### Video Transcript

In this video, we will learn how to
determine the magnitude of two-dimensional vectors. We will begin by recalling some key
facts about vectors. Any vector has two aspects, its
direction and its magnitude. The magnitude of a vector is its
size or length. There are three main ways that we
can write 2D vectors, as a column vector, in a similar way to a coordinate but with
triangular brackets, or split into 𝐢- and 𝐣-components. Each of these three vectors
represent the same thing.

We denote the magnitude of the
vector 𝐯 by using the absolute value symbol, two parallel vertical lines. We use Pythagoras’s theorem to
calculate it. The magnitude of vector 𝐯 is equal
to the square root of 𝑎 squared plus 𝑏 squared, where 𝑎 and 𝑏 are the values
five and two in this case. Our first question involves finding
the magnitude of a vector on a coordinate grid.

The vector 𝐯 is shown on the grid
of units squares below. Find the value of the magnitude of
𝐯.

We know that the magnitude of any
vector is its length. By creating a right triangle on the
grid, we can see that the vector has moved four units to the right and three units
up. The magnitude of vector 𝐯 can
therefore be found using Pythagoras’s theorem. This states that the length of the
hypotenuse is equal to the sum of the squares of the two shorter sides. The magnitude of 𝐯 is therefore
equal to the square root of 𝑎 squared plus 𝑏 squared.

Whilst it doesn’t matter which
order we substitute the four and the three, we usually do the horizontal component
first. Four squared is equal to 16, and
three squared is equal to nine. The magnitude of vector 𝐯 is equal
to the square root of 25. As 25 is a square number, we can
calculate this. The square root of 25 is equal to
positive or negative five. As we’re dealing with a length, our
answer must be positive. Therefore, the magnitude of vector
𝐯 on the grid is five.

We will now look at a couple of
questions where we need to calculate the magnitude of a vector written in different
forms.

What is the magnitude of the vector
five, 12?

We know that for any vector written
in the form 𝑎, 𝑏, the magnitude is equal to the square root of 𝑎 squared plus 𝑏
squared. As the magnitude is the length of
the vector, this can be shown on a grid. Let’s consider the vector 𝐯 as
shown. If this vector has moved a distance
𝑎 in the horizontal direction and 𝑏 in the vertical direction, we can create a
right triangle. Using Pythagoras’s theorem, the
square of the hypotenuse is equal to 𝑎 squared plus 𝑏 squared. This means that the length of the
vector will be equal to the square root of 𝑎 squared plus 𝑏 squared.

In this question, the two
components of the vector are five and 12. We can therefore calculate its
magnitude by finding the square root of five squared plus 12 squared. Five squared is equal to 25, and 12
squared is equal to 144. This means that the magnitude of
vector 𝐯 is the square root of 169. As our answer must be positive, the
magnitude of vector 𝐯 is 13.

Given that vector 𝐀 is equal to
negative five 𝐢 minus three 𝐣, where 𝐢 and 𝐣 are perpendicular unit vectors,
find the magnitude of vector 𝐀.

We can begin by drawing this on a
grid where 𝐢 and 𝐣 are perpendicular unit vectors. Our vector 𝐀 moves a distance of
negative five in the 𝐢-direction and a distance of negative three in the
𝐣-direction. Vector 𝐀 can therefore be drawn as
shown. As the magnitude of any vector is
its length, we can calculate this by drawing a right triangle as shown. The magnitude of any vector 𝐯 can
therefore be calculated using Pythagoras’s theorem, where the magnitude equals the
square root of 𝑎 squared plus 𝑏 squared.

Lower case 𝑎 and 𝑏 are the 𝐢-
and 𝐣-components, respectively. Therefore, the magnitude of vector
𝐀 is equal to the square root of negative five squared plus negative three
squared. Squaring a negative number gives a
positive answer. So, the square root of negative
five is 25, and the square root of negative three is nine. This means that the magnitude of
vector 𝐀 is equal to the square root of 34. As 34 is not a square number, we
can leave our answer in surd or radical form. If vector 𝐀 is equal to negative
five 𝐢 minus three 𝐣, then its magnitude is equal to the square root of 34.

Our next question will involve
finding the magnitude of a vector between two points.

What is the magnitude of the vector
𝐀𝐁 where 𝐴 equals 11, three and 𝐵 equals seven, three?

The magnitude of a vector is its
size or length. So in this case, we need to find
the distance or length between point 𝐴 and point 𝐵. There are several ways of
approaching this problem, we will look at two of them. Our first method will be
graphically, and we will begin by plotting the two coordinates. Point 𝐴 has coordinates 11,
three. Point 𝐵 has coordinates seven,
three. As both points have the same
𝑦-coordinate, the distance from 𝐴 to 𝐵 will be a horizontal distance. To get from 11 to seven, we need to
subtract four. As the magnitude of any vector must
be positive, then the magnitude of 𝐀𝐁 is equal to four.

