iGCSE (2021 Edition)

Lesson

What do we know about a vector so far?

Each vector has

- initial point
- terminal point
- magnitude (length)
- direction
- links to right-angled triangles

Knowing some of the above conditions will allow us to calculate the others, and because a vector also has links to trigonometry we can also use the trigonometric ratios to help us with the calculations.

Given that the vector **a**, projects from initial point $\left(1,3\right)$(1,3), at an angle of $45^\circ$45° find the terminal point if the magnitude is $3.5$3.5 units.

The information given to us here, results in the following right-angled triangle image.

The terminal point will have the coordinates $$

Using trigonometry we can see $\cos45^\circ=\frac{x}{3.5}$`c``o``s`45°=`x`3.5, so $x=3.5\times\cos45^\circ$`x`=3.5×`c``o``s`45° and similarly $\sin45^\circ=\frac{y}{3.5}$`s``i``n`45°=`y`3.5, so $y=3.5\times\sin45^\circ$`y`=3.5×`s``i``n`45°. Evaluating these to $2$2 decimal places we get $x=2.47$`x`=2.47 and $y=2.47$`y`=2.47. The fact that both x and y are equal make sense because an angle of $45^\circ$45° creates an isosceles triangle.

Now we can work out the terminal point, $$

This applet will help you to visualise the $x$`x` component and $y$`y` component. Remember that it uses the principles of right-angled trigonometry.

Consider the vector with an initial point $\left(2,5\right)$(2,5) and a terminal point $\left(4,8\right)$(4,8).

Find the $x$

`x`-component.Find the $y$

`y`-component.

Plot the vector with an $x$`x`-component $5$5 and a $y$`y`-component $9$9.

Use the origin as the starting point for the vector.

- Loading Graph...

Let $G$`G` and $H$`H` be the points $G$`G`$\left(11,3\right)$(11,3) and $H$`H`$\left(12,-2\right)$(12,−2).

Find the vector $\vec{HG}$→

`H``G`in component form:$\vec{HG}$→

`H``G`$=$=$\left(\editable{},\editable{}\right)$(,)What is the exact length of the vector $\vec{HG}$→

`H``G`?

Use vectors in any form, e.g. (a b), AB, p, ai - bj.