If x-y-a- y-z-b- x-b-y- x-y-1 and y-z-1 - vector x- is

The correct option is B γ|a×b|^2[a×b a× (a×b)]x× nbsp;y=a (i) y× nbsp;z=b (ii) x·b=γ (iii) x· nbsp;y=1 ...

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Vector Triple Product

If x⃗×y⃗=a⃗, ...

Question

If $\to x \times \to y = \to a, \to y \times \to z = \to b, \to x \cdot \to b = γ, \to x \cdot \to y = 1$ and $\to y \cdot \to z = 1$ . Vector $\to x$ is

\frac{1}{| \to a \times \to b |^{2}} [\to a \times (\to a \times \to b)]

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\frac{γ}{| \to a \times \to b |^{2}} [\to a \times \to b - \to a \times (\to a \times \to b)]

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\frac{γ}{| \to a \times \to b |^{2}} [\to a \times \to b + \to b \times (\to a \times \to b)]

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None of these

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\to x \times \to y = \to a, \to y \times \to z = \to b, \to x \cdot \to b = γ, \to x \cdot \to y = 1

\to y \cdot \to z = 1

\to y

\to x \times \to y = \to a, \to y \times \to z = \to b, \to x \cdot \to b = γ, \to x \cdot \to y = 1

\to y \cdot \to z = 1

\to z

\to x, \to y

\to z

\to x + \to y + \to z = \to a, \to x \times (\to y \times \to z) = \to b, (\to x \times \to y) \times \to z = \to c

\to a \cdot \to x = \frac{3}{2}, \to a \cdot \to y = \frac{7}{4}

| \to a | = 2

\to x

\to x, \to y

\to z,

\sqrt{2},

60^{\circ}

\to x \times (\to y \times \to z) = \to a, \to y \times (\to z \times \to x) = \to b

\to x \times \to y = \to c .

\to x, \to y, \to z

\to x + \to y + \to z = \to a

(\to x \times \to y) \times \to z = \to b

\to a \cdot \to x = \frac{3}{2}, \to a \cdot \to y = \frac{7}{4}

| \to a | = 2,

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