If a-b-c- are non-coplanar vectors- then -a-b- b-c- c-a- -

The correct option is D [#160;a #160;b #160;c]^2 [a×b #160; b×c #160; c×a] #160;=(a̅×b̅)·{(b̅×c̅)× (c̅×a̅) #160; } =(a̅×b̅)·{[b̅ ...

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Work Done as Dot Product

If a⃗,b⃗,c⃗...

Question

If $\to a, \to b, \to c$ are non-coplanar vectors, then $[\to a \times \to b$ $\to b \times \to c$ $\to c \times \to a] =$

[\to a \to b \to c]

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{[\to a \to b \to c]}^{2}

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Solution

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Similar questions

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] = 4

[\to a \times \to b

\to b \times \to c

\to c \times \to a] = ?

\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.

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