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

The correct option is C 0 We have, a × b = b × c = c × a #160; ( a + c #160; ) × b =0 a + c =t b a ×( b + c ...

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Inequalities Involving Modulus Function

If a⃗×b⃗ =b...

Question

If $\to a \times \to b = \to b \times \to c = \to c \times \to a$ , then $\to a + \to b + \to c$ is equal to

a b c

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| \to a | = 3, ∣ ∣ \to b ∣ ∣ = 4, | \to c | = 5, \to a ⊥ (\to b + \to c), \to b ⊥ (\to c + \to a)

\to c ⊥ (\to a + \to b)

\sqrt{2} ∣ ∣ \to a + \to b + \to c ∣ ∣

\to a, \to b

\to c

\to b + \to c, \to c + \to a

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| \to a + \to b | = 6, | \to b + \to c | = 8

| \to c + \to a | = 10

| \to a + \to b + \to c |

| \to a | = 1, ∣ ∣ \to b ∣ ∣ = \sqrt{2}

| \to c | = \sqrt{3}

\to a ⊥ (\to b + \to c)

\to b ⊥ (\to c + \to a)

\to c ⊥ (\to a + \to b)

∣ ∣ \to a + \to b + \to c ∣ ∣

\to A = \to b \times \to c, \to B = \to c \times \to a, \to C = \to a \times \to b

\to A \times (\to B \times \to C)

\to B \times (\to C \times \to A)

\to C \times (\to A \times \to B)

\to a, \to b, \to c

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

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