If -a-b-c-0- then a-b-b-c-c-a-

The correct option is A 0⃗ Solving =a×b×((b×c·a)c (a×c·c)a) =a×b×c(b×c·a) =((a c)b (a b)c)[abc] —–(1)Now αa+βb+γc=0 =αa×c+βb×c=0 ...

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Algebra of Limits

If αa⃗+βb⃗+...

Question

If $α \to a + β \to b + γ \to c = \to 0$ , then $(\to a \times \to b) \times {(\to b \times \to c) \times (\to c \times \to a} =$

\to 0

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Avector perpendicular to the plane of

\to a, \to b, \to c

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A scalar quantity

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2 [\to a \to b \to c]

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Solution

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Similar questions

α (\to a \times \to b) + β (\to b \times \to c) + γ (\to c \times \to a) = \to 0

α, β, γ

\to a, \to b, \to c

$\to a \neq \to 0, \to b \neq \to 0, \to a \times \to b = \to 0, \to c \times \to b = \to 0 \Rightarrow \to a \times \to c =$

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

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

\to a, \to b, \to c

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

(\to a, \to b) = \frac{π}{3}

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

\to a, \to b, \to c

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

(\to a, \to b) = \frac{π}{3}

∣ ∣ \to a \times \to b ∣ ∣ + ∣ ∣ \to b \times \to c ∣ ∣ + | \to c \times \to a | =

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