If b-c-c-a-3c- then -b-c- c-a- a-b-

The correct option is D 9 (a×c)×(c×a)=(a×c·a)c (b×c·c)a #160;3c=[abc]c 0 [abc]=3 [a×c #160;c×a #160;a×b]=(b×c×c×a)·a×b =((b×c ...

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If b⃗×c⃗×c⃗...

Question

If $(\to b \times \to c) \times (\to c \times \to a) = 3 \to c$ , then $[\to b \times \to c$ $\to c \times \to a$ $\to a \times \to b] =$

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Similar questions

\to a + 2 \to b + 3 \to c = 0

\to b \times \to c + \to c \times \to a + \to a \times \to b

\to A = \to b \times \to c, \to B = \to c \times \to a, \to C = \to a \times \to b

\to A \times (\to B \times \to C)

\to B \times (\to C \times \to A)

\to C \times (\to A \times \to B)

| \to a | = 3, ∣ ∣ \to b ∣ ∣ = 4, | \to c | = 5, \to a ⊥ (\to b + \to c), \to b ⊥ (\to c + \to a)

\to c ⊥ (\to a + \to b)

\sqrt{2} ∣ ∣ \to a + \to b + \to c ∣ ∣

| \to a | = 1, ∣ ∣ \to b ∣ ∣ = \sqrt{2}

| \to c | = \sqrt{3}

\to a ⊥ (\to b + \to c)

\to b ⊥ (\to c + \to a)

\to c ⊥ (\to a + \to b)

∣ ∣ \to a + \to b + \to c ∣ ∣

\to a \times \to b = \to b \times \to c = \to c \times \to a

\to a + \to b + \to c =

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