Hard (for me) GCSE vectors question I am unable to solve. (Given that C,D,E are on same straight line, find BE)

[erich1743]

Hi there, this is my first time on this forum so forgive me if I am not making anything very clear. This here is a maths problem on the MathsGenie GCSE revision section of grade 8-9 topics. I understand everything up until the highlighted area on the image. It would be very nice of you to help me understand this as I have my exam tomorrow and I'm desperate to understand, the original question is next to it. Thanks.

 

[blamocur]

$\overrightarrow{CE}$ is a multiple of $\frac{2}{3}(-\mathbf a + \mathbf b)$, which means it must be a multiple of $-\mathbf a + \mathbf b$. The only value of $x$ which satisfies this requirement is 4.

 

[erich1743]

$\overrightarrow{CE}$ is a multiple of $\frac{2}{3}(-\mathbf a + \mathbf b)$, which means it must be a multiple of $-\mathbf a + \mathbf b$. The only value of $x$ which satisfies this requirement is 4.

How do we know what the requirement is sorry I’m a bit confused, and how did he get 2b at the end

 

[blamocur]

How do we know what the requirement is sorry I’m a bit confused,...

C,D,E are on the same line, which means that vectors $\overrightarrow{CD}$ and $\overrightarrow{CE}$ have the same direction, which, in turn, means that one vector is a multiple of another.

and how did he get 2b at the end

$\overrightarrow{OE}=\overrightarrow{OC} + \overrightarrow{CE}= 4\mathbf a + (-4\mathbf a + 4 \mathbf b) = 4\mathbf b$ -- do you see the rest?