Question

If vectors $\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$ are coplanar, then show that
$\left|\begin{array}{ccc}\overrightarrow{a}& \overrightarrow{b}& \overrightarrow{c}\\ \overrightarrow{a}.\overrightarrow{a}& \overrightarrow{a}.\overrightarrow{b}& \overrightarrow{a}.\overrightarrow{c}\\ \overrightarrow{b}.\overrightarrow{a}& \overrightarrow{b}.\overrightarrow{b}& \overrightarrow{b}.\overrightarrow{c}\end{array}\right|=\overrightarrow{0}$

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
$x\overrightarrow{a}+y\overrightarrow{b}+z\overrightarrow{c}=\mathrm{0.....}$(i)
Taking dot with $\overrightarrow{a}$ and $\overrightarrow{b}$ respectively, we get
$x(\overrightarrow{a}\cdot \overrightarrow{a})+y(\overrightarrow{a}\cdot \overrightarrow{b})+z(\overrightarrow{a}\cdot \overrightarrow{c})=\mathrm{0.....}$(ii)
and $x(\overrightarrow{a}\cdot \overrightarrow{b})+y(\overrightarrow{b}\cdot \overrightarrow{b})+z(\overrightarrow{c}\cdot \overrightarrow{b})=\mathrm{0......}$(iii)
Since, Eqs. (i), (ii) and (iii) represent homogeneous equations with $(x,y,z)\ne (0,0,0)$.
$\Rightarrow$ Non-trivial solutions
$\therefore \Delta =0$
$\Rightarrow \left|\begin{array}{ccc}\overrightarrow{a}& \overrightarrow{b}& \overrightarrow{c}\\ \overrightarrow{a}\cdot \overrightarrow{a}& \overrightarrow{a}\cdot \overrightarrow{b}& \overrightarrow{a}\cdot \overrightarrow{c}\\ \overrightarrow{b}\cdot \overrightarrow{b}& \overrightarrow{b}\cdot \overrightarrow{b}& \overrightarrow{b}\cdot \overrightarrow{c}\end{array}\right|=\overrightarrow{0}$