### Video Transcript

In this video, we will learn how to
perform operations on vectors algebraically, such as vector addition, vector
subtraction, and scalar multiplication, in two dimensions. Let’s begin by recalling what we
mean by a vector. A vector has a size, known as its
magnitude, and a direction. A two-dimensional vector has a
horizontal and vertical component. These are often denoted by 𝑖 and
𝑗 unit vectors.

As with the coordinate grid, moving
to the right is the positive direction and moving to the left is the negative
horizontal direction. Moving up is the positive vertical
direction, and moving down is the negative vertical direction. A vector 𝐮 equal to four 𝑖 plus
three 𝑗 will move four units in the horizontal direction and three units up in the
vertical direction. This can also be shown as four,
three inside triangular brackets, as shown. When adding or subtracting two
vectors, we treat the horizontal and vertical components separately. This is also true when multiplying
a vector by a scalar.

We will now look at some questions
where we need to add and subtract vectors.

Shown on the grid of unit squares
are the vectors 𝐮, 𝐯, and 𝐮 plus 𝐯. What are the components of vector
𝐮? What are the components of vector
𝐯? What are the components of vector
𝐮 plus 𝐯?

Let’s begin by considering vector
𝐮. As with any vector, this has a
start and end point. The arrow indicates it is moving to
the right and up. The horizontal component of this
vector will be equal to four, as we move four units to the right. The vertical component will be one,
as we move up one square. The components of vector 𝐮 are
therefore equal to four, one. We always put the horizontal
component first.

We can calculate the components of
vector 𝐯 in the same way. This vector moves left and up. This means that its horizontal
component will be negative. We move five squares to the left
and one square up. Vector 𝐯 is therefore equal to
negative five, one.

There are two ways of answering the
third part of this question, to calculate the components of the vector 𝐮 plus
𝐯. We could do it directly from the
diagram, as the vector 𝐮 plus 𝐯 moves one square to the left and two squares
up. This means that it is equal to the
vector negative one, two. Alternatively, if we didn’t have
the diagram of unit squares as in this case, we could use the fact that vector 𝐮
plus 𝐯 is equal to vector 𝐮 plus vector 𝐯. We need to add the vector four, one
to the vector negative five, one.

When adding and subtracting
vectors, we treat the horizontal and vertical components separately. In this case, we need to add four
and negative five and then separately one and one. Four plus negative five is the same
as four minus five, which is equal to negative one. One plus one is equal to two. Therefore, the vertical component
of the vector 𝐮 plus 𝐯 is two. This gives us the same answer as
from the diagram, the vector negative one, two.

In our next example, we will answer
a similar question but without a diagram.

Given that vector 𝐮 is equal to
zero, four and vector 𝐯 is equal to zero, negative five, find the components of
vector 𝐮 plus 𝐯.

We recall that every
two-dimensional vector has a horizontal and vertical component. If our horizontal component is
positive, it is moving to the right, whereas if it is negative, it is moving to the
left. In a similar way, if the vertical
component is positive, our vector is moving upwards, whereas if it is negative, it
is moving downwards.

We also recall that when adding two
vectors, in this case 𝐮 and 𝐯, we add the horizontal and vertical components
separately. The vector 𝐮 plus 𝐯 is equal to
the vector 𝐮 zero, four plus the vector 𝐯 zero, negative five. Zero plus zero is equal to
zero. Therefore, the horizontal component
of vector 𝐮 plus 𝐯 is zero. Four plus negative five is equal to
negative one. Therefore, the vertical component
is negative one. The vector 𝐮 plus 𝐯 is therefore
equal to zero, negative one.

In our next question, we need to
express one vector in terms of two other vectors using scalar multiples.

Given that vector 𝐀 is equal to
negative four, negative one and vector 𝐁 is equal negative two, negative one,
express vector 𝐂 negative eight, negative one in terms of vector 𝐀 and vector
𝐁.

As we want to express vector 𝐂 in
terms of vector 𝐀 and 𝐁, we know that 𝐂 is equal to some constant 𝑝 multiplied
by vector 𝐀 plus some constant 𝑞 multiplied by vector 𝐁. Substituting in vectors 𝐀, 𝐁, and
𝐂, we have negative eight, negative one is equal to 𝑝 multiplied by negative four,
negative one plus 𝑞 multiplied by negative two, negative one.

We recall that when multiplying a
vector by any constant or scalar, we need to treat the horizontal and vertical
components separately. If we consider the horizontal
components first, we have negative eight is equal to negative four 𝑝 plus negative
two 𝑞. This is the same as negative eight
is equal to negative four 𝑝 minus two 𝑞. We can divide both sides of this
equation by negative two. This gives us four is equal to two
𝑝 plus 𝑞. We will call this equation one.

We can now repeat this process with
the vertical components. Negative one is equal to negative
𝑝 plus negative 𝑞. We can divide both sides of this
equation by negative one. This gives us one is equal to 𝑝
plus 𝑞, which we will call equation two.

