### Video Transcript

In this video, we will learn how to
perform operations on vectors in three dimensions such as addition, subtraction, and
scalar multiplication. We will begin by recalling some key
properties of vectors in 3D. We know that a vector in 3D space
has magnitude and direction. If we consider the
three-dimensional coordinate grid as shown, then the unit vectors in the π₯-, π¦-,
and π§-direction are denoted by π’, π£, and π€, respectively. Any vector can therefore be written
in the form π₯π’ plus π¦π£ plus π§ or π§π€. The values of π₯, π¦, and π§ will
be the number of units traveled in those directions. If the point π΄ has coordinates π₯,
π¦, π§, then vector π is equal to π₯π’ plus π¦π£ plus π§π€. This can also be written in
component form as shown.

We will now consider how we can add
and subtract vectors in three dimensions. If we consider two vectors π and
π with components π₯ one, π¦ one, π§ one and π₯ two, π¦ two, π§ two, respectively,
then we can add and subtract the two vectors by considering their corresponding
components. Vector π plus vector π will have
components π₯ one plus π₯ two, π¦ one plus π¦ two, and π§ one plus π§ two. Likewise, π minus π will have
components π₯ one minus π₯ two, π¦ one minus π¦ two, and π§ one minus π§ two. This method allows us to add or
subtract two or more vectors. We will now look at a couple of
examples.

If vector π is equal to negative
five π’ minus eight π£ plus six π€ and vector π is equal to four π’ minus three π£
plus 13π€, find vector π minus vector π.

We recall that if vector π is
equal to π₯ one π’ plus π¦ one π£ plus π§ one π€ and vector π is equal to π₯ two π’
plus π¦ two π£ plus π§ two π€, then vector π minus vector π is equal to π₯ one
minus π₯ two π’ plus π¦ one minus π¦ two π£ plus π§ one minus π§ two π€. In this question, we need to
subtract four π’ minus three π£ plus 13π€ from negative five π’ minus eight π£ plus
six π€. Subtracting four π’ minus three π£
plus 13π€ is the same as minus four π’ plus three π£ minus 13π€. We can then collect the like
terms. Negative five π’ minus four is
equal to negative nine π’. Negative eight π£ plus three π£ is
equal to negative five π£. Finally, six π€ minus 13π€ is equal
to negative seven π€.

Vector π minus vector π is equal
to negative nine π’ minus five π£ minus seven π€.

We will now look at an addition
problem.

Given the two vectors π is equal
to negative two, negative three, zero and π is equal to negative three, three,
negative two, find vector π plus vector π.

In order to add any two vectors in
three dimensions, we simply add their corresponding components. This means that if vector π has
components π₯ one, π¦ one, π§ one and vector π has components π₯ two, π¦ two, π§
two, then vector π plus vector π will have components π₯ one plus π₯ two, π¦ one
plus π¦ two, and π§ one plus π§ two. In this question, we add the
π₯-components negative two and negative three. We add the π¦-components negative
three and positive three. Finally, we add π§- or
π§-components zero and negative two. Negative two plus negative three is
the same as negative two minus three. This is equal to negative five. Negative three plus three is equal
to zero. Finally, zero plus negative two is
the same as zero minus two, which equals negative two.

If vector π is equal to negative
two, negative three, zero and vector π is equal to negative three, three, negative
two, then vector π plus vector π is equal to negative five, zero, negative
two.

We will now look at what happens
when we multiply a vector by a scalar. When we multiply any vector by a
scalar, we multiply each of the components of the vector by the scalar. This means that if vector π has
components π₯, π¦, and π§, then multiplying this by the scalar or constant π gives
us a vector with components ππ₯, ππ¦, and ππ§. We will now look at a question
involving scalar multiplication.

What is the vector that results
from scaling the vector π equal to negative six, negative three, negative one by a
factor of negative six?

When we multiply any vector by a
scalar, we simply multiply each of the components of the vector by the scalar. In this question, we want to
multiply the vector negative six, negative three, negative one by the scalar or
constant negative six. We recall that when multiplying two
negative numbers, we get a positive answer. This means that negative six
multiplied by negative six is positive 36. Negative six multiplied by negative
three is equal to 18. And negative six multiplied by
negative one is equal to six. If vector π is equal to negative
six, negative three, negative one, multiplying this by the scalar negative six gives
us the vector 36, 18, six.

Our next question involves both
multiplying by a scalar and subtracting two vectors.

If vector π is equal to negative
eight, nine, nine and vector π is equal to negative six, four, nine, find
two-fifths of vector π minus four-fifths of vector π.

We will begin this question by
calculating two-fifths of vector π. When multiplying any vector by a
scalar or constant, we simply multiply each of the components by that scalar. In this question, we need to
multiply the components negative eight, nine, and nine by two-fifths. Negative eight multiplied by
two-fifths is equal to negative sixteen-fifths. Nine multiplied by two-fifths is
equal to eighteen-fifths. This means that two-fifths of
vector π has components negative sixteen-fifths, eighteen-fifths, and
eighteen-fifths. We can repeat this process to
calculate four-fifths of vector π. We need to multiply the vector
negative six, four, nine by four-fifths. This gives us the vector with
components negative twenty-four fifths, sixteen-fifths, and thirty-six fifths.

