### Video Transcript

In this video, we will learn how to
multiply a vector by a scalar and how to find a unit vector in the direction of any
given vector by dividing a vector by a scalar. We will consider both of these in
both two and three dimensions.

Letβs begin by recalling what we
mean by scalar multiplication. Multiplying a vector by a scalar or
real number is called scalar multiplication. To perform scalar multiplication,
you need to multiply the scalar by each component of the vector. Letβs consider the vector π with
components π, π, and π. Now, letβs imagine we want to
multiply this vector by the constant or scalar π. We perform this in the same way as
if we were distributing parentheses or expanding brackets. The vector ππ has components
ππ, ππ, and ππ.

When multiplying a vector by a
scalar, the result is also a vector. We know that any vector has both
magnitude and direction. Multiplying a vector by a positive
number other than one changes the magnitude but does not change the direction. Multiplying a vector by negative
one reverses the direction but doesnβt change the magnitude. Multiplying by any other negative
number reverses the direction and changes the magnitude of the vector.

We will now look at a couple of
examples where we need to multiply a vector by a scalar.

Given that vector π is equal to
negative one, negative eight, find three π.

Multiplying a vector by any real
number, as in this case, is known as scalar multiplication. In order to perform scalar
multiplication, we multiply each component of the vector by the scalar. In this question, we need to
multiply the vector negative one, negative eight by three. Multiplying a positive number by a
negative number gives a negative answer. Therefore, three multiplied by
negative one is negative three. Three multiplied by negative eight
is equal to negative 24. Three multiplied by the vector π
or the vector three π is equal to negative three, negative 24.

In our next question, we will
consider what happens when we multiply a vector by a scalar graphically.

Vector π is represented by the
following graph. Which of the following graphs
represents negative two π?

Vector π goes from the origin to
the point one, one. This means that it has an
π₯-component of one and a π¦-component of one. Vector π is equal to one, one. We want to multiply vector π by
negative two. We recall that when multiplying a
vector by a scalar, we need to multiply each of the individual components by the
scalar. Multiplying negative two by one
gives us negative two. Therefore, the vector that
corresponds to negative two π is negative two, negative two.

Letβs now consider the five options
we are given and what vectors they represent. Graph (A) goes from the origin to
negative two, negative two. This means it corresponds to the
vector negative two, negative two. This is the same as negative two
π, which suggests this is the correct graph.

Graph (B) shows the vector one,
negative two. Graph (C) shows the vector one,
two. Graph (D) shows the vector one,
0.5. And graph (E) shows the vector 0.5,
0.5. This confirms that graph (A) does
indeed represent negative two π. This leads us to a key rule when
multiplying by negative scalars. When we multiply any vector by a
negative scalar other than negative one, the vector will change direction and
magnitude. Multiplying a vector by negative
two, as in this case, will double the magnitude, and the direction of the vector
will be the opposite of the original direction. This can be shown on the coordinate
plane by the green arrow.

The remainder of the questions in
this video will deal with unit vectors. We will begin by defining what we
mean by a unit vector.

A unit vector is a vector which has
a magnitude of one. We recall that a three-dimensional
vector with components π, π, π has a magnitude equal to the square root of π
squared plus π squared plus π squared. The absolute value bars denote the
magnitude of a vector. In order to find the unit vector
with the same direction as a given vector, we divide by the magnitude of the given
vector. The unit vector of π is written π
hat. This is equal to one over the
magnitude of π multiplied by vector π.

Letβs consider the vector π with
components four, three. The magnitude of vector π is equal
to the square root of four squared plus three squared. Four squared is equal to 16, and
three squared is equal to nine. This means that the magnitude of
vector π is equal to the square root of 25. We would usually leave this in
radical or surd form. But as the square root of 25 is an
integer, we can simplify so that the magnitude of vector π is equal to five.

The unit vector π is therefore
equal to one-fifth multiplied by the vector four, three. We recall that when multiplying a
vector by a scalar, we multiply each component individually by the scalar. Therefore, the unit vector π is
equal to four-fifths, three-fifths.

We will now look at some examples
where we need to calculate unit vectors.

Consider the vector π five π’
minus two π£ minus four π€. Is the unit vector in the direction
of π the same as the unit vector in the direction of three π?

We know that vector π can be
rewritten in the form five, negative two, negative four. We are also asked to consider the
vector three π. This involves multiplying vector π
by the scalar or constant three. This involves multiplying each
component by the scalar three. Three multiplied by five is 15. Three multiplied by negative two is
negative six. And three multiplied by negative
four is negative 12. This means that three π is equal
to 15, negative six, negative 12.

The unit vector π hat is equal to
one over the magnitude of vector π multiplied by vector π. This is the same as dividing the
vector by its magnitude. We know that to calculate the
magnitude of any vector, we find the square root of the sum of the squares of each
component. The magnitude of vector π is equal
to the square root of five squared plus negative two squared plus negative four
squared. This is equal to the square root of
25 plus four plus 16. This simplifies to the square root
of 45, which in turn is equal to three root five.