We now have a pair of simultaneous
equations that we can solve by elimination. We can subtract equation two from
equation one. On the left-hand side, four minus
one is equal to three, two 𝑝 minus 𝑝 is equal to 𝑝, and 𝑞 minus 𝑞 is equal to
zero. Therefore, 𝑝 is equal to
three. Substituting this value of 𝑝 back
into equation two gives us one is equal to three plus 𝑞. Subtracting three from both sides
of this equation gives us 𝑞 is equal to negative two. The vector 𝐂 negative eight,
negative one is therefore equal to three multiplied by vector 𝐀 minus two
multiplied by vector 𝐁. We have therefore expressed vector
𝐂 in terms of vector 𝐀 and vector 𝐁.

Our next question is a more
complicated problem involving scalar multiples and addition of vectors.

On a lattice where vector 𝐀𝐂 is
equal to three, three; vector 𝐁𝐂 is equal to 13, negative seven; and two
multiplied by vector 𝐂 plus two multiplied by vector 𝐀𝐁 is equal to negative
four, negative four; find the coordinates of the point C.

In this question, we are given
vector 𝐀𝐂 and vector 𝐁𝐂. We can use these to calculate the
vector 𝐀𝐁. If we consider the three points A,
B, and C as shown on the diagram, we know that vector 𝐀𝐂 is three, three. Vector 𝐁𝐂 is 13, negative
seven. We need to calculate vector
𝐀𝐁. To get from point A to point B via
point C, we can add vector 𝐀𝐂 and vector 𝐂𝐁. As vector 𝐂𝐁 is in the opposite
direction to vector 𝐁𝐂, vector 𝐀𝐁 is equal to 𝐀𝐂 minus 𝐁𝐂. We need to subtract the vector 13,
negative seven from the vector three, three.

When adding and subtracting
vectors, we treat the horizontal and vertical components separately. Three minus 13 is equal to negative
10. Three minus negative seven is the
same as three plus seven, which is equal to 10. Therefore, the vector 𝐀𝐁 is equal
to negative 10, 10. If we let the point C have
coordinates 𝑥, 𝑦, we can substitute these values into our equation. Two multiplied by 𝑥, 𝑦 plus two
multiplied by negative 10, 10 is equal to negative four, negative four.

Once again, we can treat the
horizontal and vertical components separately to create two equations. The horizontal components give us
two 𝑥 minus 20 is equal to negative four. The vertical components give us the
equation two 𝑦 plus 20 is equal to negative four. In our first equation, we add 20 to
both sides, giving us two 𝑥 is equal to 16. Dividing both sides by two gives us
a value of 𝑥 equal to eight. In our second equation, we need to
subtract 20 from both sides, giving us two 𝑦 is equal to negative 24. Once again, we can divide both
sides of this equation by two, giving us 𝑦 is equal to negative 12. As 𝑥 is equal to eight and 𝑦 is
equal to negative 12, the coordinates of point C are eight, negative 12.

Our final question involves
calculating the magnitude or modulus of a vector.

If vector 𝐀𝐁 is equal to seven 𝑖
plus six 𝑗 and vector 𝐁𝐂 is equal to 𝑖, then the magnitude of vector 𝐀𝐂 is
equal to blank.

We recall that the vector seven 𝑖
plus six 𝑗 can be written using triangular brackets as seven, six. The vector 𝑖 can be written one,
zero, as there is no vertical component 𝑗. We also recall that when dealing
with vectors, the vector 𝐀𝐂 is equal to the vector 𝐀𝐁 plus the vector 𝐁𝐂. In this question, we can calculate
the vector 𝐀𝐂 by adding the vector six, one and the vector one, zero.

When adding two vectors, we add the
horizontal and vertical components separately. Seven plus one is equal to
eight. Six plus zero is equal to six. Therefore, vector 𝐀𝐂 is equal to
eight, six. This is not the final answer to
this question though, as we need to calculate the magnitude or modulus of vector
𝐀𝐂. In order to do this, we need to
know the following rule. If vector 𝐮 is equal to 𝑥, 𝑦,
then the magnitude of vector 𝐮 is equal to the square root of 𝑥 squared plus 𝑦
squared. We use the Pythagorean theorem to
calculate the magnitude or size of vector 𝐮 by squaring the horizontal and vertical
components, finding their sum, and then square rooting the answer.

The magnitude of vector 𝐀𝐂 is
equal to the square root of eight squared plus six squared. Eight squared is equal to 64, and
six squared is 36. Adding these gives us the square
root of 100, which is equal to 10. If vector 𝐀𝐁 is equal to seven 𝑖
plus six 𝑗 and vector 𝐁𝐂 is equal to 𝑖, then the magnitude of vector 𝐀𝐂 is
10.

We will now summarize the key
points from this video. We found out in this video that a
vector has a horizontal and vertical component. The horizontal component comes
first and can be written in triangular brackets — four, negative three — or as four
𝑖 minus three 𝑗, where 𝑖 and 𝑗 are unit vectors. When adding and subtracting
vectors, we treat the horizontal and vertical components separately. We also do this when multiplying a
vector by a scalar. The magnitude of a vector is its
size. And we found out in the last
question that we can calculate this using the Pythagorean theorem.