Coplanar Vectors

Two or more vectors are coplanar if they can be drawn lying in a single plane, or in mutually parallel

planes.

Theorem (a vector coplanar with two non-collinear vectors). If $\bar a,\bar b$ are non-collinear, a vector

$\bar r$ is coplanar with $\bar a$ and $\bar b$ if and only if there exist unique scalars $t_1,t_2$ with

$\bar r=t_1\bar a+t_2\bar b$. Sketch of proof: place $\bar a$ along $\overrightarrow{OA}$ and $\bar b$ along

$\overrightarrow{OB}$; for $\overrightarrow{OP}=\bar r$, draw a line through $P$ parallel to $OB$ meeting line

$OA$ at $M$, and a line through $P$ parallel to $OA$ meeting line $OB$ at $N$; then

$\overrightarrow{OM}=t_1\bar a,\overrightarrow{ON}=t_2\bar b$ for some scalars $t_1,t_2$, and the

parallelogram/triangle law gives $\bar r=\overrightarrow{OM}+\overrightarrow{ON}=t_1\bar a+t_2\bar b$.

Uniqueness follows because if $\bar r$ also equalled $s_1\bar a+s_2\bar b$, subtracting gives

$(t_1-s_1)\bar a+(t_2-s_2)\bar b=\bar 0$, and since $\bar a,\bar b$ are non-collinear this forces

$t_1=s_1,t_2=s_2$.

A combination $m_1\bar a_1+m_2\bar a_2+\cdots+m_n\bar a_n$ (with the $m_i$ scalars, at least one non-zero) is

called a linear combination of $\bar a_1,\ldots,\bar a_n$; in particular, for $\bar c=m\bar a+n\bar b$,

the three vectors $\bar a,\bar b,\bar c$ are automatically coplanar.

Theorem (three coplanar vectors). $\bar a,\bar b,\bar c$ are coplanar if and only if there is a non-zero

combination $x\bar a+y\bar b+z\bar c=\bar 0$ (i.e. $(x,y,z)\ne(0,0,0)$). If instead $\bar a,\bar b,\bar c$ are

not coplanar, the only solution of $x\bar a+y\bar b+z\bar c=\bar 0$ is $x=y=z=0$ -- such vectors are

called linearly independent (and non-coplanar); when a non-zero solution exists, they are linearly dependent (and coplanar).

Worked examples.

• To show $4\hat i+13\hat j-18\hat k,\ 2\hat i-3\hat j+3\hat k,\ 2\hat i+3\hat j-4\hat k$ are coplanar, write the first as $m$ times the second plus $n$ times the third, match $\hat i,\hat j,\hat k$ coefficients to get three equations, solve any two for $m,n$ and check the third is automatically satisfied -- here $m=-2,n=3$ …