If a-2b-3 c- -0- then b-c-c-a-a-b- equals to-

The correct option is A 6( b× c ) a⃗+2b⃗+3c⃗=0 (a⃗+2b⃗+3c⃗)×c⃗=0c⃗×a⃗=2(b⃗×c⃗) (a⃗+2b⃗+3c⃗)×b⃗=0a⃗×b⃗=3(b⃗×c⃗) b⃗×c⃗+c⃗×a⃗+a⃗×b⃗=(b⃗×c⃗ ...

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Solving Equations with Variables on Both Sides

If a⃗+2b⃗+3...

Question

If $\to a + 2 \to b + 3 \to c = 0$ , then $\to b \times \to c + \to c \times \to a + \to a \times \to b$ equals to:

6 (b \times c)

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(a \times b)

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6 (c \times a)

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Solution

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Similar questions

\to a + 2 \to b + 3 \to c = 0

\to a \times \to b + \to b \times \to c + \to c \times \to a

λ (\to b \times \to c)

λ

(\to b \times \to c) \times (\to c \times \to a) = 3 \to c

[\to b \times \to c

\to c \times \to a

\to a \times \to b] =

[\to a + 2 \to b - \to c, \to a - \to b, \to a - \to b - \to c] =

\to a + 2 b + 3 c = 0

\to a \times \to b + \to b \times \to c + \to c \times \to a

\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 =

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