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The diagram shows two vectors, π¨ and π©.
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Each of the grid squares in the diagram has a side length of one.
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Calculate π¨ dot π©.
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In this question, weβre given two vectors, π¨ and π©, in the form of arrows drawn on a diagram.
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We are then asked to work out π¨ dot π©, the scalar product of these two vectors.
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So letβs begin by recalling the definition of the scalar product of two vectors.
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Weβll consider two general vectors, which weβll label π and π.
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And weβll suppose that both of these vectors lie in the π₯π¦-plane.
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Then we can write these vectors in component form as an π₯-component 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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Remember 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 scalar product π dot π is equal to the π₯-component of πΆ multiplied by the π₯-component of π· plus the π¦-component of πΆ multiplied by the π¦-component of π·.
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So in general, the scalar product of two vectors is given by the product of their π₯-components plus the product of their π¦-components.
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Looking at this general expression for the scalar product of two vectors, we can see that if we want to calculate the scalar product π¨ dot π©, then weβre going to need to work out the π₯- and π¦-components of our vectors π¨ and π©.
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Now the vectors π¨ and π© are given to us as arrows drawn on a diagram and the question tells us that each of the squares in this diagram has a side length of one.
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If we add a set of axes to our diagram with the origin positioned at the tail of the two vectors, then we can easily count the number of squares that each vector extends in the π₯-direction and the π¦-direction.
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And since we know that each of these squares has a side length of one, then the number of squares directly gives the π₯- and π¦-components of the vectors.
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Letβs begin by counting the squares for vector π¨.
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We see that π΄ extends two units in the positive π₯-direction, so has an π₯-component of two, and that it extends four units in the positive π¦-direction, so has a π¦-component of four.
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So we can write the vector π¨ in component form as its π₯-component, which is two, multiplied by π’ hat plus its π¦-component, four, multiplied by π£ hat.
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Now, letβs repeat this process with vector π©.
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We see that π΅ extends three squares in the negative π₯-direction, so has an π₯-component of negative three, and that it also extends three squares in the negative π¦-direction.
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So its π¦-component is also negative three.
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Then in component form, we have that π© is equal to negative three π’ hat minus three π£ hat.
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Now that we have both our vectors π¨ and π© in component form, we are ready to calculate the scalar product π¨ dot π©.
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Looking at our general expression for the scalar product of two vectors, we see that the first term is given by the product of the π₯-components of these two vectors.
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So in our case, we need the π₯-component of π¨, which is two, multiplied by the π₯-component of π©, which is negative three.
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Then we add to this a second term given by the product of the π¦-components of the vectors.
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So for us, thatβs the π¦-component of π¨, which is four, multiplied by the π¦-component of π©, which is negative three.
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The last step left to go is to evaluate this expression here.
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The first term is two multiplied by negative three, which gives negative six.
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And the second term is four multiplied by negative three, which gives negative 12.
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Then finally, we have negative six plus negative 12, which gives negative 18.
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And so, our answer to the question is that the scalar product π¨ dot π© is equal to negative 18.