Collinear vectors: parallel lines and points on one line

Two vectors other than the zero vector are collinear if and only if one of them is a number times the other:

Two uses:

• Parallel lines: if $\displaystyle {\overrightarrow{CD}}={t\cdot\overrightarrow{AB}}$, and the lines $\displaystyle AB$ and $\displaystyle CD$ are not one line, then $\displaystyle AB\parallel CD$.
• Three points on one line: if $\displaystyle {\overrightarrow{AC}}={t\cdot\overrightarrow{AB}}$, then $\displaystyle A$, $\displaystyle B$, $\displaystyle C$ are on one line: the vectors are collinear and share the point $\displaystyle A$.

If $\displaystyle M$ is on the segment $\displaystyle AB$ and $\displaystyle {AM:MB}={2:3}$, then $\displaystyle AM$ is $\displaystyle \frac{2}{5}$ of $\displaystyle AB$, in the same direction:

• $\displaystyle M$ the midpoint of $\displaystyle AB$: $\displaystyle {\overrightarrow{AM}}={\frac{1}{2}\overrightarrow{AB}}$.
• In the triangle $\displaystyle ABC$ the medians meet at a point $\displaystyle M$ that divides each median in the ratio $\displaystyle 2:1$ from the vertex (grade 10). If $\displaystyle K$ is the midpoint of $\displaystyle BC$: $\displaystyle {\overrightarrow{AM}}={\frac{2}{3}\overrightarrow{AK}}$.

In the trapezoid $\displaystyle ABCD$ ($\displaystyle AD\parallel BC$): $\displaystyle {AD}={2BC}$.
The point $\displaystyle O$ is on the diagonal $\displaystyle AC$, and $\displaystyle {OC:AO}={1:2}$.
$\displaystyle M$ and $\displaystyle N$ are the midpoints of the sides $\displaystyle BC$ and $\displaystyle AD$.
Let $\displaystyle {\overrightarrow{AB}}={\vec{u}}$, $\displaystyle {\overrightarrow{BC}}={\vec{v}}$.
Prove that $\displaystyle M$, $\displaystyle O$, $\displaystyle N$ are on one line.

The proof below expresses $\displaystyle \overrightarrow{MN}$ and $\displaystyle \overrightarrow{MO}$ in terms of $\displaystyle \vec{u}$ and $\displaystyle \vec{v}$, and shows that one is a number times the other.

• $\displaystyle {\overrightarrow{AD}}={2\vec{v}}$: $\displaystyle AD$ is parallel to $\displaystyle BC$, in the same direction and twice as long.
• $\displaystyle {\overrightarrow{AM}}={\vec{u}+\frac{1}{2}\vec{v}}$, and $\displaystyle {\overrightarrow{AN}}={\frac{1}{2}\overrightarrow{AD}}={\vec{v}}$.
• $\displaystyle {\overrightarrow{MN}}={\overrightarrow{AN}-\overrightarrow{AM}}={-\vec{u}+\frac{1}{2}\vec{v}}$.
• $\displaystyle {\overrightarrow{AO}}={\frac{2}{3}\overrightarrow{AC}}={\frac{2}{3}\vec{u}+\frac{2}{3}\vec{v}}$, so $\displaystyle {\overrightarrow{MO}}={\overrightarrow{AO}-\overrightarrow{AM}}={-\frac{1}{3}\vec{u}+\frac{1}{6}\vec{v}}$.
• $\displaystyle {\overrightarrow{MO}}={\frac{1}{3}\overrightarrow{MN}}$: the vectors are collinear and share the point $\displaystyle M$, so $\displaystyle M$, $\displaystyle O$, $\displaystyle N$ are on one line.

$\displaystyle \vec{u}$ and $\displaystyle \vec{v}$ are not collinear. Is each pair of vectors collinear? If it is, write the number $\displaystyle t$.

$\displaystyle \vec{u}$ and $\displaystyle -3\vec{u}$

$\displaystyle 2\vec{u}+\vec{v}$ and $\displaystyle 4\vec{u}+2\vec{v}$

$\displaystyle \vec{u}+\vec{v}$ and $\displaystyle \vec{u}-\vec{v}$

In the triangle $\displaystyle ABC$ the medians meet at the point $\displaystyle M$, and $\displaystyle K$ is the midpoint of the side $\displaystyle BC$.
$\displaystyle {\overrightarrow{AB}}={\vec{u}}$, $\displaystyle {\overrightarrow{AC}}={\vec{v}}$. Express in terms of $\displaystyle \vec{u}$ and $\displaystyle \vec{v}$:

$\displaystyle ABCD$ is a parallelogram, $\displaystyle {\overrightarrow{AB}}={\vec{u}}$, $\displaystyle {\overrightarrow{AD}}={\vec{v}}$.
The point $\displaystyle E$ is on $\displaystyle AB$, and $\displaystyle {AE:EB}={1:2}$.
The point $\displaystyle F$ is on the diagonal $\displaystyle AC$, and $\displaystyle {AF:FC}={1:3}$.

Are $\displaystyle D$, $\displaystyle F$, $\displaystyle E$ on one line? Explain.

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

Prove that $\displaystyle MN\parallel BD$. How many times longer than $\displaystyle MN$ is $\displaystyle BD$?