Modulus of a Vector (Edexcel IGCSE Maths)

Modulus (magnitude) of a Vector

Spec point — 8F

How do I find the modulus of a vector?

• The modulus of a vector is its length(distance)

  • It is sometimes also called the magnitude
  • This is always a positive value
  • The direction of the vector is irrelevant

• The modulus of $\overset{\rightarrow}{AB}$ is written $\left| \vec{AB} \right|$

  • The modulus of a is written |a|

• Depending on the use of the vector, the modulus of a vector represents different quantities

  • For velocity, modulus would be speed
  • For a force, modulus would be the strength of the force (in Newtons)

• In component form, the modulus is the hypotenuse of a right-angled triangle

  • Use Pythagoras' theorem to find the modulus

  • The modulus of $a = ( \begin{gathered} x \\ y \end{gathered} )$

    • is $\left| a \right| = \sqrt{x^{2 } + \mathrm{y}^{2}}$

Exam Hint If there is no diagram, sketch one!

You can sketch a vector and use it to form a right-angled triangle.

Worked Example Consider two points $A ( - 3 , 5 )$ and $B ( 7 , 1 )$.

(a) Write down the column vector $\overset{\rightarrow}{AB}$.

Answer:

Find the horizontal and vertical distances between the two points Subtract the x and y components of A from B

$\vec{AB} = ( \begin{gathered} 7 - - 3 \\ 1 - 5 \end{gathered} )$

A B with -> on top = stretchy ( | 10 | - 4 | | stretchy )

(b) Find the modulus of vector $\overset{\rightarrow}{AB}$.

Answer:

Sketching a diagram of the vector $\overset{\rightarrow}{AB}$ can help

*Apply Pythagoras' theorem to the x and y components of *$\overset{\rightarrow}{AB}$

$\begin{gathered} \left| \vec{AB} \right| & = \sqrt{10^{2} + ( - 4 )^{2}} \\ = \sqrt{100 + 16} \\ = \sqrt{116} \end{gathered}$

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(c) Briefly explain why $\left| \vec{BA} \right| = \left| \vec{AB} \right|$.

Answer:

The modulus of a vector is it's 'size'

Direction of the vector is ignored

Final answer: $\left| \vec{BA} \right| = \left| \vec{AB} \right|$since both vectors have the same length

(d) Another vector, $\overset{\rightarrow}{CD}$, has three times the modulus of vector $\overset{\rightarrow}{AB}$.

Write down a possible column vector for $\overset{\rightarrow}{CD}$.

Answer:

Being three times $\left| \vec{AB} \right|$ means the vector $\overset{\rightarrow}{AB}$ is three times longer

One way to find a vector is to multiply each component of the vector $\overset{\rightarrow}{AB}$ by 3 or -3

$\begin{gathered} \vec{CD} & = 3 \vec{AB} = 3 ( \begin{gathered} 10 \\ - 4 \end{gathered} ) \end{gathered}$

C D with -> on top = stretchy ( | 30 | - 12 | | stretchy )

Another possible answer is C D with -> on top = stretchy ( | - 30 | | 12 | stretchy )