### Video Transcript

Consider the two vectors π© equals four π’ plus five π£ and πͺ equals two π’ plus eight π£. Calculate π© cross πͺ.

Alright, so in this question, weβre given two vectors in component form, π© and πͺ. And weβre asked to calculate the vector product π© cross πͺ. Letβs start by drawing a quick sketch of these vectors.

Notice that both π© and πͺ have an π’-component and a π£-component. Recall that π’ is the unit vector in the π₯-direction and π£ is the unit vector in the π¦-direction. This means that both our vectors π© and πͺ lie in the π₯π¦-plane. Vector π© has four units in the π₯-direction and five units in the π¦-direction, meaning that the vector looks like this. Vector πͺ has two units in the π₯-direction and eight units in the π¦-direction, so it looks like this.

To answer the question, we need to evaluate the vector product π© cross πͺ. So letβs recall our definition of the vector product of two vectors. Letβs consider two general vectors that lie in the π₯π¦-plane. To distinguish them from the vectors that weβre given in the question, weβll call these vectors π and π. We can write the vectors in component form as π equals an π₯-component, π subscript π₯, multiplied by π’ plus a π¦-component, π subscript π¦, multiplied by π£, and similarly for π. Then, the vector product π cross π is defined as the π₯-component of π, thatβs π subscript π₯, multiplied by the π¦-component of π, thatβs π subscript π¦, minus the π¦-component of π, π subscript π¦, multiplied by the π₯-component of π, π subscript π₯. And then this whole thing is multiplied by unit vector π€, which points in the π§-direction.

So the vector product π cross π produces a vector with this magnitude and with a direction thatβs perpendicular to the direction of both π and π. We can use this expression to calculate the vector product of the vectors π© and πͺ that we are given in the question. We are asked to work out π© cross πͺ. So the first term in our vector product expression tells us that we need the π₯-component of π©, which is four, multiplied by the π¦-component of πͺ, which is eight. Then we subtract the second term from this. Again, looking at our expression for the vector product, we see that this second term tells us we need the π¦-component of vector π©, which is five, multiplied by the π₯-component of πͺ, which is two. Finally, we need to multiply this whole thing by the unit vector π€.

All that remains now is to evaluate this expression here. If we do the multiplications, we get that the first term gives us 32 and the second term, which we subtract, gives us 10. Subtracting 10 from 32 gives us 22. And this gives us our answer to the question that the vector product π© cross πͺ is equal to 22π€.