### Video Transcript

In this video, we will learn how to
find the magnitude of a position vector in space. We will begin by recalling what we
mean by a 3D vector.

A three-dimensional vector has an
π’-, π£-, and π€-component. If we consider the point π with
coordinates two, three, five, we can write the vector ππ in numerous ways. Firstly, we can consider the π’-,
π£-, and π€-components. The vector ππ is equal to two π’
plus three π£ plus five π€. Vectors are also sometimes written
similar to coordinates. With triangular brackets, we have
two, three, five. A third way of writing the same
vector is as a column inside parentheses, with the π’-component followed by the
π£-component and then the π€-component.

The magnitude of any vector is the
distance between two points in three-dimensional space. If we consider a vector π written
in the general form π₯π’ plus π¦π£ plus π§π€, then the magnitude of vector π can be
calculated using an application of the Pythagorean theorem. The magnitude of vector π denoted
by two vertical lines is equal to the square root of π₯ squared plus π¦ squared plus
π§ squared. We find the sum of the squares of
the π’-, π£-, and π€-components and then square root the answer.

As well as seeing our unit vectors
π’, π£, and π€ written with a hat, you may also have seen them underlined. These notations for handwritten
vectors will vary from location to location. Generally when typed, either in a
text book or on the internet, vectors will appear in bold. We will now look at some questions
where we need to calculate the magnitude of a vector.

If vector π is equal to two,
negative five, two, find the magnitude of vector π.

The π’-, π£-, and π€-components of
vector π are two, negative five, and two, respectively. Therefore, vector π could be
rewritten as two π’ minus five π£ plus two π€. We recall that the magnitude of any
vector could be calculated by square rooting π₯ squared plus π¦ squared plus π§
squared, where π₯, π¦, and π§ are the π’-, π£-, and π€-components, respectively. The magnitude of vector π is,
therefore, equal to the square root of two squared plus negative five squared plus
two squared.

Two squared is equal to four. Squaring a negative number gives a
positive answer. Therefore, negative five squared is
25. The magnitude of π is equal to the
square root of four plus 25 plus four. This is equal to the square root of
33. Whilst we could work out this
answer on the calculator, as a general rule, we will leave our answers as radicals
or surds. The magnitude of vector π is the
square root of 33.

In our next question, the vector
will be written in a different format.

If vector π is equal to two π’
plus three π£ minus π€, find the magnitude of vector π.

For any vector written in the form
π₯π’ plus π¦π£ plus π§π€, the magnitude of the vector is equal to the square root of
π₯ squared plus π¦ squared plus π§ squared. The π’-component of our vector is
equal to two, the π£-component is equal to three, and the π€-component is equal to
negative one. This means that the magnitude of
vector π is equal to the square root of two squared plus three squared plus
negative one squared.

Two squared is equal to four. Three squared is equal to nine. Squaring a negative number gives us
a positive answer. Therefore, negative one squared is
one. As four plus nine plus one equals
14, the magnitude of vector π is the square root of 14.

In our next question, we will be
given the magnitude and need to calculate one of the components of the vector.

If vector π is equal to ππ’ plus
π£ minus π€ and the magnitude of vector π is equal to the square root of six, find
all the possible values of π.

Before starting this question, it
is worth noting that the wording says find all possible values of π. This suggests there will be more
than one correct answer. We are given two pieces of
information. We are told vector π is equal to
ππ’ plus π£ minus π€ and the magnitude of vector π is the square root of six. We know that for any vector written
in the form π₯π’ plus π¦π£ plus π§π€, then its magnitude is equal to the square root
of π₯ squared plus π¦ squared plus π§ squared. In this question, the square root
of six is equal to the square root of π squared plus one squared plus negative one
squared. This is because the π’-, π£-, and
π€-components are π, one, and negative one, respectively.

We can begin to solve this equation
by squaring both sides. As squaring is the inverse or
opposite of square rooting, the square root of six squared is equal to six. In the same way, the right-hand
side becomes π squared plus one squared plus negative one squared. Both one squared and negative one
squared are equal to one. Therefore, this simplifies to six
is equal to π squared minus two. We can then subtract two from both
sides of this equation so that π squared is equal to four.

Our final step is to square root
both sides. The square root of π squared is
π. The square root of four is equal to
two. But we must take the positive or
negative of this. Therefore, π is equal to positive
or negative two. The possible values of π such that
the magnitude of vector π is the square root of six are two and negative two. This is because when we square both
of these, we get an answer of four.

Our next few questions will also
involve the addition and subtraction of vectors.

