If a- b- c- are non-coplanar non-zero vectors such that b-c-a- a-b-c- c-a-b- then

The correct options are A |a⃗|=1 B |b⃗|=1 C |a⃗|+|b⃗|+|c⃗|=3 We have b⃗×c⃗=a⃗ ⇒ #160; [a⃗b⃗c⃗]=(a⃗.a⃗)=|a⃗|^2 Similarly [a⃗b⃗c ...

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Binomial Coefficients

If a⃗, b⃗, ...

Question

If $\to a, \to b, \to c$ are non-coplanar non-zero vectors such that $\to b \times \to c = \to a$ , $\to a \times \to b = \to c$ , $\to c \times \to a = \to b$ , then

| \to a | = 1

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| \to b | = 1

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| \to a | + | \to b | + | \to c | = 3

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\to a, \to b, \to c

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| (\to a \times \to b) . \to c | = | \to a | | \to b | | \to c |

| \to a + \to b + \to c |^{2} =

\to a, \to b, \to c

[\to a \times \to b

\to b \times \to c

\to c \times \to a] =

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