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Relative to a fixed origin $O$, the points $A$ and $B$ have position vectors $\mathbf{a}$ and $\mathbf{b}$ repsectively, where $O,A$ and $B$ are not collinear.

The point $C$ has position vector $\alpha\mathbf{a}+\beta\mathbf{b}$, where $\alpha,\beta$ are positive constants with $\alpha+\beta<1$. The lines $OA$ and $BC$ intersect at the point with position vector $\mathbf{x}$.

Show that $$\mathbf{x}=\frac{\alpha}{1-\beta}\mathbf{a}.$$

My attempt:

The line segment $OA$ can be parametrized by $\mathbf{r}(t)=t\mathbf{a}, t\in[0,1]$, and the line segment $BC$ can be parametrized by $\mathbf{r}(s)=s(\alpha\mathbf{a}+(\beta-1)\mathbf{b})$. When $OA$ and $BC$ intersect, $$t\mathbf{a}=s(\alpha\mathbf{a}+(\beta-1)\mathbf{b}).$$ It's not clear to me how to proceed from here. I can take the scalar product of both sides with $\mathbf{a}$ or $\mathbf{b}$, but this doesn't seem to help.

A hint would be greatly appreciated.

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Actually, $\mathbf r(s)$ should be this:

$$\mathbf r(s)=\mathbf b+s((\alpha\mathbf a+\beta\mathbf b) - \mathbf b) = (1-s)\mathbf b +s(\alpha \mathbf a+\beta \mathbf b)$$

which, with $\mathbf r(t)=t\mathbf a$, gives you:

$$t\mathbf a=(1-s)\mathbf b +s(\alpha \mathbf a+\beta \mathbf b)$$

or, after tidying up:

$$(t-s\alpha)\mathbf a+(-(1-s)-s\beta)\mathbf b=0$$

Now, recall that $O,A,B$ are not colinear, so $\mathbf a$ and$\mathbf b$ are linearly independent, which implies:

$$\begin{align}t-s\alpha=0\\-(1-s)-s\beta=0\end{align}$$

which is a system of equations to solve for $s,t$. The second equation gives us $(1-\beta)s=1$, i.e. $s=\frac{1}{1-\beta}$. Substituting in the first, we get $t=s\alpha=(\frac{1}{1-\beta})\alpha=\frac{\alpha}{1-\beta}$, as desired.

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  • $\begingroup$ Why is $\mathbf{r}(s)=s\mathbf{b}+(1-s)(\alpha\mathbf{a}+\beta\mathbf{b})$? The direction of $BC$ is $(\alpha\mathbf{a}+\beta\mathbf{b})-\mathbf{b}$, and the line $BC$ starts at $B$, so why isn't it $\mathbf{r}(s)=\mathbf{b}+s((\alpha\mathbf{a}+\beta\mathbf{b})-\mathbf{b})$? $\endgroup$
    – A. Goodier
    Jan 1 '18 at 15:58
  • $\begingroup$ @woofy It doesn't matter. The line may start at $B$ or at $C$, and what will be $s$ in one case will be $1-s$ in the other (and vice versa). You will get a different $s$, but it will give you the same $t$ in the end. Try it out. $\endgroup$
    – user491874
    Jan 1 '18 at 16:20
  • $\begingroup$ @woofy Anyways... I've rewritten the answer to assume the line starts at $B$ rather than $C$... if it matters to you. $\endgroup$
    – user491874
    Jan 1 '18 at 17:01
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HINT: We can draw a representative image as the following and use the similarity between triangles $XDC$ and $XOB$ in order to find the magnitude of $\vec{x}$ as $$|\vec{x}| =\frac{\alpha}{1-\beta} |\vec{a}|$$

enter image description here

Notice that we draw the representative image by considering the facts $\alpha+\beta<1$ (otherwise $\vec{c}$ would intersect the line $AB$ shown as gray) and $\alpha, \beta > 0$ (otherwise, $\vec{c}$ wouldn't lie inside the triangle $AOB$) so that the representative is valid.

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