Lesson video

Hi, everyone, my name is Miss Ku, and I'm really happy that you're joining me today.

Today we'll be looking at vectors, a great topic, and chances are you've already used vectors today already, whether that be playing a game or using satellite navigation.

I hope you enjoy today's lesson.

So let's start.

Hi, everyone, and welcome to this lesson on calculating the magnitude of a vector under the unit vectors.

And by the end of the lesson, you'll be able to calculate the magnitude of a vector using Pythagoras's theorem.

So let's have a look at some keywords starting with Pythagoras's theorem.

Now, Pythagoras's theorem states the sum of the squares of the two shorter sides of a right-angled triangle is equal to the square of the hypotenuse, and we'll be looking at this a lot in our lesson today.

Today's lesson will be broken into two parts.

Firstly, we'll be calculating the magnitude of a vector and then moving on to problem solving with magnitudes of vectors.

So let's make a start calculating the magnitude of a vector.

Now vectors are utilised in various ways.

For example, computer games use vectors.

Here we have a controller which controls the movement of a character.

Here she is in the bottom left hand corner.

What we're going to do is we're gonna press one of the buttons and it will make the character jump like this.

Can you explain how the vectors are being used? Have a little look at the diagram.

Give it some thought.

Well done.

Well, hopefully you spotted that the character's jump is a combination of the following vectors.

Here we have the vector one, three.

Here we have the vector one, two.

Here we have the vector one, one.

Then we have the vector one, zero.

Then we have the vector one, minus one, followed by the vector one, negative two, and then the vector one, negative three.

So this jump is a combination of these vectors.

So once the character reaches a certain point, the computer activates the next vector.

What shape does this jump look similar to? It looks like a parabola.

Well done if you spotted this.

So let's zoom in a little bit.

We still have our controller, and what we're going to do is we're going to imagine the blue button of our controller only completes one short jump, which is simply this vector three, four.

By removing the character and highlighting the horizontal and vertical components, we have a right-angled triangle.

The magnitude of a vector is the size, it's its length, so it's always positive.

You can see it indicated here.

How do you think we can work out the magnitude of our vector? Well, it's Pythagoras's theorem, so let's work it out.

Now we know Pythagoras's theorem can be used to find the length of the hypotenuse.

In this case, the hypotenuse is the magnitude of the vector.

So let's apply Pythagoras's theorem.

Pythagoras' theorem says a squared add b squared is equal to c squared, so therefore three squared add our four squared is equal to our hypotenuse squared.

That means nine add 16 is equal to our hypotenuse squared.

So working this out, we have now worked out our hypotenuse is five.

But what does that mean with respect to our vector? Well, it means the magnitude of our vector is five units.

Therefore, we know the blue button on this controller moves the character with a horizontal component of three and a vertical component of four, giving us a magnitude of five units.

Let's have a look at a check.

Using the same controller, the green button makes the character follow this jump path, 5, 12.

Work out the magnitude of the jump.

See if you can give it a go.

Press pause if you need more time.

Well done.

Let's see how you got on.

Well, using Pythagoras's theorem, all I've done is identify that horizontal component of five and that vertical component of 12.

Then substituting into Pythagoras's theorem, I know that the hypotenuse squared is equal to 169, so therefore the hypotenuse is 13, identifying the magnitude of our vector to be 13 units.

Let's consider the magnitude of a resultant vector.

Here we have a vector a, which is one, two, and vector b, which is three, two.

And what we're going to do is we're going to use a square grid and draw the resultant vector 2a add b.

Now it's important to note we have a grid of one centimetre squares.

Now drawing our vector 2a, you can see it here.

We know the vector a is one, two, so therefore the vector 2a is two, four.

So we've drawn it here.

The vector b is simply three, two, so therefore the result vector, which is 2a add b, is indicated here.

Now let's work out the magnitude of our resultant vector 2a add b, and we're going to give our answer to one decimal place.

Identifying our horizontal and vertical component to be five and six, I can use Pythagoras's theorem and then work out that the magnitude of our vector is 7.

8 centimetres.

But without the one centimetre square grid, how can we calculate the magnitude of 2a add b given that we know the vector a is one, two and the vector b is three, two? Well, let's calculate the vector 2a add b as a column vector.

It's two multiplied by vector a add our vector b, which gives me two, four, add three, two, giving me the resultant column vector as five, six.

Now we have the horizontal component of five and the vertical component is six.

So we simply use Pythagoras again.

a squared plus b squared is equal to c squared.

Working this out, we've worked out the magnitude once again to be 7.

