For any four vectors a→,b⇀,c⇀,d⇀, the expression b→×c.a→×d+c×a.b×d+a×b.c×d is always equal to

b·ab·dc·ac·d+c·bc·da·ba·d+a·ca·db·cb·d =(a·b) (c·d)-(a·c) (b·d)+(a·d) (b·c)-(a·b) (c·d)+(a·c) (b·d)-(b·c) (a·d) =0

For any four vectors a→,b⇀,c⇀,d⇀, the expression b→×c.a→×d+c×a.b×d+a×b.c×d is always equal to

b·ab·dc·ac·d+c·bc·da·ba·d+a·ca·db·cb·d =(a·b) (c·d)-(a·c) (b·d)+(a·d) (b·c)-(a·b) (c·d)+(a·c) (b·d)-(b·c) (a·d) =0

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For any four vectors $\vec{a}, \overset{⇀}{b}, \overset{⇀}{c}, \overset{⇀}{d},$ the expression $(\vec{b} \times c) . (\vec{a} \times d) + (c \times a) . (b \times d) + (a \times b) . (c \times d)$ is always equal to

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$[a, b, c]$

$[a, c, d]$

$[b, c, d]$

answer is D.

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Detailed Solution

$|\begin{matrix} b \cdot a & b \cdot d \\ c \cdot a & c \cdot d \end{matrix}| + |\begin{matrix} c \cdot b & c \cdot d \\ a \cdot b & a \cdot d \end{matrix}| + |\begin{matrix} a \cdot c & a \cdot d \\ b \cdot c & b \cdot d \end{matrix}| = (a \cdot b) (c \cdot d) - (a \cdot c) (b \cdot d) + (a \cdot d) (b \cdot c) - (a \cdot b) (c \cdot d) + (a \cdot c) (b \cdot d) - (b \cdot c) (a \cdot d) = 0$