Question on Mechanics - Vectors

$$\vec{AC}=-\mathbf{a}+\mathbf{c}$$

$$\vec{MQ}=\frac{1}{2}\mathbf{a}+\mathbf{c}-\frac{1}{4}\mathbf{a}$$

         $$=\frac{1}{4}\mathbf{a}+\mathbf{c}$$

$$\vec{OP}=\vec{OM}+\vec{MP}$$

        $$=\vec{OA}+\vec{AP}$$

$$\vec{MP}=\lambda(\frac{1}{4}\mathbf{a}+\mathbf{c})$$

         $$=\frac{\lambda}{4}\mathbf{a}+\lambda \mathbf{c}$$

$$\vec{AP}=\mu(-\mathbf{a}+\mathbf{c})$$

        $$=-\mu\mathbf{a}+\mu\mathbf{c}$$

As $$\vec{AP}$$ and $$\vec{MP}$$ intersect, if I let them equal, I can find the $$P$$.

$$\frac{\lambda}{4}\mathbf{a}+\lambda \mathbf{c}$$ $$=-\mu\mathbf{a}+\mu\mathbf{c}$$

Therefore, $$\frac{\lambda}{4}=-\mu$$ and $$\lambda = \mu$$

Then if I subtitute $$\mu$$ to $$\frac{\lambda}{4}=-\mu$$,

$$\frac{\mu}{4}=-\mu$$

$$\mu=-4$$

Sub -4 to $$\vec{MP}=\lambda(\frac{1}{4}\mathbf{a}+\mathbf{c})$$    (As  $$\mu = \lambda$$)

$$\vec{MP}=-4(\frac{1}{4}\mathbf{a}+\mathbf{c})$$

             $$=-\mathbf{a}-4\mathbf{c}$$

Finally, 

$$\vec{OP} = \vec{OM}+\vec{MP}$$

        $$=\frac{1}{2}\mathbf{a}+(-\mathbf{a}-4\mathbf{c})$$

        $$=\frac{1}{2}\mathbf{a}-\mathbf{a}-4\mathbf{c}$$

        $$=-\frac{1}{2}\mathbf{a}-4\mathbf{c}$$

AP and MQ meet at P

I assume that you mean

AC and MQ meet at P  ??

It is a square - so 

OA=OC

so why have 2 different pronumerals?

Here you are.

It has different pronumerals, because it's vector.

vector is not only about how long or how big, it's also about the direction of it.

I don't know the answer yet, I'm gonna ask it to my teacher today.

But I think you are on right line.

Anyway, I will post the full answer here today.

Thanks for answering though 

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.

$$\\\small{\text{
The intersection of the line $\overline{MQ}$ with the line $\overline{AC}$ is
}}
$\\$\small{\text{
the point $\vec{P} = \frac{1}{2}\vec{a} + \lambda (\vec{c}+\frac{1}{4}\vec{a}) $.}}
$\\$\small{\text{
We equate:
$
\frac{1}{2}\vec{a} + \lambda (\vec{c}+\frac{1}{4}\vec{a}) = \vec{a} + \mu ( \vec{c}-\vec{a})
$
}}
$\\$\small{\text{
so
$
\lambda (\vec{c}+\frac{1}{4}\vec{a}) - \mu ( \vec{c}-\vec{a}) = \frac{1}{2}\vec{a}
$
}}
$\\$\small{\text{
The perpendicular vector of $(\vec{c}-\vec{a})$ is $(\vec{c}+\vec{a})$, because we have a square and
}}
$\\$\small{\text{
$(\vec{c}-\vec{a}) * ( \vec{c}+\vec{a} ) = 0$, because the dot-product of vectors
}}
$\\$\small{\text{
perpendicular to each other is a Zero.
}}
$\\$\small{\text{
We multiply our equation with $( \vec{c}+\vec{a} )$ and see what passed:
}}$$

$$$\\$\small{\text{
$
\lambda (\vec{c}+\frac{1}{4}\vec{a}) (\vec{c}+\vec{a}) - \mu \underbrace{( \vec{c}-\vec{a}) (\vec{c}+\vec{a})}_{dot-product = 0} = \frac{1}{2}\vec{a} (\vec{c}+\vec{a})
$
}}
$\\$\small{\text{
Now we can dissolve after $\lambda$ and receive for $\lambda$:
}}
$\\$\small{\text{
$
\lambda = \dfrac{
\frac{1}{2}\vec{a} (\vec{c}+\vec{a}) }
{(\vec{c}+\frac{1}{4}\vec{a}) (\vec{c}+\vec{a})
}
$
}}
$\\$
\boxed{
\small{\text{
$\vec{P} = \frac{1}{2}\vec{a} + \lambda (\vec{c}+\frac{1}{4}\vec{a}) $}} = \lambda \vec{c} + (\frac{1}{2}+\frac{\lambda}{4})\vec{a}
$
}}}$$

$$P.S. \qquad \lambda= \frac{ \frac{1}{2}\vec{a}(\vec{c}+\vec{a} )
} { (\vec{c}+\frac{1}{4}\vec{a}) (\vec{c}+\vec{a})
} =
\frac{ \frac{1}{2}\vec{a}\vec{c} + \frac{1}{2}a^2 }{ c^2 + \vec{c}\vec{a} + \frac{1}{4}\vec{a}\vec{c} +\frac{1}{4}a^2 }
=\frac{0 + \frac{1}{2}a^2 }{ a^2+0+0+ \frac{1}{4}a^2 } =
\frac{ \frac{1}{2}a^2 }{\frac{5}{4}a^2 } = \frac{2}{5}$$

$$\small{\text{
$ \vec{a}*\vec{c} =0 \quad \vec{a} $ and $ \vec{c} $ are perpendicular and $ a^2=c^2 $ is a square.
}}$$

$$\lambda= \frac{2}{5}$$

$$\vec{OP}=\frac{2}{5}\vec{c}+\frac{3}{5}\vec{a}$$

Sorry flflvm97 you have a missing term in your equation:

$$\vec{MA}+\vec{AP}-\vec{MP}=0 \\
\vec{MP}=\vec{MA}+\vec{AP} \\
\lambda(\frac{1}{4}\vec{a}+\vec{c})=\textcolor[rgb]{1,0,0}{\frac{1}{2}\vec{a}} +\mu(-\vec{a}+\vec{c})$$