Given a cube $ABCDA_1B_1C_1D_1$ with lower base $ABCD$ and upper base $A_1B_1C_1D_1$ and the lateral edges $AA_1,BB_1,CC_1,DD_1$ respectively. $M$ and $M_1$ are centres of the faces $ABCD$ and $A_1B_1C_1D_1$ respectively. $O$ is a point on line $MM_1$, such that $\vec{OA}+\vec{OB}+\vec{OC}+\vec{OD}=\vec{OM_1}$. If $\vec{OM}=\lambda \vec{OM_1}$. Then find the value of $\lambda$.
Could someone give me some hint about this question. I am not able to visualize it.
First thing you should do is sketch the cube and all the vectors involved.
From your picture, you should be able to see that $\vec{OA}=\vec{OM}+\vec{MA}$. Similarly, $\vec{OB}=\vec{OM}+\vec{MB},$ and the same for $\vec{OC}$ and $\vec{OD}$.
Then, plugging that into the equation given, we have: $$\vec{OA}+\vec{OB}+\vec{OC}+\vec{OD}=\vec{OM_1}$$ $$(\vec{OM}+\vec{MA})+(\vec{OM}+\vec{MB})+(\vec{OM}+\vec{MC})+(\vec{OM}+\vec{MD})=\vec{OM_1}$$ $$4\vec{OM}+\vec{MA}+\vec{MB}+\vec{MC}+\vec{MD}=\vec{OM_1}$$
Now, again looking at your picture, you should be able to see that $\vec{MA}$ and $\vec{MC}$ are opposite vectors - that is, they are parallel and of the same length, but pointing in the opposite direction (since $M$ is the centre of the bottom face). So, we can write $\vec{MA}=-\vec{MC}$. Similarly, $\vec{MB}=-\vec{MD}.$ Because of this, we get $\vec{MA}+\vec{MB}+\vec{MC}+\vec{MD}=\vec{0}$.
This gives us: $$4\vec{OM}=\vec{OM_1}.$$ Plugging in $\vec{OM}=\lambda\vec{OM_1}$, you can now solve for $\lambda$!