Let -a- b-c- - -b- c-a- and -c- a-b- - then

The correct option is B λ⃗, μ⃗, ν⃗ are coplanar λ⃗ ⃗ ⃗ ⃗#⃗1⃗6⃗0⃗;⃗ +μ⃗ ⃗ ⃗ ⃗#⃗1⃗6⃗0⃗;⃗ =a⃗×( b⃗ +c⃗ #160; ) +b⃗×( c⃗ +a⃗ #160; ) = a⃗×b⃗ +a⃗ ...

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Scalar Triple Product

Let λ⃗=a⃗× ...

Question

Let $\to λ = \to a \times (\to b + \to c)$ , $\to μ = \to b \times (\to c + \to a)$ and $\to ν = \to c \times (\to a + \to b)$ , then

\to λ + \to μ = \to ν

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\to λ, \to μ, \to ν

are coplanar

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\to λ + \to ν = 2 \to μ

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None of these

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Solution

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Similar questions

\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, \to b, \to c

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

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

\to a, \to b, \to c

(\to a \times \to b) \times (\to a \times \to c) + (\to b \times \to c) \times (\to b \times \to a) + (\to c \times \to a) \times (\to c \times \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

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

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

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

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