Lesson video

Hi everyone, my name is Ms. Coe, 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 enjoyed today's lesson, so let's start.

Hi everyone, and welcome to today's lesson on "Advanced problem solving with vectors" under the unit "Vectors." And by the end of the lesson, you'll be able to use your knowledge of vectors to solve problems. Keywords today, well, we're going to look at the word "vector." A vector can be used to describe a translation, and the vector 2, -5 shows the translation, 2 units to the right and 5 units down.

Displacement is the distance from the starting point when measured in a straight line.

And we'll also be looking at a resultant vector.

A resultant vector is the single vector that produces the same effect as a combination of other vectors.

Today's lesson will be broken into two parts.

We'll look at solving equations involving vectors and then further problem solving with vectors.

So let's make a start, "Solving equations involving vectors." We're given the vector a in column form is 4, 5.

We're given the vector b in column form is 3, -2.

I want you to work out the resultant vector of 2a + 2b, and I want you to work out the resultant vector of 3a - 2b.

Press pause as you'll need more time.

Well done, well, let's see what you've got.

Working out the resultant vector, we should have had 14, 6.

Working out the resultant vector of 3a - 2b, we should have had 6, 19.

Well done if you got this.

So if we're given the resultant vector, it's possible we may be able to work out an unknown component or scalAr multiplier.

For example, how do you think we can form an equation if we have this: 2 lots of the vector x, y add 3 lots of the vector 1, 3 gives us the final answer of 11, 13.

How do you think we can form an equation? Well, we can identify the equation from each component.

Let's look at the horizontal component first.

Looking at the horizontal components, you can see we multiply 2 by the horizontal component of x add 3 multiply by the horizontal component of 1, will equal the horizontal component of 11.

So therefore 2x + 3 = 11.

Another equation can be formed looking at the vertical components.

2 multiply by y add 3 multiply by 3 will give us the resultant vertical components of 13.

Thus we have equation 2y + 9 = 13.

From here, we can solve for X and y.

We can work out x to be 4, and we can work out y to be 2.

Knowing this, let's see if you can work out the values of x and y for each question.

See if you can give it a go, form those equations and solve, press pause as you'll need more time.

Well done, let's see how you got on.

Well, for a, looking at those horizontal components, we should have these equations: 3x + 10 = 22 and 3y + 2 = 8.

Working out x, we know x = 4 and working out y, we know y = 2.

For b, let's identify those equations.

Well, looking at the horizontal components, we have 5x - 6 = 4.

And looking at the vertical components, we have 15 + 3y = 24.

From here, we can solve.

Solving for x, we have x = 2, and solving for y, we have y = 3.

Lastly, here we have the horizontal components of 6x - 9 = 9, and the vertical components of 4x + 3 = y.

Therefore we know solving for x, x = 3, and solving for y, we have y = 15.

Great work if you've got these.

So what would happen if we formed equations with more than one variable in each? For example, we're asked to form the equations for the horizontal and vertical components.

Here we have r is multiplied by the vector 2, 3; and t is multiplied by the vector 3, 8; and that gives us the result in vector of 4, 13.

We know the horizontal equation would be 2r + 3t = 4, and the vertical equation would be 3r + 8t = 13.

So what have we just formed? Well, we've just formed simultaneous equations.

So that means we can use our knowledge on solving simultaneous equations.

This is why I love simultaneous equations 'cause there's so many different ways in which you can solve them.

For me, I'm going to look at this r term.

You might notice in equation one, we have 2r, and in equation two, we have 3r.

So what we need to do is identify the lowest common multiple of our 2 and 3.

Well the lowest common multiple of 2 and 3 is 6.

So that means if I multiply equation one by 3, I have 6r as my term in my equation.

And if I multiply equation two by 2, I also have 6r in my equation.

This is what I want because later on, I'm going to subtract these terms because I want to eliminate the r term, ensuring that we multiply the whole equation by the same value.

And remember the second equation is being multiplied by a different value to the first equation.

So what we now have is 6r + 9t = 12, and then 6r + 16t = 26.

Now, I'm going to subtract.

Subtracting these gives me 7t = 14, so I've solved for t, t is now 2.

Next, let's substitute this value of t into our equation one.

You really can substitute into any equation here, I'm just choosing equation one.

Substituting, our value of t is 2, I now have 2r + 6 = 4.

So that means I've worked out r to be -1.

This is a great question as we're using our knowledge on vectors and simultaneous equations.

So what we're gonna do is I'm going to do the question on the left and I'd like you to do the question on the right.

