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Vectors - Higher

Vectors - Higher

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Tutor: Bilal

Summary

Vectors

In a nutshell

You need to know what a vector is, how to do vector arithmetic (i.e. adding, subtracting, multiplying) and also how to use vectors to construct geometric arguments and proofs.


Definitions: vector and scalar

A vector is a quantity with both magnitude and direction. For example, "5 metres north" is a vector.

A scalar is a quantity with magnitude only. For example, "5 metres" is a scalar as you don't know in which direction.



Representing vectors

There are different ways to represent vectors:


Column vectors

Vectors are commonly represented in column form. Column vectors are written like this: (ab)\begin{pmatrix}a\\b\end{pmatrix}. This means move aa​​​ steps right and bb​ steps up.

Bold lowercase letters

Bold lowercase letters are used as a quick way to refer to a vector. For example, you may see a\textbf{a}​​ = (11)\begin{pmatrix}1\\1\end{pmatrix}​​.

Uppercase letters with an arrow

The vector from the point AAto the point BB​ can be denoted as AB\overrightarrow{AB}​​​.

On a graph

On a graph, draw vectors with an arrow to point in the direction it's moving in.



Adding and subtracting vectors

To add and subtract column vectors, add and subtract each row individually.


Example 1


If a=(21)\textbf{a}=\begin{pmatrix}2\\1\end{pmatrix} and b=(63)\textbf{b}=\begin{pmatrix}-6\\3\end{pmatrix}, work out a+b\textbf{a}+\textbf{b} and ab\textbf{a}-\textbf{b}:

a+b=(21)+(63)=(2+(6)1+3)=(44)\textbf{a}+\textbf{b}=\begin{pmatrix}2\\1\end{pmatrix}+\begin{pmatrix}-6\\3\end{pmatrix}=\begin{pmatrix}2+(-6)\\1+3\end{pmatrix}=\begin{pmatrix}-4\\4\end{pmatrix}​​


a+b=(44)\underline{\textbf{a}+\textbf{b}=\begin{pmatrix}-4\\4\end{pmatrix}}​​


ab=(21)(63)=(2(6)13)=(82)\textbf{a}-\textbf{b}=\begin{pmatrix}2\\1\end{pmatrix}-\begin{pmatrix}-6\\3\end{pmatrix}=\begin{pmatrix}2-(-6)\\1-3\end{pmatrix}=\begin{pmatrix}8\\-2\end{pmatrix}​​


ab=(82)\underline{\textbf{a}-\textbf{b}=\begin{pmatrix}8\\-2\end{pmatrix}}​​



Multiplying a vector with a scalar

To multiply a vector with a scalar, you multiply each number in the vector by the scalar.


Example 2


If a=(41)\textbf{a}=\begin{pmatrix}4\\-1\end{pmatrix}, work out 3a3\textbf{a}, and 2a-2\textbf{a}:

3a=3(41)=(3×43×1)=(123)3\textbf{a}=3\begin{pmatrix}4\\-1\end{pmatrix}=\begin{pmatrix}3\times4\\3\times-1\end{pmatrix}=\begin{pmatrix}12\\-3\end{pmatrix}​​


3a=(123)\underline{3\textbf{a}=\begin{pmatrix}12\\-3\end{pmatrix}}​​


2a=2(41)=(2×42×1)=(82)-2\textbf{a}=-2\begin{pmatrix}4\\-1\end{pmatrix}=\begin{pmatrix}-2\times4\\-2\times-1\end{pmatrix}=\begin{pmatrix}-8\\2\end{pmatrix}​​


2a=(82)\underline{-2\textbf{a}=\begin{pmatrix}-8\\2\end{pmatrix}}​​



Vector geometry: arithmetic

Adding, subtracting, and multiplying vectors each have their own geometric interpretations.


Addition

Adding two vectors a\textbf{a}​ and b\textbf{b}​ gives the vector a+b\textbf{a}+\textbf{b}​, which represents travelling along the vector a\textbf{a}​ then along the vector b\textbf{b} in one journey.


