### Video Transcript

Work out π£ cross π’.

We see that this is a question about vector products, and weβre asked to calculate the vector product between the vectors π£ and π’. Now, π£ is the unit vector in the π¦-direction, and π’ is the unit vector in the π₯-direction. Since weβre asked to calculate a vector product, letβs start by recalling the general expression for the vector product between two vectors. Weβll call these general vectors π and π and assume that both lie in the π₯π¦-plane. We can write these vectors in component form as π equals π΄ subscript π₯ times π’ plus π΄ subscript π¦ times π£ and similarly for π.

Remember that π’ is the unit vector in the π₯-direction, so π΄ subscript π₯ and π΅ subscript π₯ are the π₯-components of the vectors. Likewise, π£ is the unit vector in the π¦-direction, so π΄ subscript π¦ and π΅ subscript π¦ are the π¦-components of the vectors. Then the vector product π cross π is defined as π΄ subscript π₯ multiplied by π΅ subscript π¦ minus π΄ subscript π¦ multiplied by π΅ subscript π₯ all multiplied by the unit vector π€, which points in π§-direction. So the vector product π cross π produces a vector with this magnitude and with a direction perpendicular to the direction of both π΄ and π΅. So thatβs the general case. Now letβs apply it to our two vectors π£ and π’.

We can rewrite the vector π£ as zero multiplied by π’ plus one multiplied by π£. This says that π£, the unit vector in the π¦-direction, has zero units along the π₯-direction and one unit along the π¦-direction, which makes sense. Similarly, we can rewrite the vector π’ as one multiplied by π’ plus zero multiplied by π£. The question is asking us to calculate π£ cross π’. So in this general expression π cross π, we want to make the replacement π goes to π£ and π goes to π’. The first term in our vector cross product is π΄ subscript π₯ multiplied by π΅ subscript π¦. Now, π΄ subscript π₯ is the π₯-component of the first vector in our vector cross product which, in our calculation, is going to be the vector π£. So we look for the π₯-component of π£ and we see that that is zero.

Meanwhile, π΅ subscript π¦ is going to be the π¦-component of the second vector in our across product which, in our calculation, is the vector π’. So if we look at the π¦-component of the vector π’, we see that that is also zero. From this first term in our vector cross product equation, we then subtract a second term. The second term is given by π΄ subscript π¦ multiplied by π΅ subscript π₯. Now, π΄ subscript π¦ is the π¦-component of the first vector in our cross product. This is vector π£. So if we look to the π¦-component of π£, we see that this is one. And π΅ subscript π₯, thatβs the π₯-component of the second vector in our cross product. In our case, thatβs vector π’. So looking at the π₯-component of vector π’, we see that this is also one.

Finally, this whole expression gets multiplied by π€, the unit vector in the π§-direction. Evaluating this expression, our first term is given by zero multiplied by zero, which gives us zero. And our second term, which we subtract from this first one, is given by one multiplied by one which gives us one. And then, of course, this is all multiplied by π€, the unit vector in the π§-direction. Since zero minus one is negative one, we get our final answer that π£ cross π’ is equal to negative π€, in other words, the unit vector in the negative π§-direction.