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If the vectors $\mathbf{a}, \mathbf{b}$ and $\mathbf{c}$ are coplanar, then $\left|\begin{array}{ccc}a & b & c \\ a \cdot a & a \cdot b & a \cdot c \\ b \cdot a & b \cdot b & b \cdot c\end{array}\right|$ is equal to
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Since, $\mathbf{a}, \mathbf{b}$ and $\mathbf{c}$ are coplanar, there must exists three scalars $x, y$ and $z$ are not all zero such that
$$
x \mathbf{a}+y \mathbf{b}+z \mathbf{c}=0
$$
Multiplying both sides of Eq. (i) by a and $\mathbf{b}$ respectively, we get
$$
\begin{array}{l}
x \mathbf{a} \cdot \mathbf{a}+y \mathbf{a} \cdot \mathbf{b}+z \mathbf{a} \cdot \mathbf{c}=0 \\
x \mathbf{b} \cdot \mathbf{a}+y \mathbf{b} \cdot \mathbf{b}+z \mathbf{b} \cdot \mathbf{c}=0
\end{array}
$$
Eliminating $x, y$ and $z$ from Eqs. (i), (ii) and
(iii), we get
$$
\left|\begin{array}{ccc}
\mathbf{a} & \mathbf{b} & \mathbf{c} \\
\mathbf{a} \cdot \mathbf{a} & \mathbf{a} \cdot \mathbf{b} & \mathbf{a} \cdot \mathbf{c} \\
\mathbf{b} \cdot \mathbf{a} & \mathbf{b} \cdot \mathbf{b} & \mathbf{b} \cdot \mathbf{c}
\end{array}\right|=0
$$
$$
x \mathbf{a}+y \mathbf{b}+z \mathbf{c}=0
$$
Multiplying both sides of Eq. (i) by a and $\mathbf{b}$ respectively, we get
$$
\begin{array}{l}
x \mathbf{a} \cdot \mathbf{a}+y \mathbf{a} \cdot \mathbf{b}+z \mathbf{a} \cdot \mathbf{c}=0 \\
x \mathbf{b} \cdot \mathbf{a}+y \mathbf{b} \cdot \mathbf{b}+z \mathbf{b} \cdot \mathbf{c}=0
\end{array}
$$
Eliminating $x, y$ and $z$ from Eqs. (i), (ii) and
(iii), we get
$$
\left|\begin{array}{ccc}
\mathbf{a} & \mathbf{b} & \mathbf{c} \\
\mathbf{a} \cdot \mathbf{a} & \mathbf{a} \cdot \mathbf{b} & \mathbf{a} \cdot \mathbf{c} \\
\mathbf{b} \cdot \mathbf{a} & \mathbf{b} \cdot \mathbf{b} & \mathbf{b} \cdot \mathbf{c}
\end{array}\right|=0
$$
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