### Video Transcript

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£.