WEBVTT
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The diagram shows two vectors, π and π, in three-dimensional space.
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Both vectors lie in the π₯π¦-plane.
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Each of the squares of the grid has a side length of one.
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Calculate π cross π.
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So this is a question about vector products.
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And specifically, we are asked to calculate the vector product π cross π, where the vectors π and π are given to us in the form of arrows on a diagram.
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Letβs begin by recalling the definition of the vector product.
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Weβll consider two general vectors π and π and suppose that they lie in the π₯π¦-plane.
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Then we can write these vectors in component form as an π₯-component, which weβve labeled with a subscript π₯, multiplied by π’ hat plus a π¦-component, labeled with a subscript π¦, multiplied by π£ hat.
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Recall that π’ hat is the unit vector in the π₯-direction and π£ hat is the unit vector in the π¦-direction.
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Then, the vector product π cross π is 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.
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And this is all multiplied by π€ hat, which is a unit vector in the π§-direction.
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What this expression tells us is that in order to calculate the vector product π cross π, then weβre going to need to work out the π₯- and the π¦-components of the vectors π and π.
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Now, weβre told in the question that each of the squares of the grid in the diagram has a side length of one.
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So to get the π₯- and π¦-components of each of our vectors, we simply need to count the number of squares that they extend in each of the π₯- and π¦-directions.
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Weβll begin with vector π.
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Vector π extends one, two, three, four units in the π₯-direction and one, two, three, four, five units in the π¦-direction.
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So we can write the vector π as an π₯-component of four multiplied by π’ hat plus a π¦-component of five multiplied by π£ hat.
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Next, letβs look at the vector π.
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We see that π extends one, two, three, four units in the negative π₯-direction and one, two, three, four, five units in the negative π¦-direction.
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So we can write the vector π as negative four π’ hat minus five π£ hat.
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Now that we have each of the vectors π and π in component form, we are ready to calculate the vector product π cross π.
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Looking at our general expression for the vector product, we see that the first term is the π₯-component of the first vector in the product multiplied by the π¦-component of the second vector in the product.
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In our case, the first vector in the product is π and the second vector is π.
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So this means that we need the π₯-component of π, which is four, multiplied by the π¦-component of π, which is negative five.
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Then we subtract a second term from this.
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The second term is the π¦-component of the first vector in the product multiplied by the π₯-component of the second vector in the product.
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So, in our case, thatβs the π¦-component of π, which is five, multiplied by the π₯-component of π, which is negative four.
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And then, finally, this whole thing gets multiplied by the unit vector π€ hat.
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All thatβs left to do now is to evaluate this expression here.
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When we do this, we find that the first term, four multiplied by negative five, gives us negative 20.
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And the second term, five multiplied by negative four, also gives us negative 20.
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And so we have that the vector product π cross π is equal to negative 20 minus negative 20 multiplied by π€ hat.
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When we subtract negative 20 from negative 20, we get zero.
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And so our final answer is that the vector product π cross π is equal to zero π€ hat.