(a×b)x(a×c)-d=k[a b c] then k=

⇒a×b·c a-(a×b·a)c·d ⇒[a b c] (a·d)k=a·d

(a×b)x(a×c)-d=k[a b c] then k=

⇒a×b·c a-(a×b·a)c·d ⇒[a b c] (a·d)k=a·d

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$[(a \times b) x (a \times c)] - d = k [a b c]$ then $k =$

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$a . c$

$a . b$

$a . d$

$a . a$

answer is A.

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Detailed Solution

$\begin{array}{rcl} \Rightarrow & [(a \times b \cdot c) a - (a \times b \cdot a) c] \cdot d \\ \Rightarrow & [a b c] (a \cdot d) \\ k & = & a \cdot d \end{array}$

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(a×b)x(a×c)-d=k[a b c] then k=

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$[(a \times b) x (a \times c)] - d = k [a b c]$ then $k =$