Linear combinations of vectors

A linear combination of $\displaystyle \vec{u}$ and $\displaystyle \vec{v}$ is a vector of the form $\displaystyle s\vec{u}+t\vec{v}$, where $\displaystyle s$ and $\displaystyle t$ are numbers. All the vectors expressed so far – $\displaystyle \vec{u}+\frac{1}{2}\vec{v}$, $\displaystyle -\vec{u}+\vec{v}+\vec{w}$ – are linear combinations.

Two theorems (without proof):

• In the plane: if $\displaystyle \vec{u}$ and $\displaystyle \vec{v}$ are not collinear, every vector in their plane can be written in exactly one way as $\displaystyle s\vec{u}+t\vec{v}$. Every such combination lies in the plane of $\displaystyle \vec{u}$ and $\displaystyle \vec{v}$ or in a plane parallel to it.
• In space: if $\displaystyle \vec{u}$, $\displaystyle \vec{v}$, $\displaystyle \vec{w}$ are not in one plane, every vector in space can be written in exactly one way as $\displaystyle s\vec{u}+t\vec{v}+r\vec{w}$.

Comparing coefficients: since the way is unique, if $\displaystyle {s\vec{u}+t\vec{v}}={s'\vec{u}+t'\vec{v}}$, then $\displaystyle {s}={s'}$ and $\displaystyle {t}={t'}$.

$\displaystyle ABCD$ is a parallelogram, $\displaystyle {\overrightarrow{AB}}={\vec{u}}$, $\displaystyle {\overrightarrow{AD}}={\vec{v}}$.
$\displaystyle E$ is the midpoint of $\displaystyle BC$. The segments $\displaystyle AE$ and $\displaystyle BD$ meet at the point $\displaystyle K$.
In what ratio does $\displaystyle K$ divide $\displaystyle AE$?

The solution below expresses $\displaystyle \overrightarrow{AK}$ in two ways, and compares coefficients.

• $\displaystyle K$ is on $\displaystyle AE$: $\displaystyle {\overrightarrow{AK}}={t\cdot\overrightarrow{AE}}={t\left(\vec{u}+\frac{1}{2}\vec{v}\right)}={t\vec{u}+\frac{t}{2}\vec{v}}$.
• $\displaystyle K$ is on $\displaystyle BD$: $\displaystyle {\overrightarrow{AK}}={\overrightarrow{AB}+s\cdot\overrightarrow{BD}}={\vec{u}+s(\vec{v}-\vec{u})}={(1-s)\vec{u}+s\vec{v}}$.
• Comparing the coefficients of $\displaystyle \vec{u}$ and of $\displaystyle \vec{v}$ gives a system of equations:

So $\displaystyle {t}={1-\frac{t}{2}}$, $\displaystyle {t}={\frac{2}{3}}$. That is, $\displaystyle {\overrightarrow{AK}}={\frac{2}{3}\overrightarrow{AE}}$, and $\displaystyle {AK:KE}={2:1}$.

The plane $\displaystyle ABC$ is determined by $\displaystyle {\vec{u}}={\overrightarrow{AB}}$ and $\displaystyle {\vec{v}}={\overrightarrow{AC}}$. If $\displaystyle \overrightarrow{PQ}$ is a linear combination of $\displaystyle \vec{u}$ and $\displaystyle \vec{v}$, then $\displaystyle PQ$ lies in the plane $\displaystyle ABC$ (when $\displaystyle P$ is in the plane) or is parallel to it (when $\displaystyle P$ is off the plane).

Example. In the triangular pyramid $\displaystyle ABCD$: $\displaystyle {\overrightarrow{AB}}={\vec{u}}$, $\displaystyle {\overrightarrow{AC}}={\vec{v}}$, $\displaystyle {\overrightarrow{AD}}={\vec{w}}$.
The point $\displaystyle M$ is on $\displaystyle AD$, and $\displaystyle {AM:MD}={3:1}$.
The point $\displaystyle N$ satisfies $\displaystyle {\overrightarrow{DN}}={\frac{1}{8}\left(\overrightarrow{DB}+\overrightarrow{DC}\right)}$.
Is $\displaystyle MN$ parallel to the plane $\displaystyle ABC$?

• $\displaystyle {\overrightarrow{DB}}={\vec{u}-\vec{w}}$ and $\displaystyle {\overrightarrow{DC}}={\vec{v}-\vec{w}}$, so $\displaystyle {\overrightarrow{DN}}={\frac{1}{8}\vec{u}+\frac{1}{8}\vec{v}-\frac{1}{4}\vec{w}}$.
• $\displaystyle {\overrightarrow{MN}}={\overrightarrow{MD}+\overrightarrow{DN}}={\frac{1}{4}\vec{w}+\frac{1}{8}\vec{u}+\frac{1}{8}\vec{v}-\frac{1}{4}\vec{w}}={\frac{1}{8}\vec{u}+\frac{1}{8}\vec{v}}$.
• $\displaystyle \overrightarrow{MN}$ is a combination of $\displaystyle \vec{u}$ and $\displaystyle \vec{v}$, and $\displaystyle M$ is off the plane $\displaystyle ABC$: $\displaystyle MN$ is parallel to the plane $\displaystyle ABC$.
• More than that: if $\displaystyle F$ is the midpoint of $\displaystyle BC$, then $\displaystyle {\overrightarrow{AF}}={\frac{1}{2}\vec{u}+\frac{1}{2}\vec{v}}$, and $\displaystyle {\overrightarrow{MN}}={\frac{1}{4}\overrightarrow{AF}}$. So $\displaystyle MN\parallel AF$.

$\displaystyle ABCD$ is a parallelogram, $\displaystyle {\overrightarrow{AB}}={\vec{u}}$, $\displaystyle {\overrightarrow{AD}}={\vec{v}}$.
$\displaystyle M$ is the midpoint of $\displaystyle CD$. The segments $\displaystyle AM$ and $\displaystyle BD$ meet at the point $\displaystyle K$.

Compare coefficients, and solve the system.

In what ratio does $\displaystyle K$ divide $\displaystyle AM$?

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 $\displaystyle AA'$, and $\displaystyle N$ the midpoint of $\displaystyle CC'$.

Is $\displaystyle MN$ parallel to the base plane $\displaystyle ABCD$? Explain.

In the triangle $\displaystyle ABC$: $\displaystyle {\overrightarrow{AB}}={\vec{u}}$, $\displaystyle {\overrightarrow{AC}}={\vec{v}}$.
The point $\displaystyle D$ is on $\displaystyle BC$, and $\displaystyle {BD:DC}={1:3}$.
Write $\displaystyle \overrightarrow{AD}$ as a linear combination 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 $\displaystyle {AE}={\frac{1}{3}AB}$. The segments $\displaystyle DE$ and $\displaystyle AC$ meet at the point $\displaystyle K$.
Express $\displaystyle \overrightarrow{AK}$ in two ways, compare coefficients, and find the ratio in which $\displaystyle K$ divides $\displaystyle AC$.