### Video Transcript

Given that π is the vector two π’ plus π£ minus π€ and π is the vector π£ plus π€, find the unit vector in the direction of π minus π.

In this question, weβre given two vectors π and π, both given in terms of the unit directional vectors π’, π£, and π€. We need to determine the unit vector which points in the same direction as the vector π minus π. To answer this question, letβs start by recalling exactly what we mean by a unit vector. We call any vector with magnitude one a unit vector. So, when the question is asking us for a unit vector pointing in the direction of π minus π, we just need to find a vector pointing in the same direction as π minus π, which has a magnitude of one.

Now thereβs a few different ways of doing this. For example, we could do this graphically. We could sketch the vector π and the vector π. Then, we could find the vector π minus π on our sketch and make its magnitude equal to one because, remember, graphically, the magnitude of the vector is the length of the vector. And this would work. However, thereβs a simpler method involving what we know about unit vectors.

We recall for any vector π which is not the zero vector, we can construct a vector π hat which points in the same direction as π; however, it has magnitude one. We do this by multiplying π by one over its magnitude. We can use this formula to construct a vector pointing in the same direction as π minus π with unit magnitude. So to answer our question, all we need to do is calculate the following expression. One over the magnitude of π minus π multiplied by the vector π minus π. This will have unit magnitude and point in the same direction as π minus π. So thereβs two things we need to find. We need to find π minus π and then find its magnitude. Letβs start by finding π minus π.

To do this, we use the definitions in our question. We have the vector π is the vector π£ plus π€ and the vector π is two π’ plus π£ minus π€. And we need to subtract π from our vector π. This gives us the following expression for π minus π. And thereβs several different ways of doing this. Weβre going to distribute the negative over our parentheses. This gives us π£ plus π€ minus two π’ minus π£ plus π€. Now, all we need to do is collect like terms. For example, we can see we have π£ minus π£. This is going to be equal to zero. And we also have π€ plus π€, which is going to be equal to two π€. So by collecting like terms and simplifying, we get negative two π’ plus zero π£ plus two π€. And of course, we didnβt need to include the term zero π£. However, it can be useful to leave this in.

So weβve now found an expression for vector π minus π. However, we still need to find its magnitude. This means we want to find the magnitude of negative two π’ plus zero π£ plus two π€. And thereβs two different ways we could do this. We could sketch our vector and then remember that the magnitude of a vector weβve sketched will be the length of the vector. And we could then find this length by using the Pythagorean theorem.

However, itβs easier to just use the following formula. The magnitude of any vector will be the square root of the sum of the squares of its components. So the magnitude of the vector ππ’ plus ππ£ plus ππ€ will be the square root of π squared plus π squared plus π squared. In our case, the components of this vector are negative two, zero, and two. So its magnitude will be the square root of negative two squared plus zero squared plus two squared. And we can evaluate this. Itβs equal to the square root of eight, which we know we can simplify to give us two times the square root of two.

Now weβve found an expression for π minus π, and weβve managed to find its magnitude. So we can substitute both of these into our expression to find the unit vector pointing in the same direction as π minus π. Substituting these in, we get one over two root two multiplied by negative two π’ plus zero π£ plus two π€. And we can simplify this. First, weβre not going to include the zero π£.

Next, we have a few options. We could distribute one over two root two over our parentheses. However, we wonβt do this. Instead, all weβre going to do is rationalize our denominator. And to do this, we just need to multiply both the numerator and denominator by the square root of two. And by evaluating this and simplifying, we got our final answer of root two over four times negative two π’ plus two π€. Therefore, we were able to show that given π is the vector two π’ plus π£ minus π€ and π is the vector π£ plus π€, then the unit vector in the same direction of the vector π minus π is root two over four times negative two π’ plus two π€.