Vectors

Vectors Part 1 (O level Maths)

Contents
1 Vectors (O level Maths)
2 Vector representation
2.1 Negative vectors
2.2 Magnitude
2.3 Scalar multiplication
2.4 Addition and Subtraction
3 Position vectors
4 Examples
4.1 Question 1
4.2 Solution 1
4.3 Question 2
4.4 Solution 2

Vectors (O level Maths)

This tutorial is about O level vectors for Zimsec and Cambridge students. First we are going to define a vector, then represent it on a Cartesian plane and finally calculate problems involving vectors.

Vector representation

A vector is a quantity that has both magnitude (size, length or modulus) an direction. On a Cartesian plane, a vector is represented as a arrow joining two points.

Vectors

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The diagram above shows a vector \vec{AB} which represents the displacement from point A(1; 2) to point B(5; 3).

A vector can be written in many ways:

  • either \bar{AB}, \vec{AB}, \bf{AB}
  • or \bar{a}, \vec{a}, \bf{a}

The values of a vector are usually given as a column matrix. On the diagram above, there are dashed lines showing the horizontal displacement (in the x-axis) and vertical displacement (in the y-axis) of the vector. From the diagram we can see from the dashed lines that x = 4 units and y = 1 unit. In column form, the vector is represented as:

\vec{AB} = \begin{pmatrix} 4\\ 1 \end{pmatrix}

Negative vectors

When dealing with vectors, direction is very important. For example, \bar{AB} and \bar{BA} are not the same because they are going in opposite directions. \bar{BA} is negative of \bar{AB}.

This means that if \vec{AB} = \begin{pmatrix} 4\\ 1 \end{pmatrix}

Then \vec{BA} = \begin{pmatrix} -4\\ -1 \end{pmatrix}

Which therefore means \vec{AB} = – \vec{AB}

Magnitude

The magnitude of a vector (also known as the modulus) is the length or size of the vector.

Magnitude of \vec{AB} is usually shortened to |\vec{AB}|.

Magnitude = \sqrt{x^2+y^2}

Using our earlier example, |\vec{AB}| = \sqrt{4^2+1^2}

= \sqrt{17} (in surd form)

The modulus of a vector is always positive.

Now let us find the modulus of \vec{BA}.

|\vec{BA}| = \sqrt{(-4)^2+(-1)^2}

= \sqrt{17}

Scalar multiplication

If vector \vec{AB} is multiplied a scalar k, whereby k is any number, the resulting vector is a vector k times as big as \vec{AB} and parallel to \vec{AB}.

For example, vector 5 \vec{AB} is 5 times as big as vector \vec{AB} and also parallel to vector \vec{AB}.

Addition and Subtraction

Vectors can be added and subtracted as follows:

If \vec{a} = \begin{pmatrix} 5\\ 2 \end{pmatrix} and \vec{b} = \begin{pmatrix} 1\\ 6 \end{pmatrix}

then \vec{a} + \vec{b}= \begin{pmatrix} 5\\ 2 \end{pmatrix} + \begin{pmatrix} 1\\ 6 \end{pmatrix}

= \begin{pmatrix} 6\\ 8 \end{pmatrix}

and \vec{a} + \vec{b}= \begin{pmatrix} 5\\ 2 \end{pmatrix} - \begin{pmatrix} 1\\ 6 \end{pmatrix}

= \begin{pmatrix} 4\\ -4 \end{pmatrix}

Position vectors

A position vector is a vector that is tied to the origin. Let us illustrate it using a diagram.

Position vectors

If we have a point A, the position vector of point A is \vec{OA}. By the same token, the position vector of point B is \vec{OB}.

Now if a point has the coordinates (x; y), its position vector is \begin{pmatrix} x\\ y \end{pmatrix}.

