Video: Pack 4 β€’ Paper 1 β€’ Question 22

Pack 4 β€’ Paper 1 β€’ Question 22

04:42

Video Transcript

The diagram shows a quadrilateral 𝐴𝐡𝐢𝐷. The vector 𝐴𝐡 is equal to six π‘Ž, the vector 𝐡𝐢 is equal to three 𝑏, and the vector 𝐢𝐷 is equal to two π‘Ž minus seven 𝑏. The point 𝑋 lies on the line 𝐴𝐢 with 𝐴𝑋 to 𝑋𝐢 is equal to two to one. The point π‘Œ lies on the line 𝐴𝐷 with π΄π‘Œ to π‘Œπ· is equal to one to three. Find the vector π‘‹π‘Œ in terms of π‘Ž and 𝑏, giving your answer in its simplest form.

Let’s first consider where this point 𝑋 is. It lies somewhere along the line 𝐴𝐢, dividing it up into the ratio two to one. This means that the point 𝑋 is two-thirds of the way along the line 𝐴𝐢. Let’s consider also where the point π‘Œ is. We’re told it lies on the line 𝐴𝐷 and divides the line 𝐴𝐷 in the ratio one to three. This means that the point π‘Œ is one-quarter of the way along 𝐴𝐷.

We’re asked to find the vector π‘‹π‘Œ in terms of π‘Ž and 𝑏. We need to consider a route that we can take to get from 𝑋 to π‘Œ and then consider what each part of the route is in terms of π‘Ž and 𝑏. There are a number of different possibilities for the route that we could take. I’m going to choose to get from 𝑋 to π‘Œ by going via the vertex 𝐴 of the quadrilateral. So π‘‹π‘Œ is equal to 𝑋𝐴 plus π΄π‘Œ.

I now need to find expressions for each of these two vectors in terms of π‘Ž and 𝑏. Let’s think about the vector 𝑋𝐴 first of all. Now, we need to be careful with the direction here. So 𝑋𝐴 is equal to two-thirds of the vector 𝐢𝐴 if I was travelling almost right to left across the page. However, I don’t know what the vector 𝐢𝐴 is directly. So I need to go around the shape. Instead of going across the shape, I can get from 𝐢 to 𝐴 by going around the edge β€” so going from 𝐢 to 𝐡 and then 𝐡 to 𝐴.

I do have expressions for each of these two vectors. But as I’m travelling in the opposite direction to the vectors themselves, the signs need to be changed. As the vector 𝐡𝐢 is three 𝑏, the vector 𝐢𝐡 will be negative three 𝑏. And as the vector 𝐴𝐡 is six π‘Ž, the vector 𝐡𝐴 will be negative six π‘Ž. So we have that 𝑋𝐴 is equal to two-thirds of negative three 𝑏 minus six π‘Ž.

The numbers in this vector can be simplified. Two-thirds multiplied by negative three is negative two and two-thirds multiplied by negative six is negative four. So 𝑋𝐴 simplifies to negative two 𝑏 minus four π‘Ž. So that’s giving me the first vector in the route I want to take. And now, I need to think about how to find the vector π΄π‘Œ.

Remember π‘Œ divides the line 𝐴𝐷 in the ratio one to three. And therefore, the vector π΄π‘Œ is one-quarter of the vector 𝐴𝐷. I don’t have an expression for the vector 𝐴𝐷 directly. But I can get from 𝐴 to 𝐷 by going around the other three sides of the quadrilateral. I can go from 𝐴 to 𝐡 and then from 𝐡 to 𝐢 and then from 𝐢 to 𝐷. And I do know the vectors for all of these. I’m also travelling in the same direction as each of these vectors have been specified. And therefore, the signs are the same. I have one-quarter of six π‘Ž β€” that’s for 𝐴𝐡 β€” plus three 𝑏 β€” that’s for 𝐡𝐢 β€” plus two π‘Ž minus seven 𝑏, which is for 𝐢𝐷.

Simplifying this vector gives one-quarter of eight π‘Ž minus four 𝑏. And then dividing both of the coefficients by four, this fully simplifies to two π‘Ž minus 𝑏. I’ve now found both of the two vectors I was looking for: 𝑋𝐴 and π΄π‘Œ. So I can substitute the expressions into this sum in order to find the vector π‘‹π‘Œ. π‘‹π‘Œ is, therefore, equal to negative two 𝑏 minus four π‘Ž plus two π‘Ž minus 𝑏.

Finally, I’ve been asked to give my answer in its simplest form. Negative four π‘Ž plus two π‘Ž is negative two π‘Ž and negative two 𝑏 minus 𝑏 is negative three 𝑏. The vector π‘‹π‘Œ is, therefore, equal to negative two π‘Ž minus three 𝑏.

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