For non-zero vectors a-b- c- -a-b-c-a-b-c- holds if

The correct option is D #160;a.#160;b=#160;b.#160;c=#160;c.#160;a=0 We have #160; | (a⃗×b⃗).c⃗ |= | c⃗ | | a⃗×b⃗ |cosϕ = | c⃗ | | a⃗ | | b⃗ |si ...

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Modulus of a Complex Number

For non-zero ...

Question

For non-zero vectors $\to a, \to b, \to c$ ; $| (\to a \times \to b) . \to c | = | \to a | | \to b | | \to c |$ holds if

\to a . \to b = 0, \to b . \to c = 0

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\to b . \to c = 0, \to c . \to a = 0

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\to c . \to a = 0, \to a . \to b = 0

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\to a . \to b = \to b . \to c = \to c . \to a = 0

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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} =

[\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

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

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

\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 | = 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

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