OABC is a square. M is the mid-point of OA, and Q divides BC in the ratio 1:3. AC and MQ meet at P. If O to A = a, O to C = c . Express O to P in terms of a and c.
1. $$\small{\text{
If $ \vec{Q} = \vec{c} + \frac{3}{4}\vec{a}
$ and $ \vec{M} = \frac{1}{2}\vec{a} $
so line $ \overline{MQ} $ is $\frac{1}{2}\vec{a} + \lambda ( \vec{Q}-\vec{M}) = \frac{1}{2}\vec{a} + \lambda (\vec{c}+\frac{1}{4}\vec{a}) $
}}$$
2. $$\small{\text{
Line $ \overline{AC} $ is $\vec{a} + \mu ( \vec{c}-\vec{a}) $
}}$$
3.
$$\\\small{\text{
The intersection of the line $\overline{MQ}$ with the line $\overline{AC}$ is
}}
$\\$\small{\text{
the point $\vec{P} = \frac{1}{2}\vec{a} + \lambda (\vec{c}+\frac{1}{4}\vec{a}) $.}}
$\\$\small{\text{
We equate:
$
\frac{1}{2}\vec{a} + \lambda (\vec{c}+\frac{1}{4}\vec{a}) = \vec{a} + \mu ( \vec{c}-\vec{a})
$
}}
$\\$\small{\text{
so
$
\lambda (\vec{c}+\frac{1}{4}\vec{a}) - \mu ( \vec{c}-\vec{a}) = \frac{1}{2}\vec{a}
$
}}
$\\$\small{\text{
The perpendicular vector of $(\vec{c}-\vec{a})$ is $(\vec{c}+\vec{a})$, because we have a square and
}}
$\\$\small{\text{
$(\vec{c}-\vec{a}) * ( \vec{c}+\vec{a} ) = 0$, because the dot-product of vectors
}}
$\\$\small{\text{
perpendicular to each other is a Zero.
}}
$\\$\small{\text{
We multiply our equation with $( \vec{c}+\vec{a} )$ and see what passed:
}}$$
$$$\\$\small{\text{
$
\lambda (\vec{c}+\frac{1}{4}\vec{a}) (\vec{c}+\vec{a}) - \mu \underbrace{( \vec{c}-\vec{a}) (\vec{c}+\vec{a})}_{dot-product = 0} = \frac{1}{2}\vec{a} (\vec{c}+\vec{a})
$
}}
$\\$\small{\text{
Now we can dissolve after $\lambda$ and receive for $\lambda$:
}}
$\\$\small{\text{
$
\lambda = \dfrac{
\frac{1}{2}\vec{a} (\vec{c}+\vec{a}) }
{(\vec{c}+\frac{1}{4}\vec{a}) (\vec{c}+\vec{a})
}
$
}}
$\\$
\boxed{
\small{\text{
$\vec{P} = \frac{1}{2}\vec{a} + \lambda (\vec{c}+\frac{1}{4}\vec{a}) $}} = \lambda \vec{c} + (\frac{1}{2}+\frac{\lambda}{4})\vec{a}
$
}}}$$