Here, we're given column vectors and we're asked to solve for a and b.

Looking at our equation from our horizontal components, we have 4b - 5a = 7.

Looking at the equation formed from our vertical components, we have -5b + 5a = -5.

From here, using our knowledge on simultaneous equations, I can simply sum equation one and two to give me b to be -2.

Next, I'm gonna substitute it in into any equation.

I'm gonna choose the first one, and then solving for a to give me a = -3.

Now, it's your turn.

I want you to solve for a and b.

Take your time, and press pause if you need.

Well done, let's see how you got on.

Well, forming those equations from the horizontal component, the vertical component, we should have 2b - 6a = 16, which is formed using the horizontal components.

And we should have equation two being b + 6a = -1, which is formed using the vertical components.

And then solving our simultaneous equations should identify b to be 5.

Substituting into any equation you want really, I chose the first one, gives me a to be -1.

So notice how we've solved for a and b, well done.

Great work, everybody.

So now it's time for your task.

I want you to solve the following unknowns.

For question 1a, 2 is multiplied by the column vector x, 5 add 6 is multiplied by the column vector 2, y; giving the resultant vector of 18, 22.

For 1b, 3 is multiplied by the column vector 1, y add 5 is multiplied by the column vector x, 5; giving the resultant vector of 23, 16.

And lastly for c, 2/3 is multiplied by the column vector 9, y add 2 multiplied by the column vector x, -4; this gives the result and vector of 4, 0.

Press pause if you need more time.

Well done, let's have a look at question two.

Solve for x and y.

Here we're giving x is multiplied by the column vector 4, 6 add y is multiplied by the column vector 3, -5; giving us the resultant vector of 27, -7.

Press pause as you'll need more time.

Well done, and for question three, solve for a and b where a is multiplied by the column vector 3, 4 and we're adding it to b being multiplied by the column vector 2, 3; giving us the resultant vector of 330, 445.

Press pause as you'll need more time.

Great work, let's have a look at these answers.

Well, hopefully you've identified the horizontal and vertical components formed an equation where the equation formed in the horizontal component is 2x + 12 = 18, and the equation formed from the vertical component is 10 + 6y = 22, and solve for x and y, where x = 3 and y = 2.

For question b, forming those equations using the horizontal and vertical components from the horizontal components and from the vertical components.

So the equation you should have formed using the horizontal components is 3 + 5x = 23.

And the equation formed from the vertical components is 3y + 25 = 16.

And solve them simultaneously to give you x = 4 and y = -3.

And for c, once again, forming those equations using those horizontal and vertical components, we can form those equations.

Using the horizontal components, we simply have 6 + 2x = 4.

And forming an equation from the vertical components, we have 2/3y - 8 = 0.

We can solve using our knowledge on simultaneous equations to find x = -1 and y = 12.

For question two, I've structured it a little bit more for you to see.

So you can see forming our equations, we have this which can be solved simultaneously, giving me an answer for y to be 5, and giving me an answer for x to be 3, well done.

Next, let's form our equations using the horizontal and vertical components.

You should have 3a + 2b = 330, and our second equation is 4a + 3b = 445.

And then from here, one method to solve this pair of simultaneous equations is to look at the b term, and make the coefficient of the b term the same.

Here, you can see I have a b term of 2b and a b term of 3b.

So identifying the lowest common multiple of 2b and 3b, it's simply 6b.

So that means I need to multiply equation one by 3 to get me 9a + 6b = 990, and I need to multiply equation two by 2, giving me 8a + 6b = 890.

Notice how both equations now have 6b as a term in their equation.

We can then subtract, giving me a to be 100.

This is why I like simultaneous equations 'cause there's lots of different ways in which you could have solved for b.

For me, I'm choosing to substitute that value of a to be 100 into equation one.

You could have substituted it into equation two if you wanted as well.

From here, I've simply solved to give me b = 15.

Great work, everybody.

So let's move on to the second part of our lesson, "Further problem solving with vectors." Now, we can use our knowledge of parallel vectors to work out unknown vector components, and one approach is to equate the corresponding components to form and then solve the subsequent equations.

For example, when the vector a given in column form is 4, 5 and the vector b in column form is x, -2.

And we know 2a + 3b is parallel to 34, 8; we're asked to work out the value of x.

Well, firstly, let's work out 2a + 3b as a column vector.

2a + 3b, I've written it here as a calculation, works out to be 8 + 3x as its horizontal component and 4 as the vertical component.

