If b- is a unit vector then a-b-b-b-a-b- is equal to

The correct option is C #160;a #10; (#160;a.#160;b)#160;b+#160;b×(#160;a×#160;b) =(#160;a.#160;b)#160;b+#160;a(#160;b.#160;b) (#160;a.#160;b)# ...

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Multiplication with Vectors

If b⃗ is a ...

Question

If $\to b$ is a unit vector then $(\to a . \to b) \to b + \to b \times (\to a \times \to b)$ is equal to

| \to a |^{2} \to b

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(\to a . \to b) \to a

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\to a

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(\to a . \to b) \to b

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\to b

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

\to a, \to b

\to α = \to a - (\to a . \to b) \to b, \to β = \to a \times \to b

\to b

\to c

\to a

(\to a . \to b) \to b + (\to a . \to c) \to c + \frac{\to a . (\to b \times \to c)}{| \to b \times \to c |^{2}} (\to b \times \to c) =

\to a

\to b

\to u = \to a - (\to a . \to b) \to b

\to v = \to a \times \to b,

| \to v |

\to a, \to b, \to c

\to d

∣ ∣ (\to a . \to b) (\to b \times \to c) + (\to b . \to d) (\to c \times \to a) + (\to c . \to d) (\to a \times \to b) ∣ ∣

\to d

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