### Video Transcript

Given that π is the vector five over two, two, express the vector π in terms of the unit vectors π’ and π£, and find its norm the magnitude of π.

In this question, weβre given a vector π and weβre given it in terms of its components. We need to express this vector in terms of the unit directional vectors π’ and π£. And we also need to find the norm of our vector π. And weβre given the notation for this. Itβs written the absolute value of vector π. To answer this question, weβre first going to need to recall what it means to express the vector in terms of its components. And because vectors can represent a lot of different things, there are a lot of different ways of thinking about this. For example, we often think of vectors graphically representing a distance traveled. However, vectors donβt always represent this. Itβs just an easy way to think about them.

Vectors are just an object with magnitude and direction. And the components of our vector help us determine the magnitude and direction of our vector. Each component of our vector will represent a different direction. And the value of this component will tell us the magnitude in that direction. In this case, the vector π, π will be ππ’ plus ππ£. And sometimes, we hear the value of π called the π’-component of our vector. And sometimes we hear the value of π called π£-component of our vector.

We can directly apply this to express the vector π in terms of π’ and π£. The π’-component of vector π will be five over two and the π£-component will be two. So the coefficient of π’ should be five over two and the coefficient of π£ should be two. So vector π is five over two π’ plus two π£. So that gives us the answer to the first part of our question. But we also need to find the norm of our vector π. And thereβs a few different ways of doing this.

For example, we know if we graphically represent our vector, then the norm of our vector will be the length. And this would work. However, because weβve given π in terms of its components, thereβs actually an easier way. We know the norm or magnitude of the vector π, π is going to be equal to the square root of the sum of the squares of its components.

The norm of the vector π, π is the square root of π squared plus π squared. So we can instead just use this to find the norm of our vector π. We need to find the square root of the sum of the squares of its components. The norm of vector π is the square root of five over two all squared plus two squared. Simplifying this, we get the square root of 25 over four plus four. Next, by writing four as 16 divided by four, we can add the two terms inside of our square root together. We get the square root of 41 over four.

Now, instead of taking the square root of both the numerator and the denominator together, weβll take the square root of each separately. This gives us the square root of 41 divided by the square root of four. And of course, we can evaluate the square root of four; itβs equal to two. And this gives us that the norm of vector π is equal to root 41 divided by two. Therefore, we were able to show if π is the vector five over two, two, then in terms of the unit directional vectors π’ and π£, π is equal to five over two π’ plus two π£ and the norm of vector π is root 41 over two.