\vec{AB} as ab\mathbf{a} - \mathbf{b} instead of the correct ba\mathbf{b} - \mathbf{a}.
  • Making arithmetic errors, especially with negative signs, during vector addition, subtraction, or scalar multiplication.
  • Assuming vectors are parallel if only one component is proportional, rather than all corresponding components.

Exam tips

  • Always draw a clear diagram for geometry problems involving vectors; it can help visualise the problem and identify relationships.
  • Show all steps in your vector calculations, especially for magnitude and unit vectors, to gain full marks.
  • Pay close attention to positive and negative signs when performing vector operations, as a single sign error can lead to an incorrect answer.
  • Ensure your final answer is in the required format (e.g., column vector, i,j,k\mathbf{i}, \mathbf{j}, \mathbf{k} form, or a scalar magnitude) as specified in the question.

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