If (a×b)×(c×d)=λc+μd then λ, μ=

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

If (a×b)×(c×d)=λc+μd then λ, μ=

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

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If $(a \times b) \times (c \times d) = λ c + μd$ then $λ, μ =$

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$[a b c], 1$

$[a b c], [b c d]$

$[a b d], - [a b c]$

1, 0

answer is A.

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

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If (a×b)×(c×d)=λc+μd then λ, μ=

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If $(a \times b) \times (c \times d) = λ c + μd$ then $λ, μ =$