If $a, b, c$ are non-coplanar vectors, then $|\begin{matrix} a \cdot a & a \cdot b & a \cdot c \\ b \cdot a & b \cdot b & b \cdot c \\ c \cdot a & c \cdot b & c \cdot c \end{matrix}| =$

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$[a b c]$

${[a b c]}^{2}$

answer is B.

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Detailed Solution

$a = a_{1} i + a_{2} j + a_{3} k, b = b_{1} i + b_{2} j + b_{3} k and c = c_{1} i + c_{2} j + c_{3} k = |\begin{array}{lll} a \cdot a a \cdot b a \cdot c \\ b \cdot a b \cdot b b \cdot c \\ c \cdot a c \cdot b c \cdot c \end{array}| = |\begin{array}{ccc} {a_{1}}^{2} + {a_{2}}^{2} + {a_{3}}^{2} & a_{1} b_{1} + a_{2} b_{2} + a_{3} b_{3} & a_{1} c_{1} + a_{2} c_{2} + a_{3} c_{3} \\ b_{1} a_{1} + b_{2} a_{2} + b_{3} a_{3} & {b_{1}}^{2} + {b_{2}}^{2} + {b_{3}}^{2} & b_{1} c_{1} + b_{2} c_{2} + b_{3} c_{3} \\ c_{1} a_{1} + c_{2} a_{2} + c_{3} a_{3} & c_{1} b_{1} + c_{2} b_{2} + c_{3} b_{3} & c_{1}^{2} + {c_{2}}^{2} + {c_{3}}^{2} \end{array}| = |\begin{array}{lll} a_{1} a_{2} a_{3} \\ b_{1} b_{2} b_{3} \\ c_{1} c_{2} c_{3} \end{array}| \times |\begin{array}{lll} a_{1} a_{2} a_{3} \\ b_{1} b_{2} b_{3} \\ c_{1} c_{2} c_{3} \end{array}| = [a b c] [a b c]$