Given that vector π plus vector π
is equal to negative two, four, three and vector π is equal to three, five, three,
determine the magnitude of vector π.

We recall that when adding two
vectors, we simply add the π’-, π£-, and π€-components separately. The vector π plus π is equal to
the vector π plus the vector π. If we let vector π have π’-, π£-,
and π€-components π₯, π¦, and π§, respectively, then negative two, four, three is
equal to three, five, three plus π₯, π¦, π§. We can then subtract vector π from
both sides of this equation. The left-hand side becomes negative
two, four, three minus three, five, three.

Negative two minus three is equal
to negative five. Therefore, our π’-component of
vector π is negative five. Four minus five is equal to
negative one, so the π£-component is negative one. Finally, three minus three is equal
to zero. Vector π is, therefore, equal to
negative five, negative one, zero.

We can calculate the magnitude of
this vector by squaring each of the components, finding their sum, and then square
rooting. The magnitude of vector π is equal
to the square root of negative five squared plus negative one squared plus zero
squared. Negative five squared is 25,
negative one squared is one, and zero squared is equal to zero. The magnitude of vector π is,
therefore, equal to the square root of 26.

In our next question, weβll find
the magnitude of a vector joining the endpoints of two other vectors.

Given that vector ππ is equal to
negative five π’ plus two π£ minus four π€ and vector ππ is equal to four π’ plus
four π£ plus six π€, determine the magnitude of vector ππ.

In this type of question, it is
worth drawing a diagram first. This will hopefully ensure that our
direction and signs are correct. We are given three points π΄, π΅,
and πΆ. We can join these to form a
triangle. In the question, we are given the
value of vector ππ. We are also given the value of
vector ππ. Our aim is to calculate the
magnitude of vector ππ. Therefore, our first step is to
work out vector ππ.

We can see that one way to get from
point π΄ to point πΆ is via point π΅. Therefore, vector ππ is equal to
vector ππ plus vector ππ. Vector ππ is, therefore, equal to
negative five π’ plus two π£ minus four π€ plus four π’ plus four π£ plus six
π€.

We can add two vectors by adding
the individual components separately. Negative five π’ plus four π’ is
equal to negative π’. Two π£ plus four π£ is equal to six
π£. Finally, negative four π€ plus six
π€ is equal to two π€. Vector ππ is equal to negative π’
plus six π£ plus two π€.

The magnitude of any vector can be
found by squaring the π’-, π£-, and π€-components, finding their sum, and then
square rooting the answer. This means that the magnitude of
vector ππ is the square root of negative one squared plus six squared plus two
squared. Negative one squared is equal to
one, six squared is 36, and two squared is equal to four. One, 36, and four sum to 41. Therefore, the magnitude of vector
ππ is the square root of 41.

In our final question, we will find
the magnitude of the difference of two vectors.

If vector π is equal to four π’
plus four π£ minus five π€ and vector π is equal to three π’ minus π€, determine
the magnitude of vector π minus vector π.

Our first step in this question is
to calculate vector π minus vector π. This is important as a common
mistake here would be to think that the magnitude of vector π minus vector π is
equal to the magnitude of vector π minus the magnitude of vector π. This, however, is not true.

To work out vector π minus vector
π, we simply subtract the individual components separately. Four π’ minus three π’ is equal to
one π’, or just π’. There is no π£-component in vector
π. Therefore, we are left with four
π£. Negative five π€ minus negative π€
is equal to negative four π€. This is because negative five minus
negative one is the same as negative five plus one, which is equal to negative
four.

We now need to calculate the
magnitude of this vector. We know that for any vector written
in the form π₯π’ plus π¦π£ plus π§π€, its magnitude is equal to the square root of
π₯ squared plus π¦ squared plus π§ squared. This means that the magnitude of
vector π minus vector π is the square root of one squared plus four squared plus
negative four squared.

Both four squared and negative four
squared are equal to 16, so we are left with the square root of one plus 16 plus
16. This is equal to the square root of
33. The magnitude of vector π minus
vector π is the square root of 33.

We will now summarize the key
points from this video. For any vector written in the form
π₯π’ plus π¦π£ plus π§π€, the magnitude of the vector is the square root of π₯
squared plus π¦ squared plus π§ squared. We square the individual components
of π’, π£, and π€, find their sum, and then square root the answer. The answer for the magnitude will
always be positive.

When adding or subtracting two
vectors, we add or subtract the individual components separately. We also saw that the magnitude of
the sum or difference of two vectors is not the same as the magnitude of their
individual parts. The magnitude of vector π plus
vector π is not equal to the magnitude of vector π plus the magnitude of vector
π.