Therefore, using the diagram above we can see that:

  • the position vector of A = OA = \begin{pmatrix} 1\\ 2 \end{pmatrix} since A has the coordinates (1; 2)
  • the position vector of B = OB = \begin{pmatrix} 5\\ 3 \end{pmatrix} since B has the coordinates (5; 3)
  • \vec{b} = \vec{OA}
  • \vec{c} = \vec{OB}
  • \vec{a} = \vec{AB}

Now here is the thing about vectors, a displacement of \vec{AO} followed by \vec{OB} is equivalent to a resultant displacement of \vec{AB}.

Therefore \vec{AB} = \vec{AO} + \vec{OB}

\vec{AB} = \vec{AO} + \vec{OB}

\vec{AB} = \begin{pmatrix} -1\\ -2 \end{pmatrix} + \begin{pmatrix} 5\\ 3 \end{pmatrix}

\vec{AB} = \begin{pmatrix} 4\\ 1 \end{pmatrix}

Examples

Question 1

Given points A(7; 8) and B(2; 1), find:

a) \bf{AB}

b) \bf{BA}

Solution 1

To make it easy when solving vectors you first convert all the given points to position vectors: \bf{OA} = \begin{pmatrix} 7\\ 8 \end{pmatrix} and \bf{OB} = \begin{pmatrix} 2\\ -1 \end{pmatrix}

a) \bf{AB} = \bf{AO}+\bf{OB}

  • \bf{AB} = \begin{pmatrix} -7\\ -8 \end{pmatrix} + \begin{pmatrix} 2\\ -1 \end{pmatrix}

  • \bf{AB} = \begin{pmatrix} -5\\ -9 \end{pmatrix}

b) \bf{BA} = \bf{-AB}

  • \bf{AB} = -\begin{pmatrix} -5\\ -9 \end{pmatrix}

  • \bf{AB} = \begin{pmatrix} 5\\ 9 \end{pmatrix}

Question 2

The points O, P, Q, R, S have coordinates (0; 0), (1; 5), (3; 8), (7; 10), (10; 3) respectively. Express each of the following as a column vector.

a) \bf{OQ}

b) \bf{OS}

c) \bf{PQ}

d) \bf{QR}

e) \bf{QS}

f) \bf{RP}

Solution 2

First convert the points to position vectors

  • \bf{OP} = \begin{pmatrix} 1\\ 5 \end{pmatrix}
  • \bf{OQ} = \begin{pmatrix} 3\\ 8 \end{pmatrix}
  • \bf{OR} = \begin{pmatrix} 7\\ 10 \end{pmatrix}
  • \bf{OS} = \begin{pmatrix} 10\\ 3 \end{pmatrix}

a) \bf{OQ} = \begin{pmatrix} 3\\ 8 \end{pmatrix}

b) \bf{OS} = \begin{pmatrix} 10\\ 3 \end{pmatrix}

c) \bf{PQ} = \bf{PO}+\bf{OQ}

  • \bf{PQ} = \begin{pmatrix} -1\\ -5 \end{pmatrix} + \begin{pmatrix} 3\\ 8 \end{pmatrix}

  • \bf{PQ} = \begin{pmatrix} 2\\ 3 \end{pmatrix}

d) \bf{QR} = \bf{QO}+\bf{OR}

  • \bf{QR} = \begin{pmatrix} -3\\ -8 \end{pmatrix} + \begin{pmatrix} 7\\ 10 \end{pmatrix}

  • \bf{QR} = \begin{pmatrix} 4\\ 2 \end{pmatrix}

e) \bf{QS} = \bf{QO}+\bf{OS}

  • \bf{QS} = \begin{pmatrix} -3\\ -8 \end{pmatrix} + \begin{pmatrix} 10\\ 3 \end{pmatrix}

  • \bf{QS} = \begin{pmatrix} 7\\ -5 \end{pmatrix}

f) \bf{RP} = \bf{RO}+\bf{OP}

  • \bf{RP} = \begin{pmatrix} -7\\ -10 \end{pmatrix} + \begin{pmatrix} 1\\ 5 \end{pmatrix}

  • \bf{RP} = \begin{pmatrix} -6\\ -5 \end{pmatrix}