If -a-b- b-c- c-a- then - isA- 1B- 3C- 2D- 0

The correct option is A 1[a⃗×b⃗, b⃗×c⃗, c⃗×a⃗]=(a⃗×b⃗). {(b⃗×c⃗) × (c⃗×a⃗)} (1) Let (b⃗×c⃗)=p⃗, then (b⃗×c⃗) × (c⃗×a⃗)=p⃗× (c⃗×a⃗) =(a⃗. p⃗) c⃗ (c⃗. p⃗)a ...

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Byju's Answer

Standard XII

Mathematics

Permutation: n Different Things Taken r at a Time

If [a⃗×b⃗, b⃗...

Question

If $[\to a \times \to b, \to b \times \to c, \to c \times \to a]$ \ then $λ$ is

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Solution

The correct option is A 1
$[\to a \times \to b, \to b \times \to c, \to c \times \to a] = (\to a \times \to b) . {(\to b \times \to c) \times (\to c \times \to a)} (1) L e t (\to b \times \to c) = \to p,$ then
$(\to b \times \to c) \times (\to c \times \to a) = \to p \times (\to c \times \to a) = (\to a . \to p) \to c - (\to c . \to p) \to a$ [This identity for vector triple product]
Now, $\to a . \to p = \to a . (\to b \times \to c) = [\to a . \to b . \to c] \to c . \to p = \to c (\to b \times \to c) = [\to c, \to b, \to c] = 0$
$∴$ (2) becomes
$(\to b \times \to c) \times (\to c \times \to a) = [\to a . \to b . \to c] \to c - 0$
$∴$ (1) becomes
$(\to a \times \to b) . [\to a . \to b . \to c] \to c = [\to a . \to b . \to c] \to c {(\to a \times \to b) . \to c} = [\to a . \to b . \to c]^{2} ∴ λ = 1$

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If [→a×→b,→b×→c,→c×→a]\ then λ is

If $[\to a \times \to b, \to b \times \to c, \to c \times \to a]$ \ then $λ$ is