1
$\begingroup$

$\textbf{Question:}$ $OAB$ is a triangle.

$Q$ is the point on $AB$ such that $OQP$ is a straight line, where $P$ is a point outside of the triangle.

$$\overrightarrow{OA}=4\bf{a},$$ $$\overrightarrow{OB}=6\bf{b},$$ $$\overrightarrow{AP}=2\bf{a}+\bf{8}b$$

Using a vector method, find the ratio, $AQ:QB$

$\textbf{Attempted solution:}$ I thought I had solved it, but then I realised I had made various assumptions, including the incorrect fact that $Q$ is the midpoint of $OP$. However, I think this much is correct: $$\overrightarrow{AQ}=-\bf{a}+4\bf{b}$$ $$\overrightarrow{QB}=-3\bf{a}-2\bf{b}$$ But I can't write these as factors of one another, so I've either made a mistake or there's something I'm not seeing. I also realised that I can't use Pythagoras' theorem because the vectors aren't necessarily orthogonal.

Diagram

$\endgroup$
5
  • $\begingroup$ You’ve made some mistake. Check whether $\vec{OA}+\vec{AQ} +\vec{QB}=\vec{OB}$ $\endgroup$ Jun 13, 2022 at 16:50
  • $\begingroup$ In situations like these, I often find it useful to turn the question into something a bit more familiar, solve that question, and then try to work backwards to get to the solution that I want. In this case, try drawing all of these as line segments on $\mathbb{R}^2$ and solve for the point of intersection between the line segment from $(0,6)$ to $(4,0)$ and the line segment from the origin to the point $(2,8)$. From there, it should only take a minute to find the ratios, and then you can try to find a way apply the vector methods that you know to get to that same answer $\endgroup$
    – peabody
    Jun 13, 2022 at 16:59
  • $\begingroup$ Is the answer 8:9? Just checking the answer to avoid posting a wrong answer. $\endgroup$ Jun 13, 2022 at 17:31
  • $\begingroup$ I don't know what the answer is. That sounds plausible though. It's from an exam, and I doubt it's a really awful ratio. I'm trying to do as was suggested above, but I can't do so without making assumptions about the nature of the vectors (e.g. that they are perpendicular, which they definitely aren't in the diagram that comes with the paper). $\endgroup$ Jun 13, 2022 at 17:39
  • $\begingroup$ I hope it isn't an ongoing exam :) just confirming. $\endgroup$ Jun 13, 2022 at 17:46

1 Answer 1

1
$\begingroup$

enter image description here

All the lengths written in the image can be obtained via triangle law of vector addition by considering various triangles. (Left as an exercise for the reader 😅)

Let $\vec {AC}=\lambda\cdot \vec {AB}$ and $\vec {CB}=\mu \cdot \vec {AB}$. Now, $\vec {AC}+\vec{CB}=\vec{AB}$ so $\lambda + \mu =1$.
Similarly, $\vec{OC}= \phi \vec {AP}$.
Now, $$\vec{OA}+ \vec{AC}+ \vec{CO}= \vec{0}$$ so that $$4 \vec{a}+\lambda (6\vec{b}-4 \vec{a})+\phi (-8\vec{b}-6 \vec{a})= \vec{0}$$ or $$(4-4\lambda-6\phi) \vec{a}+(6\lambda-8\phi) \vec{b}= \vec{0}.$$
Since $\vec{a}$ and $\vec{b}$ are non collinear, so each of their coefficients is equal to 0; i.e.$$4-4\lambda-6\phi=0$$ and $$6\lambda-8\phi=0$$ from which $\lambda = \frac{8}{17}$. Obviously, $\mu=1-\lambda=\frac{9}{17}$ so that $$\frac{AQ}{QB}=\frac{\lambda|\vec{AB}|}{\mu |\vec{AB}|}=\frac{\lambda}{\mu}=\frac89.$$

$\endgroup$
2
  • $\begingroup$ Amazing! Many thanks. Not an ongoing exam, but a past paper I'm helping a friend prepare for, and we were both super stuck! $\endgroup$ Jun 13, 2022 at 19:34
  • $\begingroup$ @Penguinking14 :) $\endgroup$ Jun 13, 2022 at 19:35

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .