a-b-a-b-a-b-a-b- is equal to

The correct option is B 2(a⃗·b⃗)b⃗×a⃗ Given, { (a⃗×b⃗ )×a⃗ #160;}×b⃗ +a⃗×{b⃗× (a⃗×b⃗ ) } =[ ( a⃗ .a⃗ #160;) b⃗ ( a⃗ .b⃗ #160;) a⃗ #160; ...

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Sigma n2

a⃗×b⃗×a⃗×b⃗+a...

Question

${(\to a \times \to b) \times \to a} \times \to b + \to a \times {\to b \times (\to a \times \to b)}$ is equal to

2 (\to a \cdot \to b) \to b \times \to a

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\to 0

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2 \to a \times \to b

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(\to a \times \to b)^{2}

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Solution

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Similar questions

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

(\to A - \to B) \times (\to A + \to B)

2 (\to A \times \to B)

\to a

\to b

(\to a - \to b) \times [(\to b + \to a) \times (2 \to a + \to b)] = \to a + \to b

θ

\to a

\to b

(\to a - \to b) \times (\to a + \to b) = 2 (\to a \times \to b)

\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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