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 $\vec{A B}$ is written $|\vec{A B}|$
- 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 $a = x i + y j$ is
  - $|a| = \sqrt{x^{2} + y^{2}}$
  - where $a = (\begin{matrix} x \\ y \end{matrix})$

Finding the magnitude of a vector

How do I find harder magnitudes?

The magnitude of a sum of vectors is not equal to the sum of the magnitudes
- - 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 $x = \pm 3$

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 $a = \vec{O A} = x_{1} i + y_{1} j = (\begin{matrix} x_{1} \\ y_{1} \end{matrix})$ , then point A has coordinates $(x_{1}, y_{1})$
  - If $b = \vec{O B} = x_{2} i + y_{2} j = (\begin{matrix} x_{2} \\ y_{2} \end{matrix})$ , then point B has coordinates $(x_{2}, y_{2})$
- The distance between two points is given by
  - So $|\vec{A B}| = \sqrt{{(x_{1} - x_{2})}^{2} + {(y_{1} - y_{2})}^{2}}$
- For example, if points A and B have position vectors and respectively
  - then $|\vec{A B}| = \sqrt{{(5 - 3)}^{2} + {(3 - (- 6))}^{2}} = \sqrt{85} = 9.22 (3 s . f .)$

Alternatively, you could find by
- first using to find in vector form
  - and then calculating its magnitude directly
- See the Worked Example below

Exam Tip

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 $\vec{O A} = 2 i + m j$ and $\vec{O B} = - i + 4 j$ .

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

Use $\vec{A B} = \vec{A O} + \vec{O B} = - \vec{O A} + \vec{O B}$
Simplify and collect components

$\begin{array}{rcl} \vec{A B} & = & - (2 i + m j) + (- i + 4 j) \\ = & - 2 i - m j - i + 4 j \\ = & - 3 i + (4 - m) j \end{array}$

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

$\begin{array}{rcl} |\vec{A B}| & = & \sqrt{{(- 3)}^{2} + {(4 - m)}^{2}} \end{array}$

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

$\begin{array}{rcl} |\vec{A B}| & = & \sqrt{9 + 16 - 8 m + m^{2}} \\ = & \sqrt{25 - 8 m + m^{2}} \end{array}$

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

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

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

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

Exam Tip

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 $\vec{O A} = - 7 i + j$

Find a unit vector in the direction of $\vec{O A}$ , giving your answer in exact form.

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

$\begin{array}{rcl} | \vec{O A} | & = & \sqrt{{(- 7)}^{2} + 1^{2}} \\ = & \sqrt{49 + 1} \\ = & \sqrt{50} \\ = & \sqrt{25 \times 2} \\ = & 5 \sqrt{2} \end{array}$

Divide each component in $\vec{O A}$ by $\begin{array}{rcl} 5 \sqrt{2} \end{array}$

$\frac{\vec{O A}}{| \vec{O A} |} = - \frac{7}{5 \sqrt{2}} i + \frac{1}{5 \sqrt{2}} j$

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, $- \frac{7 \sqrt{2}}{10} i + \frac{\sqrt{2}}{10} j$

Magnitude of a Vector (Edexcel IGCSE Further Maths)

Revision Note