# Video: Pack 4 • Paper 3 • Question 17

Pack 4 • Paper 3 • Question 17

04:56

### Video Transcript

Consider the diagram. 𝐴𝐵𝐷, 𝐴𝑀𝐶, and 𝐵𝑁𝐶 are straight lines, where 𝐵 is the midpoint of 𝐴𝐷 and 𝑁 is the midpoint of 𝐵𝐶. The vector 𝐶𝐴 is equal to nine 𝑎, the vector 𝐶𝐵 is equal to six 𝑏, and the vector 𝐶𝑀 is equal to 𝑘𝑎, where 𝑘 is a constant. 𝑀𝑁𝐷 is a straight line. Find the value of 𝑘.

Let’s begin by adding the information we’re given about the vectors 𝐶𝐴, 𝐶𝐵, and 𝐶𝑀 onto the diagram. So we have 𝐶𝐴 is equal to nine 𝑎, 𝐶𝑀 is equal to 𝑘𝑎, and 𝐶𝐵 is equal to six 𝑏. We’re also told in the question that 𝑀𝑁𝐷 is a straight line. This means that the vectors 𝑁𝐷 and 𝑀𝑁 must be parallel or phrased another way 𝑁𝐷 is a scalar multiple of 𝑀𝑁.

Let’s see if we can find expressions for these two vectors, starting with 𝑀𝑁. We don’t have a vector for going directly from 𝑀 to 𝑁, but we can go via the point 𝐶. The vector 𝑀𝑁 is, therefore, equal to 𝑀𝐶 plus 𝐶𝑁.

Now, we haven’t been given the vector 𝑀𝐶, but we have been given the vector 𝐶𝑀. And as we’re travelling in the opposite direction, this means that we change the sign. So we have that 𝑀𝑁 is equal to negative 𝐶𝑀 plus 𝐶𝑁.

We’re also told in the question that 𝑁 is the midpoint of 𝐵𝐶. This means that the vectors 𝐶𝑁 and 𝑁𝐵 are each half of the vector 𝐶𝐵. As the vector 𝐶𝐵 is equal to six 𝑏, this means that the vectors 𝐶𝑁 and 𝑁𝐵 are each equal to three 𝑏. Substituting the vectors for 𝐶𝑀 and 𝐶𝑁 into our expression for 𝑀𝑁, we have that 𝑀𝑁 is equal to negative 𝑘𝑎 plus three 𝑏.

Now, let’s consider the vector 𝑁𝐷. We don’t have a vector that describes how to go directly from 𝑁 to 𝐷. But we can go via the point 𝐵. The vector 𝑁𝐷 is equal to 𝑁𝐵 plus 𝐵𝐷. Now, we already have an expression for the vector 𝑁𝐵, it’s three 𝑏. So we just need to find the vector 𝐵𝐷.

As 𝐵 is the midpoint of the line 𝐴𝐷, the vector 𝐵𝐷 will be equal to the vector 𝐴𝐵. We can find the vector 𝐴𝐵 by going via the point 𝐶. The vector 𝐴𝐵 is equal to 𝐴𝐶 plus 𝐶𝐵, both of which we have expressions for. The vector 𝐶𝐵 is six 𝑏. And as we’re travelling in the opposite direction from 𝐴 to 𝐶, the vector 𝐴𝐶 is negative nine 𝑎. So this gives an expression for 𝐴𝐵. And as 𝐴𝐵 is equal to 𝐵𝐷, we also have an expression for 𝐵𝐷.

Substituting the vectors for 𝑁𝐵 and 𝐵𝐷 into our expression for 𝑁𝐷, we have three 𝑏 minus nine 𝑎 plus six 𝑏. This simplifies to negative nine 𝑎 plus nine 𝑏. And we have our vector for 𝑁𝐷.

Now, remember we said that 𝑁𝐷 is a scalar multiple of 𝑀𝑁, which means that the vector 𝑁𝐷 is equal to 𝑝 multiplied by the vector 𝑀𝑁 for some number 𝑝. Substituting the expressions we found for 𝑁𝐷 and 𝑀𝑁 gives negative nine 𝑎 plus nine 𝑏 is equal to 𝑝 multiplied by negative 𝑘𝑎 plus three 𝑏. We can expand the bracket on the right side of this equation to give negative nine 𝑎 plus nine 𝑏 is equal to negative 𝑝𝑘𝑎 plus three 𝑝𝑏.

Now, remember we’re looking to find the value of 𝑘. And to do so, we need to equate the coefficients of 𝑎 and 𝑏 on the two sides of this equation. Equating the coefficients of 𝑎, we have negative nine is equal to negative 𝑝𝑘. And equating the coefficients of 𝑏, we have that nine is equal to three 𝑝.

We can solve the second equation to find the value of 𝑝 by dividing both sides by three. This tells us that 𝑝 is equal to three.

Finally, we can substitute this value into our first equation, which tells us that negative nine is equal to negative three 𝑘. The negatives on each side of the equation will cancel out. And to find the value of 𝑘, we need to divide both sides by three. This gives three is equal to 𝑘.

The value of 𝑘 is three.