If A-B- and C- are three non-coplanar vectors then A-B-C- - - A-B- - A-C- - equals

The correct option is D [ A⃗B⃗C⃗ ] nbsp; ( A⃗+B⃗+C⃗ ). [ (A⃗+B⃗ )× ( A⃗+C⃗ ) ] = ( A⃗+B⃗+C⃗).[A⃗×C⃗+B⃗×A⃗+B⃗×C⃗] =A⃗·(B⃗×C⃗)+B⃗·(A⃗ ...

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Arithmetic Progression

If A⃗,B⃗ an...

Question

If $\to A, \to B$ and $\to C$ are three non-coplanar vectors then $(\to A + \to B + \to C) . [(\to A + \to B) \times (\to A + \to C)]$ equals

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[\to A \to B \to C]

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2 [\to A \to B \to C]

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- [\to A \to B \to C]

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Solution

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If $\to a, \to b$ and $\to c$ are three non-coplanar vectors, then $(\to a + \to b + \to c) \cdot [(\to a + \to b) \times (\to a + \to c)]$ equals

\to a, \to b, \to c

\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]}

(2 \to a + 3 \to b + 4 \to c) . \to p + (2 \to b + 3 \to c + 4 \to a) . \to q +

(2 \to c + 3 \to a + 4 \to b) . \to r =

\to a, \to b, \to c

\to a, \to b, \to c

\to a \times \to b = \to b \times \to c = \to c \times \to a

[\to a \to b \to c] = 0.

\to a, \to b, \to c

\to a + \to b + \to c = \to 0,

\to a \times \to b = \to b \times \to c = \to c \times \to a,

[\to a \to b \to c] = 0.

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