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Area of a Triangle

a⃗, b⃗, c...

Question

$\to a$ , $\to b$ , $\to c$ are these consecutive vectors forming a triangle, then $\to a$ + $\to b$ + $\to c$ is

A

\infty

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B

0

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C

1

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D

- 1

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Solution

The correct option is B $0$
$\begin{matrix} f r o m t r i a n g l e r u l e \to a + \to b + \to c = 0 H e n c e, o p t i o n B i s c o r r e c t a n s w e r . \end{matrix}$

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Similar questions

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

, then the vectors

\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

are three vectors such that

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

then prove that

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

\to a, \to b, \to c

are non coplanar non zero vectors, then the value of

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

\to a, \to b, \to c

are three vectors and

| \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

are unit vectors such that

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

(\to a, \to b) = \frac{π}{3}

| \to a \times \to b | + | \to b \times \to c | + | \to c \times \to a | =

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Area of a Triangle

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