If r- a unit vector such that r-xb-c-yc-a-za-b- then -r-a-b-c-r-b-c-a-r-c-a-b- is equal to

The correct option is B |[#160;a#160;b#160;c]| r⃗ =x(b⃗×c⃗ )+y(c⃗×a⃗ )+z(a⃗×b⃗ ) |(r⃗.a⃗)(b⃗ #215;c⃗)+(r⃗.b⃗)(c⃗ #215;a⃗)+(r⃗.c⃗)(a⃗ #215; ...

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Properties of Inequalities

If r⃗ a uni...

Question

If $\to r$ a unit vector such that $\to r = x (\to b \times \to c) + y (\to c \times \to a) + z (\to a \times \to b)$ , then $| (\to r . \to a) (\to b \times \to c) + (\to r . \to b) (\to c \times \to a) + (\to r . \to c) (\to a \times \to b) |$ is equal to

| \to a \to b \to c |

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1

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| [\to a \to b \to c] |

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0

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[\to a \times \to b \to b \times \to c \to c \times \to a] = λ {[\to a \to b \to c]}^{2}

λ

\to a

\to b

\to c

\to c = \to a + \to b

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

\to a, \to b, \to c

\to d

(\to a \times \to b) \times (\to c \times \to d) = [\to a \to b \to d] \to c - [\to a \to b \to c] \to d

\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 d

(\to a \times \to b) \times (\to c \times \to d) = [\to a \to c \to d] \to b - [\to b \to c \to d] \to a

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