Hi, I'm stuck on this vectors question (from Cambridge):
Suppose that OABC is a parallelogram. Let M be the midpoint of OA and let P be the point of intersection of MC and OB. Prove, using vectors, that OP = 1/3 * OB.
This question drove me insane for the past hour because I genuinely cannot think of any MX1 way of doing it. Even the MX2 approach I took feels outside of the scope of the HSC. I have no idea what the intended method the authors of the Cambridge textbook had on my mind.
Regardless, this was the closest to MX1 scope I managed to find. It's adapted from other sources. Let \(\overrightarrow{OA} = \mathbf a\) and \(\overrightarrow{OC} = \mathbf c\). Then \(\overrightarrow{OB} = \mathbf a + \mathbf c\).
Note that \(P\) obviously lies on \(OB\), so \(\overrightarrow{OP}\) is automatically parallel to \(\overrightarrow{OB}\).
Hence, we can write \(\overrightarrow{OP} = \lambda (\mathbf{a}+\mathbf{c})\) for some scalar \(\lambda>0\).
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\[ \text{Inevitably, }\overrightarrow{MP}\text{ and }\overrightarrow{CP}\text{ have to be involved.}\\ \text{But we can just note that }\overrightarrow{MP}\text{ and }\overrightarrow{PC}\text{ point in the same direction, so we can write}\\ \overrightarrow{MP} = \mu\overrightarrow{PC}\text{ for some scalar }\mu > 0. \]
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We now derive expressions for \(\overrightarrow{MP}\) and \(\overrightarrow{PC}\) in terms of \(\mathbf a\) and \(\mathbf c\).
\begin{align*}
\overrightarrow{MP} &= \overrightarrow{MO} + \overrightarrow{OP}\\ &= -\frac12 \mathbf{a} + \lambda(\mathbf{a}+\mathbf{c})\\
&= \left( \lambda - \frac12 \right) \mathbf{a} + \lambda \mathbf{c}
\end{align*}
\begin{align*}
\overrightarrow{PC} &= \overrightarrow{PO}+\overrightarrow{OC}\\
&= -\lambda(\mathbf{a}+\mathbf{c}) + \mathbf{c}\\
&= -\lambda \mathbf{a} + (1-\lambda)\mathbf{c}.
\end{align*}
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So by subbing these expressions in, we now obtain
\[ \left( \lambda - \frac12 \right) \mathbf{a} + \lambda \mathbf{c} = -\lambda\mu \mathbf{a} + (1-\lambda)\mu\mathbf{c}. \]
This is where I believe the problem goes outside of MX1. How we
can continue here is supposedly by equating the scalars on each of the vectors:
\begin{align*}
\lambda - \frac12 &= -\lambda\mu\\
\lambda &= (1-\lambda) \mu
\end{align*}
In theory, to justify this rigorously we'd need to use linear independence. As a theorem, two vectors \(\mathbf{v}\) and \(\mathbf{w}\) are 'linear independent' when they are non-parallel, as it turns out. This is obviously true in the case of the parallelogram.
But the actual definition is that two vectors are linear independent, if whenever \(\lambda \mathbf{v} +\mu \mathbf{w}=\mathbf{0}\), the scalars \(\lambda\) and \(\mu\) must be equal to \(0\) themselves. The consequence of this result is that when two vectors are linearly independent, as they are here, extracting and equating coefficients suddenly becomes acceptable.
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Assuming that ramble above, we can then solve. The second equation becomes \( \mu = \frac{\lambda}{1-\lambda} \), which when subbed in gives
\begin{align*}
\lambda-\frac12 &= -\frac{\lambda^2}{1-\lambda}\\
(2\lambda-1)(1-\lambda) &= -2\lambda^2\\
2\lambda-1-2\lambda^2 +\lambda &= -2\lambda^2\\
\lambda &= \frac13.
\end{align*}
So \(\overrightarrow{OP} = \frac13 (\mathbf a+\mathbf b) = \frac13 \overrightarrow{OB}\) as required.
Obviously, I'm highly unconvinced that this was the intended approach. I am watchful for if there's any more MX1-like responses that show up, but can't make any promises.Edit: So I've asked some of my friends on this problem as well. They're not too sure how this can be done either. (At least, not without linear independence.) I'm somewhat convinced now that this question shouldn't be in the exercise altogether, since linear independence is not a part of the HSC syllabus. I'm not sure what the author's intended methods were here.
(It may be worth that with similar triangles, the problem gets made a lot easier. But obviously, Euclidean geometry has been wiped from the syllabus.)