We could also have calculated the
distance between point 𝐴 and point 𝐵 using one of our coordinate geometry
formulas. The distance between any two points
is equal to the square root of 𝑥 one minus 𝑥 two squared plus 𝑦 one minus 𝑦 two
squared, where our two points have coordinates 𝑥 one, 𝑦 one and 𝑥 two, 𝑦
two. Substituting in our values gives us
𝑑 is equal to the square root of 11 minus seven squared plus three minus three
squared.

It doesn’t matter which coordinate
is 𝑥 one, 𝑦 one and which one is 𝑥 two, 𝑦 two. 11 minus seven is equal to four,
and three minus three is zero. As zero squared is equal to zero,
𝑑 is equal to the square root of four squared. As our distance must be positive,
this is equal to four. Once again, we have calculated that
the magnitude of the vector 𝐀𝐁 is four.

Our final question will involve
finding the magnitude of two separate vectors and their sum.

Consider the vectors 𝐮 equals two,
three and 𝐯 equals four, six. What is the magnitude of vector
𝐮? What is the magnitude of vector
𝐯? What is the magnitude of vector 𝐮
plus vector 𝐯? In all three questions we need to
give our answer to two decimal places where appropriate.

We recall that the magnitude of any
vector 𝐰 can be found by square rooting 𝑎 squared plus 𝑏 squared, where 𝑎 and 𝑏
are the two components of the vector. In vector 𝐮, 𝑎 is equal to two
and 𝑏 is equal to three. Whereas in vector 𝐯, 𝑎 is equal
to four and 𝑏 is equal to six. The magnitude of vector 𝐮 is
therefore equal to the square root of two squared plus three squared. As two squared is equal to four and
three squared is equal to nine, the magnitude of vector 𝐮 is equal to the square
root of 13. We would often leave this in surd
or radical form. However, in this case, we’re asked
to give our answer to two decimal places. The square root of 13 is equal to
3.605551 and so on.

To round to two decimal places, our
key or deciding number will be the first five. This will round our answer up. The magnitude of vector 𝐮 to two
decimal places is 3.61. We can repeat this process to
calculate the magnitude of vector 𝐯. Four squared is equal to 16, and
six squared is equal to 36. Therefore, the magnitude of vector
𝐯 is the square root of 52. Typing this into the calculator
gives us 7.211102 and so on. This time our deciding number is a
one. As this is less than five, we’ll
round down. The magnitude of vector 𝐯 is
therefore equal to 7.21.

The final part of our question asks
us to work out the magnitude of 𝐮 plus 𝐯. Our first step here will be to
calculate the vector 𝐮 plus 𝐯. We do this by adding the
corresponding components. Two plus four is equal to six, and
three plus six is equal to nine. We can then calculate the magnitude
of 𝐮 plus 𝐯 in the same way. This is equal to the square root of
six squared plus nine squared. Six squared is equal to 36, and
nine squared is equal to 81. Therefore, the magnitude of 𝐮 plus
𝐯 is equal to the square root of 117. Typing this into the calculator
gives us 10.816653. The deciding number here is a six,
and anything five or greater means that we round up. The magnitude of 𝐮 plus 𝐯 is
equal to 10.82.

When looking at our three answers,
you might think you’ve spotted a pattern, as 3.61 plus 7.21 is equal to 10.82. This suggests that the magnitude of
𝐮 plus 𝐯 is equal to the magnitude of 𝐮 plus the magnitude of 𝐯. This, however, is not normally the
case. The only reason this works in this
question is that vector 𝐯 is actually a multiple of vector 𝐮. Two multiplied by two is equal to
four. And three multiplied by two is
equal to six.

Therefore, vector 𝐯 is actually
two lots or two multiplied by vector 𝐮. This in turn means that the
magnitude of vector 𝐯 is twice the magnitude of vector 𝐮. Root 52 is equal to two root
13. The magnitude of 𝐮 plus 𝐯 in this
question becomes three times the magnitude of vector 𝐮. It is important to note, however,
as previously mentioned, this will not hold for the majority of vector
questions.

We will now summarize the key
points from this video. The magnitude of a vector is its
length. We can calculate the magnitude of
any vector in two dimensions using Pythagoras’s theorem. The magnitude of vector 𝐯 is equal
to the square root of 𝑎 squared plus 𝑏 squared, where 𝑎 and 𝑏 are the two
components of the vector. Whilst our answer is usually
written as a surd or radical, we can calculate the decimal value.

Finally, we found out that in the
majority of cases, the magnitude of vector 𝐮 plus vector 𝐯 is not equal to the
magnitude of vector 𝐮 plus the magnitude of vector 𝐯. Everything we have used in this
video can also be applied to three-dimensional vectors. These would be written in the form
𝑎, 𝑏, 𝑐, the column vector 𝑎, 𝑏, 𝑐, or 𝑎𝐢 plus 𝑏𝐣 plus 𝑐𝐤.