### Video Transcript

Two vectors π and π have the same length and are perpendicular to each other. The vector product of π and π has a magnitude of nine. What is the length of each vector?

Okay, so this is a question about vector products. In this question, weβre told the magnitude of the vector product of two vectors π and π. And weβre told that these vectors both have the same length and that they are perpendicular to each other. Weβre asked to find the length of each vector. Letβs start by taking the information that the question gives us about the vectors π and π and trying to use this to write expressions for π and π in component form.

We arenβt told anything about the absolute orientation of either vector, only that they are perpendicular to each other. So, we are free to orient them as we wish. The absolute direction in space isnβt going to affect our answer. And generally, in this case, itβs easiest to choose to orient the vectors along coordinate axes for the purpose of doing the calculation. So, letβs imagine that our vector π points along the π₯-axis and our vector π points along the π¦-axis. This fulfills the requirement from the question that the two vectors are perpendicular to each other.

We can then write our vectors π and π in component form. Recall that π’ is the unit vector in the π₯-direction, so the vector π can be written as the magnitude of π multiplied by the unit vector π’, and π£ is the unit vector in the π¦-direction, so the vector π can be written as the magnitude of π multiplied by the unit vector π£. But actually, the questionβs also given us another piece of information about the vectors π and π. Weβve been told that these two vectors have the same length.

Now, the length and the magnitude of a vector are equivalent. So, this means that the magnitude of vector π must equal the magnitude of vector π. So, letβs set the magnitude of π and the magnitude of π equal to the same value, which weβll call π for magnitude. Then we have that vector π is equal to π multiplied by unit vector π’, and we have that π is equal to π multiplied by unit vector π£. So, we now have expressions for both vector π and vector π in component form. So, weβve achieved our step one.

So, what next? Well, the other bit of information that the question has given us is about the vector product of π and π. Weβre told that this vector product has a magnitude of nine. So, a sensible next step would be to calculate the vector product of π and π. And then from there, our final step will be to use the vector product to find the length π of each of the vectors π and π. So, letβs recall the definition of the vector product of two vectors.

First, weβll define two general vectors in the π₯π¦-plane, capital π and capital π. We can write these in component form as follows. Capital π is equal to an π₯-component capital π΄ subscript π₯ multiplied by the unit vector π’ plus a π¦-component capital π΄ subscript π¦ multiplied by the unit vector π£ and likewise, for capital π, with an π₯-component capital π΅ subscript π₯ and a π¦-component capital π΅ subscript π¦. Then, the vector product π cross π is equal to the π₯-component of π΄ multiplied by the π¦-component of π΅ minus the π¦-component of π΄ multiplied by the π₯-component of π΅ all multiplied by π€, which is the unit vector in the π§-direction.

So, letβs apply this to our vectors from the question, lowercase π and π. We can make it a little more clear whatβs going on if we explicitly include both π₯- and π¦-components for both vectors. Vector π only has an π₯-component. But we can rewrite this as π equals π multiplied by π’ plus zero multiplied by π£ to explicitly include the π¦-component of zero. And we can do the same for vector π which only has a π¦-component. But we can explicitly include an π₯-component of zero and rewrite it as zero multiplied by π’ plus π multiplied by π£.

So now, we want to calculate the vector product of π and π. But should we be calculating π cross π or π cross π? It actually turns out that it doesnβt matter which one of these we choose to calculate. To see why, letβs look back to our general expression for the vector cross product. If instead we wrote π cross π, our first term would be the π₯-component of π΅ multiplied by the π¦-component of π΄. And our second term would be the π¦-component of π΅ multiplied by the π₯-component of π΄.

And then this whole thing will be multiplied by our unit vector π€ in the π§-direction. Taking out a factor of minus one from the brackets gives us this expression here. Then, these two minus signs cancel out to give us a plus. And if we swap over these two terms in the brackets, we get that the vector product π cross π is equal to minus one multiplied by π΅ subscript π¦ multiplied by π΄ subscript π₯ minus π΅ subscript π₯ multiplied by π΄ subscript π¦ again multiplied by the unit vector π€ in the π§-direction.

Letβs look at our first term in the brackets π΅ subscript π¦ multiplied by π΄ subscript π₯. We should recall that when weβre multiplying together two numbers, the order of those two numbers in the multiplication doesnβt matter. So, this term is equivalent to π΄ subscript π₯ multiplied by π΅ subscript π¦. So, letβs rewrite it. And similarly, for our second term in the brackets, π΅ subscript π₯ multiplied by π΄ subscript π¦, we can rewrite this as π΄ subscript π¦ multiplied by π΅ subscript π₯.

Then, if we compare our expression for π cross π with our definition of π cross π, we see that these bits in brackets are exactly the same in each case. And the only difference is that in our π cross π, we have this factor of minus one out the front. This factor just gives us an overall minus sign. So, we can say that π cross π is equal to negative π cross π. And since in this question, weβre only interested in the magnitude of the vector product, then it doesnβt matter whether we calculate π cross π or π cross π because they both have the same magnitudes. They just have opposite signs, so the vectors point in opposite directions.

So getting back to the question and our two vectors lowercase π and lowercase π, letβs choose to calculate the vector product π cross π. This vector product is given by the π₯-component of π, which is π, multiplied by the π¦-component of π, which is also π, minus the π¦-component of π, which is zero, multiplied by the π₯-component of π, which is also zero. And then this whole thing is multiplied by the unit vector π€. Since the unit vector π€ by definition has a magnitude of one, then the magnitude of the vector product π cross π is given by this part of the expression here. We can write this as π squared. And so, we can say that the magnitude of π cross π is equal to π squared.

Okay, by this point, weβve definitely achieved our step two and calculated the vector product of π and π. All that remains is step three, which is to use this magnitude of the vector product in order to find the length π of each vector. The question tells us that the magnitude of the vector product is nine. So, we can set our magnitude π cross π equals π squared equal to nine.

Now, we were asked to find the length or magnitude of each of the vectors π and π. And we know that this length is equal to π. So, we want to solve this equation here for π. To do this, we take the square root of both sides of the equation. The square root of π squared is simply π and the square root of nine is three. So, we have that π equals three. And so finally, weβve achieved our third step using the vector products to find the length π of each vector. And we have that our answer to the question the length of both vector π and vector π is three.