WEBVTT
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The diagram shows a quadrilateral π΄π΅πΆπ·.
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The vector π΄π΅ is equal to six π, the vector π΅πΆ is equal to three π, and the vector πΆπ· is equal to two π minus seven π.
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The point π lies on the line π΄πΆ with π΄π to ππΆ is equal to two to one.
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The point π lies on the line π΄π· with π΄π to ππ· is equal to one to three.
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Find the vector ππ in terms of π and π, giving your answer in its simplest form.
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Letβs first consider where this point π is.
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It lies somewhere along the line π΄πΆ, dividing it up into the ratio two to one.
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This means that the point π is two-thirds of the way along the line π΄πΆ.
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Letβs consider also where the point π is.
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Weβre told it lies on the line π΄π· and divides the line π΄π· in the ratio one to three.
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This means that the point π is one-quarter of the way along π΄π·.
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Weβre asked to find the vector ππ in terms of π and π.
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We need to consider a route that we can take to get from π to π and then consider what each part of the route is in terms of π and π.
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There are a number of different possibilities for the route that we could take.
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Iβm going to choose to get from π to π by going via the vertex π΄ of the quadrilateral.
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So ππ is equal to ππ΄ plus π΄π.
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I now need to find expressions for each of these two vectors in terms of π and π.
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Letβs think about the vector ππ΄ first of all.
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Now, we need to be careful with the direction here.
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So ππ΄ is equal to two-thirds of the vector πΆπ΄ if I was travelling almost right to left across the page.
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However, I donβt know what the vector πΆπ΄ is directly.
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So I need to go around the shape.
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Instead of going across the shape, I can get from πΆ to π΄ by going around the edge β so going from πΆ to π΅ and then π΅ to π΄.
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I do have expressions for each of these two vectors.
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But as Iβm travelling in the opposite direction to the vectors themselves, the signs need to be changed.
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As the vector π΅πΆ is three π, the vector πΆπ΅ will be negative three π.
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And as the vector π΄π΅ is six π, the vector π΅π΄ will be negative six π.
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So we have that ππ΄ is equal to two-thirds of negative three π minus six π.
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The numbers in this vector can be simplified.
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Two-thirds multiplied by negative three is negative two and two-thirds multiplied by negative six is negative four.
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So ππ΄ simplifies to negative two π minus four π.
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So thatβs giving me the first vector in the route I want to take.
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And now, I need to think about how to find the vector π΄π.
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Remember π divides the line π΄π· in the ratio one to three.
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And therefore, the vector π΄π is one-quarter of the vector π΄π·.
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I donβt have an expression for the vector π΄π· directly.
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But I can get from π΄ to π· by going around the other three sides of the quadrilateral.
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I can go from π΄ to π΅ and then from π΅ to πΆ and then from πΆ to π·.
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And I do know the vectors for all of these.
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Iβm also travelling in the same direction as each of these vectors have been specified.
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And therefore, the signs are the same.
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I have one-quarter of six π β thatβs for π΄π΅ β plus three π β thatβs for π΅πΆ β plus two π minus seven π, which is for πΆπ·.
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Simplifying this vector gives one-quarter of eight π minus four π.
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And then dividing both of the coefficients by four, this fully simplifies to two π minus π.
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Iβve now found both of the two vectors I was looking for: ππ΄ and π΄π.
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So I can substitute the expressions into this sum in order to find the vector ππ.
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ππ is, therefore, equal to negative two π minus four π plus two π minus π.
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Finally, Iβve been asked to give my answer in its simplest form.
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Negative four π plus two π is negative two π and negative two π minus π is negative three π.
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The vector ππ is, therefore, equal to negative two π minus three π.