# Question Video: Calculating the Vector Product of Two Vectors Shown on a Grid Physics

The diagram shows two vectors, π and π. Each of the grid squares in the diagram has a side length of 1. Calculate π Γ π.

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### Video Transcript

The diagram shows two vectors, π and π. Each of the grid squares in the diagram has a side length of one. Calculate π cross π.

Okay, so this is a question about vector products. We have a diagram showing two vectors, which are labeled π and π. We are told that each of the grid squares has a side length of one. We are asked to calculate the vector product π cross π. Letβs begin by writing out the vectors in component form. In order to do this, we need to find the π₯- and π¦-components of each vector using the diagram. Weβll add an π₯- and a π¦-axis to this diagram to make this process a little clearer. We see that vector π extends positive two units in the π₯-direction and positive six units in the π¦-direction.

Now, recall that the unit vector in the π₯-direction is labeled π’ and the unit vector in the π¦-direction is π£. So we can write the vector π as its π₯-component, which is two, multiplied by π’ plus its π¦-component, which is six, multiplied by π£. For vector π, we see that it extends positive five units in the π₯-direction and positive one unit in the π¦-direction. So we can write that vector π equals its π₯-component, five multiplied by π’, plus its π¦-component, one multiplied by π£.

We now have expressions for both vector π and vector π in component form. The question is asking us to calculate the vector product π cross π. So letβs recall the definition of the vector product of two vectors. Weβll define two general vectors that lie in the π₯π¦-plane. And weβll label these vectors lowercase π and lowercase π, where we use the lowercase letters to distinguish this general case from our two specific vectors from the question. We can write these general vectors in component form, labeling the π₯-components with a subscript π₯ and the π¦-components with a subscript π¦.

Then, the vector product π cross π is defined as 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. And this is all multiplied by π€, which is the unit vector in the π§-direction. We can use this definition to calculate the vector product of the two vectors from the question, capital π and capital π.

We want to work out π cross π. Using our general expression for the vector product, we see that the first term tells us that we need the π₯-component of π, which is two, multiplied by the π¦-component of π, which is one. And from this first term we then subtract a second term. Looking again at our general expression for the vector product, we see that for the second term, we need the π¦-component of vector π, which is six, multiplied by the π₯-component of vector π, which is five. And finally, we see that we need to multiply this whole thing by π€, the unit vector in the π§-direction.

All thatβs left to do is evaluating this expression here. If we do the multiplications, we see that the first term, two multiplied by one, gives us two and the second term, six multiplied by five, gives us 30. Finally, subtracting 30 from two, we get our answer to the question that the vector product π cross π is equal to negative 28π€.