If a-b-c-0- then a-a-b-b-2a-b-c-c-

The correct option is C 0 We have, a⃗+b⃗+c⃗=0 .........(1) ∴a⃗·a⃗+b⃗·b⃗+2a⃗·b⃗ c⃗·c⃗ =a⃗· ( a⃗+b⃗)+b⃗· (b⃗+a⃗) c⃗·c⃗ =a⃗· ( c⃗)+b⃗· ( ...

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Unit Vectors

If a⃗+b⃗+c⃗...

Question

If $\to a + \to b + \to c = 0$ , then $\to a . \to a + \to b . \to b + 2 \to a . \to b - \to c . \to c =$

2 (\to b . \to c + \to c . \to a)

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- \frac{3}{2}

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- 2 (\to b . \to c + \to c . \to a)

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\to a

\to b

\to c

\to c = \to a + \to b

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

[\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} =

\to a, \to b, \to c

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

| \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

[\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}

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