The magnitude of vector π is three
root five. The unit vector in the direction of
vector π is therefore equal to one over three root five multiplied by five,
negative two, negative four. We could rewrite this as the vector
five over three root five, negative two over three root five, and negative four over
three root five.

Letβs now consider the unit vector
in the direction of three π. The magnitude of vector three π is
equal to the square root of 15 squared plus negative six squared plus negative 12
squared. This is equal to the square root of
405, which in turn simplifies to nine root five. The magnitude of vector three π is
equal to nine root five.

We might notice at this point that
this is three times the magnitude of vector π. This leads us to a general
rule. The magnitude of ππ is equal to
π multiplied by the magnitude of vector π. This means that three multiplied by
the magnitude of vector π is equal to three multiplied by three root five. The unit vector in the direction of
three π is therefore equal to one over nine root five multiplied by the vector 15,
negative six, negative 12.

Factoring out a three from the
vector gives us three over nine root five multiplied by the vector five, negative
two, negative four. This is the same as one over three
root five multiplied by the vector five, negative two, negative four. We can therefore conclude that the
answer is yes, the unit vector in the direction of π is the same as the unit vector
in the direction of three π.

This will be true of any vector
multiplied by a positive scalar. As long as our value of π is
positive, then the unit vector in the direction of π will be the same as the unit
vector in the direction of ππ.

In our next question, we will find
the unit vector in the same direction as a two-dimensional vector.

Find the unit vector in the same
direction as the vector negative three π’ plus five π£.

We know that the unit vector π hat
is equal to one over the magnitude of vector π multiplied by vector π, where the
magnitude of a two-dimensional vector with components π and π is equal to the
square root of π squared plus π squared.

In this question, we have a vector
with π’ and π£ components negative three and five. The magnitude of this vector is
equal to the square root of negative three squared plus five squared. Negative three squared is equal to
nine, and five squared is equal to 25. This means that the magnitude of
vector π is equal to root 34. The unit vector π is therefore
equal to one over root 34 multiplied by negative three, five.

When multiplying any vector by a
scalar, we multiply each individual component by the scalar. This gives us negative three over
root 34, five over root 34. We can rationalize the denominator
of one over root 34 by multiplying the numerator and denominator by root 34. This means that one over root 34 is
equal to root 34 over 34. This is true of any radical. One over root π is equal to root
π over π.

We can therefore rewrite our two
components as negative three root 34 over 34 and five root 34 over 34. Rewriting this in terms of π’ and
π£, the unit vector in the same direction as the vector negative three π’ plus five
π£ is equal to negative three root 34 over 34 π’ plus five root 34 over 34 π£.

In our final question, we will use
a combination of scalar multiplication and unit vectors.

Given vector π is equal to two,
zero, negative two and vector π is equal to one, negative one, one, determine the
unit vector in the direction two π minus π.

Our first step here is to calculate
the vector two π minus π. We do this by multiplying vector π
by the scalar two and then subtracting vector π. When multiplying a vector by a
scalar, we multiply each of the components by the scalar. Therefore, two π is equal to two,
negative two, two.

We need to subtract the vector two,
zero, negative two. We do this by subtracting each
component individually. Two minus two is equal to zero. Negative two minus zero is equal to
negative two. Finally, two minus negative two is
equal to four, as subtracting negative two is the same as adding two.

We need to find the unit vector of
this. We know that the unit vector π hat
is equal to one over the magnitude of π multiplied by vector π. The magnitude of two π minus π is
equal to the square root of zero squared plus negative two squared plus four
squared. This is equal to the square root of
20, which simplifies to two root five. The unit vector is therefore equal
to one over two root five multiplied by zero, negative two, four. We then multiply each individual
component by one over two root five. Multiplying this by zero gives us
zero. Negative two multiplied by one over
two root five is equal to negative one over root five.

Rationalizing the denominator by
multiplying the numerator and denominator by root five gives us negative root five
over five. Finally multiplying four by one
over two root five gives us two root five over five. The unit vector in the direction
two π minus π has component zero, negative root five over five, and two root five
over five.

We will now summarize the key
points from this video. When multiplying a vector by a
scalar, we multiply each component of the vector by the scalar. This means that π multiplied by
the vector π, π is equal to the vector ππ, ππ. This works in both two and three
dimensions. The magnitude of any vector is
denoted by absolute value bars. The magnitude of a two-dimensional
vector π, π is equal to the square root of π squared plus π squared.

We also found that a unit vector
has magnitude one in the same direction as the given vector. The unit vector denoted π hat is
equal to one over the magnitude of π multiplied by vector π. Another way of saying this is that
the unit vector is equal to the given vector divided by its magnitude.