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

The correct option is D 0⃗ Let us take #160; a× #160;b #160;as #160; x #160; The question becomes #160; x×[ c× #160;d] = ( ...

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Theorems for Differentiability

If a⃗,b⃗, c...

Question

If $\to a, \to b, \to c, \to d$ are coplanar vectors then ${(\to a \times \to b) \times (\to c \times \to d)} \times (\to a - \to b) =$

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

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

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[\to a \to b \to c]

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\to 0

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

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 ∣= 1, ∣ \to b ∣= 2

∣ \to c ∣= 3

\to a, \to b, \to c

∣ \to d ∣= 4

∣ \frac{[\to d \to b \to c] \to a + [\to d \to c \to a] \to b + [\to d \to a \to b] \to c}{[\to a \to b \to c]} ∣

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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