If three vectors a , b and c are coplanar (lie in one plane…

If three vectors $\vec{a}$, $\vec{b}$ and $\vec{c}$ are coplanar (lie in one plane or are parallel to one plane), then there exist constants $\alpha$ and $\beta$ such that $\vec{c} = \alpha\vec{a}+\beta\vec{b}$. The converse also holds. Prove both directions.

Step 1: Preliminary: a non-collinear basis

Throughout we assume that $\vec{a}$ and $\vec{b}$ are non-collinear. If $\vec{a} \parallel \vec{b}$, every combination $\alpha\vec{a}+\beta\vec{b}$ is collinear with $\vec{a}$ and the statement reduces to the collinear case. So we consider only $\vec{a} \nparallel \vec{b}$.

Step 2: Coplanar $\Rightarrow$ $\vec{c} = \alpha\vec{a}+\beta\vec{b}$

Hypothesis

$\vec{a}$, $\vec{b}$, $\vec{c}$ lie in a common plane $\pi$ (or are parallel to it).

Construction

Translate the three vectors so that their tails meet at a common point $O$ of $\pi$; let $P$ be the tip of $\vec c$, so $\vec c=\overrightarrow{OP}$ and $P\in\pi$.

• Through $P$ draw a line parallel to $\vec{b}$. Because $\vec a\nparallel\vec b$, it meets the line of $\vec{a}$ in exactly one point $Q$. Since $\overrightarrow{OQ}$ lies on the line of $\vec a$, $\overrightarrow{OQ} = \alpha\,\vec{a}$ for some $\alpha \in \mathbb{R}$.
• Through $P$ draw a line parallel to $\vec{a}$. It meets the line of $\vec{b}$ in exactly one point $S$, and $\overrightarrow{OS} = \beta\,\vec{b}$ for some $\beta \in \mathbb{R}$.

$OQPS$ is a parallelogram, so $\overrightarrow{QP} = \overrightarrow{OS} = \beta\vec b$. By the triangle rule

\[ \vec c=\overrightarrow{OP} = \overrightarrow{OQ} + \overrightarrow{QP} = \alpha\,\vec{a} + \beta\,\vec{b}. \]

Uniqueness of $\alpha$ and $\beta$

Suppose $\vec{c} = \alpha_1\vec{a}+\beta_1\vec{b} = \alpha_2\vec{a}+\beta_2\vec{b}$. Then

\[ (\alpha_1-\alpha_2)\vec{a} + (\beta_1-\beta_2)\vec{b} = \vec{0}. \]

If $\alpha_1\neq\alpha_2$ we could solve for $\vec a$ as a multiple of $\vec b$, contradicting $\vec a\nparallel\vec b$. Hence $\alpha_1 = \alpha_2$, and then $\beta_1 = \beta_2$. The decomposition is unique.

Step 3: $\vec{c} = \alpha\vec{a}+\beta\vec{b}$ $\Rightarrow$ coplanar

Hypothesis

There are real numbers $\alpha$, $\beta$ with $\vec{c} = \alpha\vec{a}+\beta\vec{b}$.

Proof

Place all vectors at a common origin $O$. The non-collinear vectors $\vec{a}$ and $\vec{b}$ determine a unique plane $\pi$ through $O$. The vector $\alpha\vec{a}$ lies on the line of $\vec{a}$, hence in $\pi$; likewise $\beta\vec{b}$ lies in $\pi$. Their sum, constructed by the parallelogram rule from two vectors in $\pi$, is the diagonal of a parallelogram lying in $\pi$. Therefore $\vec c$ lies in $\pi$, and $\vec{a}$, $\vec{b}$, $\vec{c}$ are coplanar. $\blacksquare$

Step 4: Complete statement

Theorem (coplanar decomposition). Let $\vec{a}$ and $\vec{b}$ be non-collinear vectors. A vector $\vec{c}$ is coplanar with $\vec{a}$ and $\vec{b}$ if and only if there exist unique real numbers $\alpha$, $\beta$ such that

\[\vec{c} = \alpha\,\vec{a} + \beta\,\vec{b}\]