- a- -b- b- -c- c- -a- - a- b- c- then

The correct option is C a⃗ , b⃗ ,c⃗ are coplanarGiven, [ #8594;a+ #8594;b #8594;b+ #8594;c #8594;c+ #8594;a]=[ #8594;a #8594;b #8594;c] ...

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Logarithmic Differentiation

[ a⃗ +b⃗ b⃗ +...

Question

$[\to a + \to b \to b + \to c \to c + \to a] = [\to a \to b \to c]$ , then

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

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\to a, \to b, \to c

are coplanar

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

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\to a, \to b, \to c

are mutually perpendicular

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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 a + \to b) (\to b + \to c) (\to c + \to a)] = 2 [\to a \to b \to c] .

\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] = ?

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

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