If a - b - c be three non-coplanar vectors and r be any arbitrary vector- then a - b - r - c - b - c - r - a - c - a - r - b is equal to

The correct option is C 2[ a b c #160;] r We have, ( a × b #160;) ×( r × c #160;) =( ( a × b #160;) . c #160;) r ( ( a × ...

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Distance vs Displacement

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Question

If $\to a, \to b, \to c$ be three non-coplanar vectors and $\to r$ be any arbitrary vector, then $(\to a \times \to b) \times (\to r \times \to c) + (\to b \times \to c) \times (\to r \times \to a) + (\to c \times \to a) \times (\to r \times \to b)$
is equal to

0

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[\to a \to b \to c] \to r

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2 [\to a \to b \to c] \to r

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3 [\to a \to b \to c] \to r

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\to a, \to b, \to c

\to r

\to r . \to a = 1, \to r . \to b = 2

\to r . \to c = 3.

[\to a, \to b, \to c] = 1

\to r

\to a, \to b, \to c

\to r

\to a \times {(\to r - \to b) \times \to a} + \to b \times {(\to r - \to c) \times \to b} + \to c \times {(\to r - \to a) \times \to c} = \to 0

\to r

\to a, \to b, \to c

\to r

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

λ [\to a \to b \to c]

λ

\to a, \to b, \to c

\to p, \to q, \to r

\to p = \frac{\to b \times \to c}{[\to a \to b \to c]}, \to q = \frac{\to c \times \to a}{[\to a \to b \to c]}, \to r = \frac{\to a \times \to b}{[\to a \to b \to c]}

(\to a + \to b) . \to p + (\to b + \to c) . \to q + (\to c + \to a) . \to r

\to a, \to b, \to c

\to p = \frac{\to b \times \to a}{[\to a \to b \to c]}, \to q = \frac{\to c \times \to a}{[\to a \to b \to c]}, \to r = \frac{\to a \times \to b}{[\to a \to b \to c]}

(\to a + \to b) . \to p + (\to b + \to c) . \to q + (\to c + \to a) . \to r =

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