If the three points a- b- c- are collinear- then b-c-c-a-a-b-

The correct option is A 0 a⃗=kb⃗=k_1c⃗ So k_1kc⃗×c⃗+c⃗× kc⃗+kb⃗×b⃗ =0+0+0 =0 ...

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Theorems for Differentiability

If the three ...

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If the three points $\to a, \to b, \to c$ are collinear, then $\to b \times \to c + \to c \times \to a + \to a \times \to b =$

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\to a, \to b, \to c

\to a, \to b, \to c

\to a \times \to b = \to b \times \to c = \to c \times \to a

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

\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

| \to a | = | \to b | = | \to c |

\to a, \to b, \to c

| \to b | = | \to c |

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

\to a, \to b, \to c

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