8 centimetres, but notice how we did it without that square centimetre grid.

Now it's time for a check.

Laura says, "I've got the resultant vector of negative six, eight." And then she shows this working out, the square root of negative six all squared add eight squared is equal to plus or minus the square root of 100, so that equals plus or minus 10.

Now does this mean the magnitude is negative? Who gives the correct explanation? Alex says, "Yes, because you have a negative component." Sophia says, "Yes, as the square root and plus or minus the square root means the same thing." And Andeep says, "No, because the magnitude is its size so will always be positive." Which response do you think is correct? Well done.

It's Andeep.

He's correct because the magnitude is size and size will always be positive.

Next, I want you to work out the magnitude of the following vectors to one decimal place where appropriate, given that a is four, one and b is negative two, negative three.

See if you can give it a go.

Press pause if you need more time.

Well done.

Let's see how you got on.

Well, the vector 2a add 3b.

My working out is here giving me the resultant column vector of two, negative seven.

Using Pythagoras's theorem, two squared add negative seven all squared gives me 53.

The square root of 53 is 7.

3 units.

Well done if you got this.

3a subtract 2b.

My working out is here, giving me the resultant vector of 16, nine.

Using Pythagoras's theorem, I have 16 squared add nine squared giving me 337, so therefore working out the magnitude to be 18.

4 units.

Well done if you got this.

Great work, everybody.

So now it's time for your task.

I want you to work out the magnitude of these vectors given a is 24, 10, b is 20, 21, and c is negative 28, 45.

See if you can give this a go.

Press pause if you need more time.

And for question two, you need to work on the magnitude of the resultant vectors to one decimal place where appropriate, given that a is four, five, b is three, negative two.

See if you can give this a go.

Press pause if you need more time.

Well done.

Let's move on to question three.

The character in a game has jumped from one platform to another as you can see in the diagram.

The jump is made of lots of vectors.

Now the final position on the platform is the resultant of all these vectors.

I want you to work out the straight line distance from platform A to platform B giving your answer to one decimal place.

Once again, each square length is one unit.

See if you can give it a go.

Press pause if you need more time.

Well done.

Let's go through these answers.

Well for question 1a, it should have been 26 units.

For question b, it should have been 29 units.

And for question 1c, it should have been 53 units.

Really well done if you've got that.

For question 2a, we should have had 21.

1 units, and for 2b, we should had 26.

9 units.

Really well done.

For question three, well, we know this is our final resultant vector, identifying our horizontal and vertical components to be six and three.

That means we can apply Pythagoras's theorem.

Using Pythagoras's theorem, we can work out that the magnitude of our vector is 6.

7 units.

Well done if you got this.

Great work, everybody.

So let's move on to problem solving with magnitudes of vectors.

Now understanding how to find the magnitude of a vector using Pythagoras's theorem allows us to solve more complex problems. For example, if we're given the vector OA is 18, 21 and OB is 26, 27, work out the vector AB.

Well, we can draw a vector diagram to help us visualise this if it helps.

Here I've drawn a triangle, and as you can see I've got the vector OB, and my vector OA, and the resultant vector AB.

Now we can see to work out vector AB it is vector OB subtract vector OA.

So let's substitute our column vectors.

26, 27 subtract 18, 21 means our vector AB is eight, six.

Now we know vector AB is eight, six, we can use Pythagoras's theorem to find the magnitude of this vector.

Using Pythagoras's theorem, we know the hypotenuse squared is 100, so therefore the magnitude must be the square root of 100, which is 10 units.

Now let's do a check.

Two drones are released from the docking port at O.

The first drone travels O to X and then X to Y.

The second drone travels directly to Y.

Now given the numbers in the vectors are metres, we know the vector OX is eight, nine and the vector OY is 12, 15.

The question asks how much further did the first drone travel than the second drone? And we're ask to give our answer to two decimal places.

See if you can give it a go.

Press pause if you need more time.

Well done.

Let's see how you got on.

Firstly, let's work out that distance O to X.

Well, we know the vector O to X was eight, nine.

So using Pythagoras's theorem, that means we know O to X is 12.

04 metres.

We've labelled it on our diagram.

Now let's work out the distance from X to Y.

Now we know the vector XY can only be found using the vector path OY subtract OX.

So 12, 15 subtract eight, nine gives me the vector XY in column form to be four, six.

Applying Pythagoras's theorem allows me to work out the magnitude of vector XY to be 7.

21 metres.

Now let's work out the distance or the magnitude from O to Y.

