Question

If vectors $ \overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$ are coplanar, then show that
$\begin{vmatrix}\overrightarrow{a} & \overrightarrow{b} & \overrightarrow{c} \\ \overrightarrow{a}.\overrightarrow{a} & \overrightarrow{a}.\overrightarrow{b} & \overrightarrow{a}.\overrightarrow{c} \\ \overrightarrow{b}.\overrightarrow{a} & \overrightarrow{b}.\overrightarrow{b} & \overrightarrow{b}.\overrightarrow{c} \end{vmatrix}=\overrightarrow{0}$

Answer 1

Given that, $\overrightarrow{ a } \overrightarrow{ b } \overrightarrow{ c }$ are coplanar vectors.
$\therefore$ There exists scalars $x, y, z$ not all zero, such that
$x \overrightarrow{ a }+y \overrightarrow{ b }+z \overrightarrow{ c }=0 .....$(i)
Taking dot with $\overrightarrow{ a }$ and $\overrightarrow{ b }$ respectively, we get
$x(\overrightarrow{ a } \cdot \overrightarrow{ a })+y(\overrightarrow{ a } \cdot \overrightarrow{ b })+z(\overrightarrow{ a } \cdot \overrightarrow{ c })=0 .....$(ii)
and $x(\overrightarrow{ a } \cdot \overrightarrow{ b })+y(\overrightarrow{ b } \cdot \overrightarrow{ b })+z(\overrightarrow{ c } \cdot \overrightarrow{ b })=0 ......$(iii)
Since, Eqs. (i), (ii) and (iii) represent homogeneous equations with $(x, y, z) \neq(0,0,0)$.
$\Rightarrow$ Non-trivial solutions
$\therefore \Delta = 0$
$\Rightarrow \begin{vmatrix}\overrightarrow{ a } & \overrightarrow{ b } & \overrightarrow{ c } \\ \overrightarrow{ a} \cdot \overrightarrow{ a } & \overrightarrow{ a } \cdot \overrightarrow{ b } & \overrightarrow{ a } \cdot \overrightarrow{ c } \\ \overrightarrow{ b } \cdot \overrightarrow{ b } & \overrightarrow{ b } \cdot \overrightarrow{ b } & \overrightarrow{ b } \cdot \overrightarrow{ c }\end{vmatrix}=\overrightarrow{0}$