Well, what does that mean? Well, given the fact that we know 2a + 3b is parallel to 34, 8; we know there must be a scalar and multiplier given that they are parallel.

So that means I'm going to label it k.

So I know 2a + 3b = k multiplied by the vector 34, 8.

So that means I know 2a + 3b, which is written in column vector form, is 8 + 3x, 4 is equal to k 34, 8.

Now, we can equate the horizontal components and form an equation.

Equating the horizontal components, I know 8 + 3x has got to be equal to 34k.

And equating the vertical components, I have 4 has got to be equal to 8k.

So now, looking at these equations, which one can be solved easily? Well, we can solve for k easily on the first one.

So let's solve for k.

We know k = 1/2.

Now we know k is 1/2, We can substitute it back into the other equation and solve for x.

8 + 3x = 17.

Solving for x, I know x = 3.

Now, let's move on to a check.

When a is given as 3, 1 in column vector form and b is given as -3, x in column vector form, the question says 4a - 3b is parallel to 63, -33; and we're asked to calculate the value of x.

See if you can give it a go, and press pause if you need more time.

Well, first of all, let's identify what 4a - 3b is in column vector form.

Working this out, it's 21, 4 -3x.

Because we know it's parallel to 63, -33; that means there is a scalar multiplier here.

So 21, 4 -3x has got to be equal to k multiplied by that vector 63, -33.

Now, equating those horizontal components, we have 21 = 63k.

So we can solve for k, so k must be 1/3.

Equating the vertical components, we have 4 - 3x which is equal to -33k.

4 - 3x = -11.

So solving for x, we have x = 5.

Moving on further, here we have a diagram, and it's not drawn accurately, and it states that X is the midpoint of OA and W is the midpoint of OB.

Y divides AW in the ratio of AY:AW = 2:1.

We know the vector OX is 3a, and we know the vector OW is 3B.

X, Y, and B are collinear, such that k multiplied by vector XY gives you YB, and the question wants us to work out the value of k.

So, let's identify all these wonderful vectors and points on our diagram.

From here, we can work out vector XY.

I've chosen this vector path here and work out vector Y to be -a + 2b.

Now, let's work out the vector YB.

I've chosen this vector pathway and identified the vector YB = -2a + 4b.

From here, you can spot a relationship between these two vectors.

2 lots of vector XY is equal to vector YB.

So therefore we know k has got to be 2.

Great work, everybody, so let's have a look at a check.

We have another diagram, not drawn accurately, and we know OBA and OBC are triangles.

A is the midpoint of OC and Y is the midpoint of AB, and X divides OB in the ratio of OX:XB is in the ratio of 2:1.

The question says XY:YC is in the ratio of 1:n, and we're asked to work out the value of n.

This is a tricky question, take your time.

My little hint to you is always label anything that you need on that vector diagram.

So you can give it a go, press pause if you need.

Well done, let's see how you got on.

Well, working out vector XY, it's 1/2p - 1/2q.

Working out vector YC, it's 3/2p - 3/2q.

That means 3 lots of vector XY = YC.

So therefore, n has to be 3.

Great work if you've got this.

Fantastic work, everybody.

So let's move on to your task.

When m is equal to the vector x, -3 and y is equal to the vector 7, -3; and we know 10y - 5m is parallel to the vector 4, -1; you are asked to work out x.

See if you can give it a go, press pause if you need more time.

Well done, let's move on to question two.

AOBC is a parallelogram and ACD is a straight line.

N divides AB in the ratio AN:NB = 2:1.

ON:ND is in the ratio of n:1, and you're asked to work out n.

See if you can give it a go, press pause if you need.

Well done, let's go through these answers.

Well for question one, you should have formed the resultant vector for 10y - 5m to be 70 - 5x, -15.

We know it's parallel to 4, -1; so that means I've put a scalar multiplier of k right there.

From here, we can form an equation by equating vertically.

<v ->15 = -k, so that means we know k must be 15.

</v> And then equating horizontally, we can work out the value of x to be 2.

Fantastic work if you got this.

For question two, wonderful question.

Label any vectors that you need.

From here, we can work out vector ON to be 2/3a + 4/3b.

We can work out vector ND to be 4/3a + 8/3b, so we can spot a relationship.

2 lots of vector ON is equal to vector ND, so that means n = 2.

Great work, everybody.

So in summary, equations can be formed and solved using the horizontal and vertical components of vectors.

Simultaneous equations can be formed to help us work out the unknown multiplier or unknown multipliers, and/or horizontal and vertical components.

Great work everybody, it was wonderful learning with you.