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

I dont know how to prove it.I read from wikipedia page the condition for three vectors $\vec{a},\vec{b},\vec{c}$ to be coplanar is $(\vec{c}\cdot\hat{a})\hat{a}+(\vec{c}\cdot\hat{b})\hat{b}=\vec{c}$,but this is not proving helpful here.
Please help me.Thanks.
 A: I use $\langle \cdot,\cdot\rangle$ for the dot product.
From:
$$
\begin{vmatrix}
\vec{a}&\vec{b}&\vec{c}\\
\langle\vec{a},\vec{a}\rangle&\langle\vec{a},\vec{b}\rangle&\langle\vec{a},\vec{c}\rangle\\
\langle\vec{b},\vec{a}\rangle&\langle\vec{b},\vec{b}\rangle&\langle\vec{b},\vec{c}\rangle
\end{vmatrix}=\vec{0}
$$
we have:
$$
\vec a 
\begin{vmatrix}
\langle\vec{a},\vec{b}\rangle&\langle\vec{a},\vec{c}\rangle\\
\langle\vec{b},\vec{b}\rangle&\langle\vec{b},\vec{c}\rangle
\end{vmatrix}
- \vec b
\begin{vmatrix}
\langle\vec{a},\vec{a}\rangle&\langle\vec{a},\vec{c}\rangle\\
\langle\vec{b},\vec{a}\rangle&\langle\vec{b},\vec{c}\rangle
\end{vmatrix}
+\vec c
\begin{vmatrix}
\langle\vec{a},\vec{a}\rangle&\langle\vec{a},\vec{b}\rangle\\
\langle\vec{b},\vec{a}\rangle&\langle\vec{b},\vec{b}\rangle
\end{vmatrix}=\vec a|A|-\vec b|B|+\vec c|C| =0 \qquad \Rightarrow
$$
$$
\Rightarrow
\vec c =\dfrac{\vec b |B|}{|C|}-\dfrac{\vec a |A|}{|C|}
$$
so, by linearity of the inner product, we have:
$$
\langle\vec c,\vec c\rangle =\dfrac{\langle\vec b,\vec c\rangle |B|}{|C|}-\dfrac{\langle\vec a,\vec c\rangle |A|}{|C|}
$$
i.e.
$$
\langle \vec c,\vec c\rangle |C|+\langle \vec c,\vec a\rangle |A|-\langle \vec c,\vec b\rangle |B|=0
$$
and, back substituting $|A|,|B|,|C|$ we see that this is the Gramian of the three vectors that is null iff the vectors are coplanar.
A: Let $\vec{a}=a_1\hat{\imath}+ a_2\hat{\jmath}+a_3\hat{k}$ , $\vec{b}=b_1\hat{\imath}+ b_2\hat{\jmath}+b_3\hat{k}$ and $\vec{c}=c_1\hat{\imath}+ c_2\hat{\jmath}+c_3\hat{k}$
Then given determinant can be written as the product of the two determinants in this manner:
$$\begin{vmatrix}\vec{a}&\vec{b}&\vec{c}\\\vec{a}\cdot\vec{a}&\vec{a}\cdot\vec{b}&\vec{a}\cdot\vec{c}\\\vec{b}\cdot\vec{a}&\vec{b}\cdot\vec{b}&\vec{b}\cdot\vec{c}\end{vmatrix}
=\begin{vmatrix} \hat{\imath}&\hat{\jmath}&\hat{k}\\\ a_1&a_2&a_3 \\\ b_1&b_2&b_3\end{vmatrix}\cdot\begin{vmatrix} a_1&b_1&c_1\\\ a_2 & b_2&c_2 \\\ a_3&b_3&c_3 \end{vmatrix}
$$
I don't suppose there is need for any further explanation...
A: Let $\vec a\cdot \vec a=d$, $\vec a\cdot \vec b=e$, and $\vec a\cdot \vec c=f$.
Then the determinant is $\vec a(ef-fe)-\vec b(df-fd)+\vec c(de-ed)=\vec 0$.
