Magnitude of a Vector (Edexcel IGCSE Further Maths)

Spec point — 7C

Magnitude of a Vector

How do I find the magnitude of a vector?

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

  • It is also called the modulus

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

  • The magnitude of a is written |a|

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

  • Use Pythagoras' theorem to find it

  • The magnitude of $a=\xi+yj$ is

    • $\left| a \right| = \sqrt{x^{2 } + \mathrm{y}^{2}}$
    • where $a = ( \begin{gathered} x \\ y \end{gathered} )$

How do I find harder magnitudes?

• The magnitude of a sum of vectors is not equal to the sum of the magnitudes

  • $\vert a \pm b \vert \text{eq} \vert a \vert \pm \vert b \vert$

    • Work out a + b (or a - b) first, then find its magnitude

• You may need to form an equation

  • For example, |a| = 5 where a = 4i + *xj

    • $\sqrt{4^{2}+x^{2}}=5$
    • This solves to give $x=\pm3$

How do I find the magnitude of a displacement vector?

• You can use coordinate geometry to find magnitudes of displacement vectors from A to B

  • From the position vectors of A and B you know their coordinates

    • If $a = \vec{OA} = x_{1} i + y_{1} j = ( \begin{gathered} x_{1} \\ y_{1} \end{gathered} )$, then point A has coordinates $( x_{1} , y_{1} )$
    • If $b = \vec{OB} = x_{2} i + y_{2} j = ( \begin{gathered} x_{2} \\ y_{2} \end{gathered} )$, then point B has coordinates $( x_{2} , y_{2} )$

  • The distance between two points is given by $d = \sqrt{( x_{1} - x_{2} )^{2} + ( y_{1} - y_{2} )^{2}}$

    • So $\left| \vec{AB} \right| = \sqrt{( x_{1} - x_{2} )^{2} + ( y_{1} - y_{2} )^{2}}$

  • For example, if points A and B have position vectors $5i+3j$ and $3i-6j$ respectively

    • then $\left| \vec{AB} \right| = \sqrt{( 5 - 3 )^{2} + ( 3 - ( - 6 ) )^{2}} = \sqrt{85} = 9.22 ( 3 \mathrm{s} . \mathrm{f} . )$

• Alternatively, you could find $\left| \vec{AB} \right|$ by

  • first using $\overset{\rightarrow}{AB}=-\overset{\rightarrow}{OA}+\overset{\rightarrow}{OB}$ to find $\overset{\rightarrow}{AB}$ in vector form

    • and then calculating its magnitude directly

  • See the Worked Example below

Exam Hint

• When magnitudes involve algebra, it helps to square both sides to get rid of the square root sign!

Worked Example $O$, $A$and $B$ are fixed points, where $\overset{\rightarrow}{OA}=2i+mj$ and $\overset{\rightarrow}{OB}=-i+4j$.

Given that $\vert \vec{AB} \vert = 3 \sqrt{2}$ and $m>5$, find the value of $m$.

Use $\overset{\rightarrow}{AB}=\overset{\rightarrow}{AO}+\overset{\rightarrow}{OB}=-\overset{\rightarrow}{OA}+\overset{\rightarrow}{OB}$ Simplify and collect components

$\begin{gathered} \vec{AB} & = - ( 2 i + m j ) + ( - i + 4 j ) \\ = - 2 i - m j - i + 4 j \\ = - 3 i + ( 4 - m ) j \end{gathered}$

Use $\vert x i + y j \vert = \sqrt{x^{2} + y^{2}}$* *

$\begin{gathered} \left| \vec{AB} \right| & = \sqrt{( - 3 )^{2} + ( 4 - m )^{2}} \end{gathered}$

Expand and simplify inside the square root You cannot just take the square root of the individual terms!

$\begin{gathered} \left| \vec{AB} \right| & = \sqrt{9 + 16 - 8 m + m^{2 } } \\ = \sqrt{25 - 8 m + m^{2 } } \end{gathered}$

Substitute in $\left| \vec{AB} \right| = 3 \sqrt{2}$ from the question Square both sides and form a quadratic in $m$

$\begin{gathered} 3 \sqrt{2} & = \sqrt{25} - 8 m + m^{2} \\ ( 3 \sqrt{2} )^{2} & = 25 - 8 m + m^{2} \\ 18 = 25 - 8 m + m^{2} \\ 0 = m^{2} - 8 m + 7 \end{gathered}$

Factorise and solve

$0 = ( m - 1 ) ( m - 7 ) m = 1 , m = 7$

You are given that $m>5$ in the question This means $m=7$ is the only valid answer

$m=7$

Unit Vectors

Spec point — 7E

What is a unit vector?

• A unit vector is a vector of length 1

• Any vector, a, can be made a unit vector by dividing it by its magnitude,$\frac{a}{\left| a \right|}$

  • This will result in a vector of length 1

    • It still points in the same direction as a

• For example, to make 3i – 4j a unit vector

  • find its magnitude, $\sqrt{3^{2} + ( - 4 )^{2}} = \sqrt{25} = 5$

  • divide by its magnitude

    • $\frac{(3i-4j)}{5}=\frac{3}{5}i-\frac{4}{5}j$
    • Each component is divided by the magnitude

Exam Hint

• Read vector questions carefully

  • Some may want you to give your final answer as a unit vector!

Worked Example The position vector of a point,* $A$, relative to the origin, *$O$, is given by $\overset{\rightarrow}{OA}=-7i+j$

Find a unit vector in the direction of $\overset{\rightarrow}{OA}$, giving your answer in exact form.

To find a unit vector, divide the vector by its magnitude Find and simplify the magnitude, $\vert \vec{OA} \vert$, using $\sqrt{x^{2}+y^{2}}$

$\begin{gathered} \vert \vec{OA} \vert & = \sqrt{( - 7 )^{2} + 1^{2}} \\ = \sqrt{49 + 1} \\ = \sqrt{50 \\ = \sqrt{25} \times 2} \\ = 5 \sqrt{2} \end{gathered}$

Divide each component in $\overset{\rightarrow}{OA}$ by $\begin{array}{rcl} 5 \sqrt{2} \end{array}$

$\frac{\vec{OA}}{\vert \vec{OA} \vert} = - \frac{7}{5 \sqrt{2} } i + \frac{1}{5 \sqrt{2} } j$

Final answer: The unit vector is$-\frac{7}{5\sqrt{2}}i+\frac{1}{5\sqrt{2}}j$

$-\frac{7}{\sqrt{50}}i+\frac{1}{\sqrt{50}}j$would be accepted Decimals would not be accepted Rationalised denominators would be accepted,$-\frac{7\sqrt{2}}{10}i+\frac{\sqrt{2}}{10}j$