Video Transcript
In this video, weβll learn how to
write vectors in component form using fundamental unit vectors.
We know that in two dimensions,
there are two components to a vector, called the π₯- and π¦-components. And thatβs where π₯ is the
horizontal component and π¦ is the vertical component. Given the components of a vector in
two dimensions, one way we can express it is in component form. For example, if the π₯- and
π¦-components are π and π, respectively, the component form of the vector is π,
π. And we can represent the vector in
the Cartesian plane by an arrow starting at the origin and ending at the point π,
π, as shown.
But there is another way to express
a vector in two dimensions, and thatβs in terms of fundamental unit vectors. These are unit vectors denoted π’
and π£ with nonnegative components and length one, where π’ is the unit vector in
the horizontal direction and π£ is the unit vector in the vertical direction. π’ has components one, zero, and π£
has components zero, one.
We write π’ and π£ with hats on to
distinguish them as the fundamental unit vectors. But you may also see them written
with an arrow above or emphasized in bold. Note that the fundamental unit
vectors have one nonzero component, which for both vectors is equal to one. From the component form of π’, we
can represent it by an arrow starting from a given origin and ending at the point
one, zero, which is on the positive π₯-axis. Similarly, from the component form
of π£, we can represent it in the Cartesian plane as a vertical arrow of length one
starting at a given origin. So, π’ and π£ lie parallel to the
π₯- and π¦-axes, respectively, pointing in the positive directions for each
respective axis.
Letβs look at an example now of how
we might express a vertical vector in terms of the fundamental unit vectors.
Given that the vector π has
components zero and two, express the vector π in terms of the unit vectors π’
and π£.
Weβre given a vector in
component form, which we need to express in terms of the fundamental unit
vectors π’ and π£. To do this, we begin by
recalling that we can represent a vector with components π and π by an arrow
starting from the origin and ending at the point with coordinates π, π. Since the component form of our
vector π is zero, two, we represent it by an arrow from the origin to the point
zero, two. So how do we express this in
terms of our fundamental unit vectors π’ and π£?
Well, recall the component
forms of π’ and π£. π’ is equal to one, zero and π£
is equal to zero, one. And theyβre represented in the
Cartesian plane as shown. In particular, π’ is a
horizontal vector of length one, while π£ is a vertical vector also of length
one. Since our vector π is purely
vertical, the π₯-component being zero, to express π in terms of the fundamental
unit vectors, we need only use the vector π£. Since the π¦-component of π is
two, we can achieve this by stacking two copies of π£ on top of each other to
produce the vector π. This tells us that π is equal
to π£ plus π£, which is two π£. Hence, in terms of the
fundamental unit vectors π’ and π£, the vector π is equal to two π£.
In this example, we expressed a
vertical vector in two dimensions in terms of the fundamental unit vector π£. Now letβs see how to express a
vector thatβs neither vertical nor horizontal in terms of the unit vectors.
The given figure shows a vector
π in a plane. Express this vector in terms of
the unit vectors π’ and π£.
Weβre asked to express the
vector shown in the graph in terms of the fundamental unit vectors π’ and
π£. So letβs remind ourselves what
these look like in the Cartesian plane. π’ is a horizontal vector and
π£ is a vertical vector, and they both start at a given origin traveling in the
positive π₯- and π¦-directions, respectively, with length equal to one.
Now, to express the vector π
in terms of π’ and π£, we need to consider its π₯- and π¦-components
separately. So letβs begin with the
π₯-component. From the graph, we see that the
π₯-component is equal to negative three. Using the horizontal unit
vector π’, which is going in the positive π₯-direction, we flip this unit vector
so it becomes negative, as itβs now pointing in the negative π₯-direction. And to reach π₯ is negative
three, weβre adding three copies of negative π’. This tells us that the
π₯-component of π in terms of the fundamental unit vector π’ is negative three
π’.
Next, considering the
π¦-component of π, we see that this is positive two. And we can produce this by
adding two copies of π£. Hence, the π¦-component of
vector π in terms of the fundamental unit vector π£ is two π£.
So now adding the two
components together, where by convention we write the π₯-component first, we
have π equals negative three π’ plus two π£.
We can apply the method we used in
this example to any vector with integer-valued components. This leads to a general formula for
expressing a vector with components π, π in terms of the fundamental unit
vectors. Thatβs π times π’ plus π times
π£. Note that this conversion also
works in reverse. Starting from a vector in terms of
the fundamental unit vectors, so given ππ’ plus ππ£, we can covert this to the
vector with components π and π.
Weβve shown how to convert between
component and unit vector form for integer-valued components of vectors. But actually, this works for any
real-valued components. In our next example, weβll use this
to express a two-dimensional vector in terms of fundamental unit vectors.
Express the vector π equals
negative five over two, negative 19 using the unit vectors π’ and π£.
Weβre given a vector in
component form, and we want to express this in terms of the fundamental unit
vectors π’ and π£. To do this, we recall that a
two-dimensional vector with components π and π can be written in terms of the
fundamental unit vectors as π times π’ plus π times π£. Since our given vector π has
components π equal to negative five over two and π equal to negative 19, we
can write π as negative five over two π’ plus negative 19π£. Hence, in terms of the
fundamental unit vectors π’ and π£, the vector π is equal to negative five over
two π’ minus 19π£.