Our next step is to subtract
four-fifths of vector π from two-fifths of vector π. When subtracting two vectors, we
simply subtract their corresponding components. We begin by subtracting negative
twenty-four fifths from negative sixteen-fifths. Negative 16 minus negative 24 is
equal to positive eight. Therefore, the π₯-component is
eight-fifths. Subtracting sixteen-fifths from
eighteen-fifths gives us two-fifths. The π¦-component of our vector is
two-fifths. Finally, we need to subtract
thirty-six fifths from eighteen-fifths. As 18 minus 36 is negative 18, the
π§ or π§-component is negative eighteen-fifths. Two-fifths of vector π minus
four-fifths of actor π is equal to eight-fifths, two-fifths, negative
eighteen-fifths.

Whilst we donβt need to in this
question, we could factor out one-fifth, giving us the scalar one-fifth multiplied
by the vector eight, two, negative 18.

In our next question, we will find
the missing vector in the vector expression.

If vector π is equal to negative
one, one, one and vector π is equal to one, one, negative two, determine the vector
π for which two π plus five π is equal to five π.

There are lots of ways of
approaching this problem. One way would be to begin by
rearranging our equation to make vector π the subject. We can do this by firstly
subtracting five π from both sides. This gives us two π is equal to
five π minus five π. We can then divide both sides of
this equation by two such that vector π is equal to five π minus five π divided
by two. The right-hand side can be
rewritten as one-half of five π minus five π.

We can now calculate five
multiplied by vector π and five multiplied by vector π. When multiplying any vector by a
scalar, we simply multiply each of the individual components by that scalar. To calculate five π, we multiply
the vector one, one, negative two by five. This gives us the vector five,
five, negative 10. In the same way, five multiplied by
vector π is equal to five multiplied by the vector negative one, one, one. This is equal to negative five,
five, five.

We can now subtract these two
vectors. And we do this by subtracting the
corresponding components. Five minus negative five is equal
to 10. Five minus five is equal to
zero. And negative 10 minus five is equal
to negative 15. Five π minus five π is equal to
10, zero, negative 15. We know that vector π is equal to
a half of this. It is equal to a half of the vector
10, zero, negative 15. A half of 10 is equal to five, a
half of zero is zero, and a half of negative 15 is negative 15 over two or negative
fifteen-halves. The vector π for which two π plus
five π is equal to five π is five, zero, negative 15 over two.

In our final question, we will also
consider the magnitude of vectors in three dimensions.

π and π are two vectors, where
vector π is equal to negative one, five, negative two and vector π is equal to
three, one, one. Comparing the magnitude of vector
π minus vector π and the magnitude of vector π minus the magnitude of vector π,
which quantity is larger?

We recall that the magnitude of any
vector π with components π₯, π¦, and π§ is equal to the square root of π₯ squared
plus π¦ squared plus π§ squared. We find the sum of the squares of
the individual components and then square root the answer. This means that the magnitude of
vector π is equal to the square root of negative one squared plus five squared plus
negative two squared. This is equal to the square root of
one plus 25 plus four. The magnitude of vector π is equal
to the square root of 30. We can repeat this process for
vector π. This is equal to the square root of
three squared plus one squared plus one squared, which simplifies to the square root
of nine plus one plus one, which is the square root of 11. The magnitude of vector π is the
square root of 11.

Before calculating a numerical
value for the magnitude of vector π minus the magnitude of vector π, we will
firstly work out vector π minus vector π. We know that to subtract two
vectors, we subtract the corresponding components. Negative one minus three is equal
to negative four. Five minus one is equal to positive
four. And negative two minus one is equal
to negative three. Vector π minus vector π is equal
to negative four, four, negative three. The magnitude of this is equal to
the square root of negative four squared plus four squared plus negative three
squared. This simplifies to the square root
of 16 plus 16 plus nine. The magnitude of vector π minus
vector π is therefore equal to the square root of 41.

We now need to calculate the
decimal values of the magnitude of vector π minus vector π and the magnitude of
vector π minus the magnitude of vector π. The square root of 41 is equal to
6.4031 and so on. The square root of 30 minus the
square root of 11 is equal to 2.1606 and so on. This means that the square root of
41 is greater than the square root of 30 minus the square root of 11. We can therefore conclude that the
magnitude of vector π minus vector π is greater than the magnitude of vector π
minus the magnitude of vector π. The larger quantity is the
magnitude of vector π minus vector π.

We will now summarize the key
points from this video. We saw in this video that to add or
subtract vectors in three dimensions, we simply add or subtract the corresponding
components. If vector π has components π₯ one,
π¦ one, π§ one and vector π has components π₯ two, π¦ two, π§ two, then to
calculate vector π plus vector π, we add the components and to calculate vector π
minus vector π, we subtract the corresponding components. We also saw that to multiply a
vector by a scalar, we multiply each component of the vector by the scalar. If vector π has components π₯, π¦,
and π§, then multiplying this by the scalar π gives us the vector with components
ππ₯, ππ¦, and ππ§.