The scalar triple product - a- - b- - c- b- - c- - a- c- - a- - b- is equal to

The correct option is D 4[a⃗b⃗c⃗] (b⃗ + c⃗ a⃗) × (c⃗ + a⃗ b⃗) = 2 b⃗×c⃗ + 2 c⃗×a⃗ Thus required scalar triple product = 2 (a⃗ + b⃗ c⃗) . (b ...

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Arithmetic Progression

The scalar tr...

Question

The scalar triple product
$[\to a + \to b - \to c \to b + \to c - \to a \to c + \to a - \to b$ ] is equal to

0

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

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

k \neq 0

[(\to b \times \to c) \times (\to c \times \to a) (\to c \times \to a) \times (\to a \times \to b) (\to a \times \to b) \times (\to b \times \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

\to a \times (\to a \times \to b) = \to b \times (\to b \times \to c)

\to a . \to b \neq 0

[\begin{matrix} \to a & \to b & \to c \end{matrix}]

\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 \times (\to a \times \to b) = \to b \times (\to b \times \to c)

(\to a . \to b) \neq 0

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

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