In this example, we expressed a
two-dimensional vector given in component form in terms of the fundamental unit
vectors. Letβs see now how to achieve this
when our vector is specified by its beginning and endpoints in the Cartesian
plane. One way to write this vector in
terms of the fundamental unit vectors would be to first find the coordinate form of
the vector and then convert the form. But we donβt have to go through the
coordinate form to achieve this. Instead, we can just identify the
horizontal and vertical components of this vector so that we can write it as the sum
of horizontal and vertical vectors. Letβs see how this works in an
example.
The given figure shows a vector
in a plane. Express this vector in terms of
the unit vectors π’ and π£.
Weβre asked to write the vector
shown in the graph in terms of the fundamental unit vectors π’ and π£. And we recall that the unit
vectors are defined by π’ equals the vector with components one, zero and π£ is
the vector with components zero, one. In other words, these are unit
vectors pointing in the positive horizontal and vertical directions of the π₯-
and π¦-axes, respectively. These are the fundamental unit
vectors.
Note that unit vectors need not
necessarily start at the origin. They describe moving a distance
of one in either the horizontal or vertical direction from a given initial
point. So we need to express the given
vector as a sum of horizontal and vertical unit vectors. Letβs first identify the
relevant horizontal and vertical vectors on our graph.
First, our initial point is the
point two, negative two. And starting from here, we see
that the horizontal vector spans two grid lengths in the positive
π₯-direction. So the vectorβs horizontal
component is equal to positive two, which we can write as the vector with
components two, zero. And this can be written as two
times the vector one, zero, which is two times the fundamental unit vector
π’. Similarly, the vertical vector
spans 10 grid lengths and points in the positive π¦-direction from our initial
point so that its vertical component is positive 10. This gives us that the vertical
vector is 10 times the fundamental unit vector π£. Adding these two vectors
together will produce the given vector. Hence, the given vector is
equal to two π’ plus 10π£.
In this example, we expressed a
graphed vector on a grid in terms of the fundamental unit vectors. This is a perfectly valid
method. But to use it, we need to graph the
points on the coordinate plane. Itβs useful to know the formula for
achieving this when we only have the coordinates of the initial point and the
endpoint of the vector.
Letβs consider a vector from
initial point π΄ with coordinates π₯ one, π¦ one to endpoint π΅ with coordinates π₯
two, π¦ two. In this case, the π₯-component of
the vector ππ is π₯ two minus π₯ one and the π¦-component is π¦ two minus π¦
one. We can then write the vector ππ
from initial point π΄ to endpoint π΅ in terms of the fundamental unit vectors as π₯
two minus π₯ one π’ plus π¦ two minus π¦ one π£.
Itβs worth noting here that itβs
very important weβre clear which is the initial and which is the endpoint of a
vector. Here, our initial point is π΄ and
the endpoint is π΅. However, if we start from π΅ and go
in the opposite direction from π΅ to π΄, the vector ππ has components π₯ one minus
π₯ two and π¦ one minus π¦ two, which are not the same as those of vector ππ in
the opposite direction.
In our final example, letβs apply
this formula to write a two-dimensional vector in terms of the fundamental unit
vectors π’ and π£.
Given that π΄ has coordinates
two, three and π΅ has coordinates five, nine, express the vector ππ in terms
of the unit vectors π’ and π£.
In this example, given the
initial and endpoint of a vector in two dimensions, we want to express the
vector in terms of the fundamental unit vectors π’ and π£. To do this, we recall that a
vector ππ with initial point π΄ π₯ one, π¦ one and endpoint π΅ π₯ two, π¦ two
can be written in terms of the fundamental unit vectors as ππ is equal to π₯
two minus π₯ one times π’ plus π¦ two minus π¦ one times π£.
Weβre given the points π΄ and
π΅. And since π΄ is the initial
point and π΅ the endpoint, we have π₯ one equals two and π₯ two equals five, π¦
one equals three and π¦ two equals nine. Using the formula gives us ππ
is equal to five minus two times π’ plus nine minus three times π£, that is,
three π’ plus six π£. Hence, in terms of the
fundamental unit vectors, the vector with initial point π΄ two, three and
endpoint π΅ five, nine is ππ equals three π’ plus six π£.
Letβs now complete this video by
recapping a few important points weβve covered. The fundamental unit vectors π’ and
π£ are horizontal and vertical vectors of length one pointing in the positive π₯-
and π¦-directions, respectively. The component form of π’ is the
vector one, zero, and the component form of π£ is the vector zero, one. A vector in component form π, π
can be written in terms of unit vectors π’ and π£ as π times π’ plus π times
π£. Similarly, given a vector in terms
of the fundamental unit vectors, we can write this in component form π, π. And finally, a vector ππ with
initial point π΄ π₯ one, π¦ one and endpoint π΅ π₯ two, π¦ two can be written in
terms of the fundamental unit vectors as ππ equals π₯ two minus π₯ one times π’
plus π¦ two minus π¦ one times π£.