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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 $\to y$ is

A

\frac{\to a \times \to b}{γ}

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B

\to a + \frac{\to a \times \to b}{γ}

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C

\to a + \to b + \frac{\to a \times \to b}{γ}

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D

None of these.

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Solution

The correct option is A $\frac{\to a \times \to b}{γ}$
$\to x \times \to y = \to a (i)$
$\to y \times \to z = \to b (i i)$
$\to x \cdot \to b = γ (i i i)$
$\to x \cdot \to y = 1 (i v)$
$\to y \cdot \to z = 1 (v)$

From $(i i)$ , $\to x \cdot (\to y \times \to z) = \to x \cdot \to b = γ$
$\Rightarrow [\to x \to y \to z] = γ$

From $(i)$ and $(i i)$ ,
$(\to x \times \to y) \times (\to y \times \to z) = \to a \times \to b$
$∴ [\to x \to y \to z] \to y - [\to y \to y \to z] \to x = \to a \times \to b$
or $\to y = \frac{\to a \times \to b}{γ}$

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Similar questions

\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 x

\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

be unit vectors such that

\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

.

Then which of the following option(s) is/are CORRECT ?

\to x

\to x, \to y

\to z,

each of magnitude

\sqrt{2},

make an angle of

60^{\circ}

with each other.

\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

are units vectors such that

\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,

then which of the following is FALSE?

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