Linear combinations and dividing a segment

A linear combination of the vectors $\displaystyle \underline{u}$ and $\displaystyle \underline{v}$ is a vector of the form $\displaystyle s\underline{u}+t\underline{v}$, where $\displaystyle s$ and $\displaystyle t$ are numbers. In the same way $\displaystyle s\underline{u}+t\underline{v}+r\underline{w}$ is a linear combination of three vectors.

To express a vector this way, go to it along a path that runs only along the given vectors – and each piece of the path is a multiple of one of them.

The point $\displaystyle M$ on the segment $\displaystyle AB$ divides it in the ratio $\displaystyle {AM:MB}={m:n}$. Then $\displaystyle {\overrightarrow{AM}}={\frac{m}{m+n}\overrightarrow{AB}}$, and for any point $\displaystyle O$:

• $\displaystyle {\overrightarrow{OM}}={\overrightarrow{OA}+\frac{m}{m+n}\left(\overrightarrow{OB}-\overrightarrow{OA}\right)}$, that is:

Note: the coefficient of $\displaystyle \overrightarrow{OA}$ is the part far from $\displaystyle A$. When $\displaystyle M$ is the midpoint of $\displaystyle AB$: $\displaystyle {\overrightarrow{OM}}={\frac{1}{2}\left(\overrightarrow{OA}+\overrightarrow{OB}\right)}$.

In the triangle $\displaystyle ABC$: $\displaystyle {\overrightarrow{AB}}={\underline{u}}$, $\displaystyle {\overrightarrow{AC}}={\underline{v}}$.
The point $\displaystyle D$ is on $\displaystyle BC$, and $\displaystyle {BD:DC}={1:3}$.
Express $\displaystyle \overrightarrow{AD}$ with $\displaystyle \underline{u}$ and $\displaystyle \underline{v}$.

• By dividing a segment, from the point $\displaystyle A$: $\displaystyle {\overrightarrow{AD}}={\frac{3}{4}\overrightarrow{AB}+\frac{1}{4}\overrightarrow{AC}}={\frac{3}{4}\underline{u}+\frac{1}{4}\underline{v}}$.
• Another way: $\displaystyle {\overrightarrow{AD}}={\overrightarrow{AB}+\frac{1}{4}\overrightarrow{BC}}={\underline{u}+\frac{1}{4}(\underline{v}-\underline{u})}$ – and after simplifying, the same result.

When $\displaystyle D$ is the midpoint of $\displaystyle BC$, $\displaystyle AD$ is the median: $\displaystyle {\overrightarrow{AD}}={\frac{1}{2}(\underline{u}+\underline{v})}$. The centroid $\displaystyle G$, where the medians meet, is $\displaystyle \frac{2}{3}$ of the way from $\displaystyle A$, so $\displaystyle {\overrightarrow{AG}}={\frac{1}{3}(\underline{u}+\underline{v})}$.

In the box $\displaystyle ABCDA'B'C'D'$: $\displaystyle {\overrightarrow{AB}}={\underline{u}}$, $\displaystyle {\overrightarrow{AD}}={\underline{v}}$, $\displaystyle {\overrightarrow{AA'}}={\underline{w}}$.

• $\displaystyle {\overrightarrow{AC'}}={\underline{u}+\underline{v}+\underline{w}}$: along $\displaystyle AB$, then $\displaystyle BC$ (which equals $\displaystyle \underline{v}$), then $\displaystyle CC'$ (which equals $\displaystyle \underline{w}$).
• $\displaystyle {\overrightarrow{BD'}}={-\underline{u}+\underline{v}+\underline{w}}$.
• $\displaystyle M$ is the centre of the top face, the midpoint of the diagonal $\displaystyle A'C'$: $\displaystyle {\overrightarrow{AM}}={\underline{w}+\frac{1}{2}(\underline{u}+\underline{v})}$.

In the triangle $\displaystyle ABC$: $\displaystyle {\overrightarrow{AB}}={\underline{u}}$, $\displaystyle {\overrightarrow{AC}}={\underline{v}}$.
The point $\displaystyle D$ is on $\displaystyle BC$, and $\displaystyle {BD:DC}={2:1}$.
Express $\displaystyle \overrightarrow{AD}$ with $\displaystyle \underline{u}$ and $\displaystyle \underline{v}$.

In the same box $\displaystyle ABCDA'B'C'D'$: $\displaystyle {\overrightarrow{AB}}={\underline{u}}$, $\displaystyle {\overrightarrow{AD}}={\underline{v}}$, $\displaystyle {\overrightarrow{AA'}}={\underline{w}}$.
$\displaystyle N$ is the midpoint of the edge $\displaystyle B'C'$.
Express with $\displaystyle \underline{u}$, $\displaystyle \underline{v}$, $\displaystyle \underline{w}$:

In the tetrahedron $\displaystyle ABCD$: $\displaystyle {\overrightarrow{AB}}={\underline{a}}$, $\displaystyle {\overrightarrow{BC}}={\underline{b}}$, $\displaystyle {\overrightarrow{AD}}={\underline{c}}$.

Prove that $\displaystyle {\overrightarrow{AC}+\overrightarrow{BD}}={\overrightarrow{AD}+\overrightarrow{BC}}$. Does the proof also hold when the points $\displaystyle A$, $\displaystyle B$, $\displaystyle C$, $\displaystyle D$ lie in one plane?

The base of the pyramid $\displaystyle SABCD$ is a parallelogram. $\displaystyle {\overrightarrow{AB}}={\underline{u}}$, $\displaystyle {\overrightarrow{AD}}={\underline{v}}$, $\displaystyle {\overrightarrow{AS}}={\underline{w}}$.
$\displaystyle M$ is the midpoint of $\displaystyle SC$. The point $\displaystyle N$ is on $\displaystyle SD$, and $\displaystyle {SN:ND}={1:2}$.