Maths; Trigonometry; KS4 Year 10; Vectors - Higher

Multiplication by a positive scalar

Multiplying a vector a\textbf{a}​ by a positive scalar (number) kk gives the vector kak\textbf{a}​, which is a vector in the same direction as a\textbf{a}​, but its length is multiplied by a factor of kk​.

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Maths; Trigonometry; KS4 Year 10; Vectors - Higher
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Negative of a vector/multiplication by 1-1

The negative of a vector a\textbf{a}​ is denoted as (a-\textbf{a}), which means it is the same length as a\textbf{a}​, but going in the opposite direction.

Maths; Trigonometry; KS4 Year 10; Vectors - Higher

Subtraction

ab=a+(b)\textbf{a}-\textbf{b}=\textbf{a}+(-\textbf{b})​. So subtracting the vector b\textbf{b}​ from a\textbf{a}​ represents going along the vector a\textbf{a}​, and then going along the vector b-\textbf{b}.

Maths; Trigonometry; KS4 Year 10; Vectors - Higher


Example 3

In the diagram, find the vector AB\overrightarrow{AB}:


Maths; Trigonometry; KS4 Year 10; Vectors - Higher


To get from the point AA​ to the point BB​, You go from AA​ to OO​, then from OO​ to BB​. Mathematically:

AB=AO+OB\overrightarrow{AB}=\overrightarrow{AO}+\overrightarrow{OB}​​


From the diagram, OB=b\overrightarrow{OB}=\textbf{b}.

If OA=a\overrightarrow{OA}=\textbf{a}, then AO=a\overrightarrow{AO}=-\textbf{a} because it's going in the opposite direction.

Therefore:

 AB=AO+OB=(a)+b=ba\overrightarrow{AB}=\overrightarrow{AO}+\overrightarrow{OB}=(-\textbf{a})+\textbf{b}=\textbf{b}-\textbf{a}.


AB=ba\underline{\overrightarrow{AB}=\textbf{b}-\textbf{a}}​​



Vector geometry: parallel vectors and lines

Parallel vectors

Two vectors are parallel if one is a multiple of the other.


Example 4
  • (12)\begin{pmatrix}1\\2\end{pmatrix} is parallel to (36)\begin{pmatrix}3\\6\end{pmatrix} because (36)=3×(12)\begin{pmatrix}3\\6\end{pmatrix}=3\times\begin{pmatrix}1\\2\end{pmatrix}
  • a\textbf{a}is parallel to 32a-\frac{3}{2}\textbf{a}
  • a+2b\textbf{a}+2\textbf{b}is parallel to 0.5a+b0.5\textbf{a}+\textbf{b} because a+2b=2×(0.5a+b)\textbf{a}+2\textbf{b}=2\times(\textbf{0.5a}+\textbf{b})


Straight lines

Suppose there are three positions AA​, BB​ and CC​. Then, ABCABC​ is a straight line if ABAB​ is parallel to BCBC​ since BB​ is a shared point.


Example 5

The vector AB\overrightarrow{AB} is given to be 2a3b2\textbf{a}-3\textbf{b}. Similarly, the vector BC\overrightarrow{BC} = 9b6a9\textbf{b}-6\textbf{a}. Prove that ABC is a straight line:


This is equivalent to showing that AB\overrightarrow{AB} is parallel to BC\overrightarrow{BC}. To do this, show that one is a multiple of another:

BC=9b6a=3(3b2a)=3(2a+3b)=3(2a3b)=3AB\overrightarrow{BC}=9\textbf{b}-6\textbf{a}=3(3\textbf{b}-2\textbf{a})=3(-2\textbf{a}+3\textbf{b})=-3(2\textbf{a}-3\textbf{b})=-3\overrightarrow{AB}


BC=3×AB\therefore \overrightarrow{BC}=-3\times\overrightarrow{AB}​​​

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Always conclude your proof with a sentence that explains what you have done and how it answers the question:


Hence, the lines AB\underline{AB} and BC\underline{BC} are parallel. But, since B\underline{B}​ is a common point, it can be deduced that ABC\underline{ABC}​ is indeed a straight line.​



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FAQs - Frequently Asked Questions

How do you show that two vectors are parallel?

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What is a vector?

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