Well, it would be simply 12 squared and 15 squared.

Working this out gives me a final magnitude of 19.

21 metres.

So let's find out how much each drone travelled.

The first drone has travelled 19.

25 metres and the second drone has travelled 19.

21 metres.

So therefore the first drone has travelled 0.

04 metres more than the second drone.

Well done if you got this.

Sometimes we may need to combine our knowledge of vectors with other topics.

For example, given that the vector OA is four, five, and the vector OB is seven, nine, and the vector OC is eight, two, the question asks what type of triangle is ABC? And we need to explain how we know.

Well, to do this, we need to work out the magnitude of each vector.

So let's work out vector AB first as this is one of our sides of our triangle.

To work out vector AB, it's OB subtract OA, giving me the column vector of three, four.

Therefore using Pythagoras's theorem, I've just worked out that vector AB is five units in length.

Now let's work out vector AC as this is another length of our triangle.

To work out vector AC, it's OC subtract OA, giving me the vector in column form to be four, negative three.

From here, applying Pythagoras's theorem, I've just worked out the vector AC is five units.

Now let's work out the vector BC.

Vector BC is OC subtract OB, which gives me the column vector of one, negative seven.

Then using Pythagoras's theorem, I have five root two units, which I've labelled here.

So what type of triangle is ABC? Well, you can see we have two equal lengths, so it's an isosceles triangle.

Great work, everybody.

So let's move on to another check.

We have vector OE is two, three, OG is nine, two, and OF is five, six.

I want you to show that triangle EFG is a right-angled triangle.

This is quite a tough question.

Do what you can.

See if you can give it a go.

Press pause if you need more time.

Well done.

Let's see how you got on.

Well, let's work out the length of EG.

Using the vector EG, we can identify in column form it's seven, negative one, giving me a length of five root two because I've used Pythagoras's theorem.

Then I'm going to work out the length FG, which is four, negative four.

Using Pythagoras's theorem again, I have the magnitude of FG to be four root two.

Then I know EF is simply three, three, giving me the magnitude of vector EF to be two root two.

But how do these three lengths show it's a right-angled triangle? Well, if it was a right-angled triangle, then we would know EF squared add FG squared is equal to our hypotenuse squared.

Remember that definition at the very start of our lesson.

Pythagoras's theorem states the sum of the squares of the two shorter sides of a right-angled triangle is equal to the square of the hypotenuse.

So we can substitute this in, and we can concur that this triangle is a right-angled triangle.

Well done.

Great work, everybody.

So now it's time for your task.

The diagram shows a map with three landing stations, A, B, and C.

We know the vector AB is 10, negative two in column vector form, and the vector BC is one, five.

A helicopter flies from A to B and then B to C, and a plane flies directly from A to C.

We're asked to work out the difference in distance travelled by the plane and the helicopter, giving our answer to one decimal place.

See if you can give it a go.

Press pause for more time.

Well done.

Let's move on to question two.

Question two has a quadrilateral ABCD, and it's drawn using the following.

We know vector O to A is 11, 10, the vector OB is eight, eight, the vector OC is 11, six, and the vector OD is 14, eight.

Now using the vectors, what type of quadrilateral could ABCD be? See if you can give it a go.

Press pause if you need more time.

Well done.

Let's see how you got on.

Well, for question one, let's work out that length AB.

AB is simply 10.

20 kilometres.

Notice how we use the vector in its column form 10 negative, two and inserted it into Pythagoras's theorem.

Now let's work out vector BC.

Same again, working out vector BC, it's already in column form, so we simply substitute it into Pythagoras's theorem to give me 5.

10 kilometres.

Lastly, vector AC.

The vector AC has a magnitude of 11.

40 kilometres, so I've labelled it here.

So the helicopter travelled 15.

30 kilometres and the plane travelled 11.

40 kilometres.

The difference is 3.

9 kilometres.

Great work if you've got this.

Okay, for question two, let's identify what type of quadrilateral we have.

Well, to work this out we have to work out the following vectors.

We have to work out vector AD, vector BA, vector BC, and vector CD.

And working this out, you'll notice we have all of our magnitudes of our vectors are root 13.

So that means the quadrilateral could be a rhombus or a square.

Great work if you've got this.

Fantastic work, everybody.

So in summary, the magnitude of a vector is its size, so it's always positive.

Pythagoras's theorem can be used to find the magnitude of the vector given the horizontal and vertical components.

And remember understanding how to find the magnitude of a vector using Pythagoras's theorem allows us to solve more complex problems. Really well done, everybody.