If - a- b- c- -1- then a-b-c-c-a-b-b-c-a-a-b-c-c-a-b-b-c-a- is equal to

The correct option is A 3 [a,b,c] =a.(b×c) =b.(c×a) =c.(a×b) Hence a.(b×c)b.(c×a)+b.(c×a)c.(a×b)+c.(a×b)a.(b×c) =1+1+1 = ...

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Transpose of a Matrix

If [ a⃗ b⃗ ...

Question

If $[\begin{matrix} \to a & \to b & \to c \end{matrix}] = 1$ , then $\frac{\to a . \to b \times \to c}{\to c \times \to a . \to b} + \frac{\to b . \to c \times \to a}{\to a \times \to b . \to c} + \frac{\to c . \to a \times \to b}{\to b \times \to c . \to a}$ is equal to

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[\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 | = 1, | \to b | = 2, | \to c | = 3

\to a + \to b + \to c = 0

\to a . \to b + \to b . \to c + \to c . \to a

(\to b \times \to c) . (\to a \times \to d) + (\to c \times \to a) . (\to b \times \to d)

+ (\to a \times \to b) . (\to c \times \to d) =

\to a, \to b, \to c

\to a + \to b + \to c = 0

μ = \to a . \to b + \to b . \to c + \to c . \to a

| \to a | = 1; ∣ ∣ \to b ∣ ∣ = 4; | \to c | = 2

\to a, \to b, \to c

\to a + \to b + \to c = 0

\to a . \to b + \to b . \to c + \to c . \to a

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