If a-a-b-b-b-c- and a-b- 0- then -a- b- c-

The correct option is D 0 Let solve LHS â× (a⃗×b⃗)=(a⃗.b⃗)a⃗ | a |^2 b⃗ #160; #160; #160; #160; #160;–(1)RHS = b⃗× (b⃗×c⃗ ...

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How to Find the Inverse of a Function

If a⃗×a⃗×b⃗...

Question

If $\to a \times (\to a \times \to b) = \to b \times (\to b \times \to c)$ and $(\to a . \to b) \neq 0$ , then $[\to a \to b \to c] =$

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Similar questions

\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}]

[\begin{matrix} \to a & \to b & \to c \end{matrix}] = 1

\frac{\to a . \to b \times \to c}{\to c \times \to a . \to b} + \frac{\to b . \to c \times \to a}{\to a \times \to b . \to c} + \frac{\to c . \to a \times \to b}{\to b \times \to c . \to a}

[\to a \to b \to c] = 2

\frac{\to a . (\to b \times \to c)}{(\to c \times \to a) . \to b} + \frac{\to b . (\to c \times \to a)}{(\to a \times \to b) . \to c} + \frac{\to c . (\to a \times \to b)}{(\to b \times \to c) . \to a} =

\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.

[\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)] =

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