Position vectors of A- B- C are given by a-b-c- and they are collinear- prove that a-b-b-c-c-a-0-

If A, B, C are collinear A⃗B⃗ and B⃗C⃗ are parallel ⇒(b⃗ a⃗)×(c⃗ b⃗)=0 ⇒(b⃗ a⃗)×c⃗ (b⃗ a⃗)×b⃗=0 ⇒(b⃗×c⃗) (a⃗×c⃗) (b⃗×b⃗)+(a⃗×b⃗)= ...

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Collinear Vectors

Position vect...

Question

Position vectors of $A, B, C$ are given by $\to a, \to b, \to c$ and they are collinear, prove that $(\to a \times \to b) + (\to b \times \to c) + (\to c \times \to a) = 0$ .

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A, B, C

\to a, \to b, \to c

[\to b \times \to c + \to c \times \to a + \to a \times \to b] = \to 0

\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) = 0

\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) = 0

\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 = 0

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

\to a, \to b, \to c

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