If a- b- and c- are three unit vectors- then -a-b-2-b-c-2-c-a-2 does not exceed

The correct option is B 9 [ | a b | ^ 2 +| b c | ^ 2 +| c a | ^ 2; a^ 2 + b^ 2 2a.b+ b^ 2 + c^ 2 2b.c+ c^ 2 + a^ 2 +2a.c; ...

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Multiplication with Vectors

If a⃗, b⃗ a...

Question

If $\to a, \to b$ and $\to c$ are three unit vectors, then $| \to a - \to b |^{2} + | \to b - \to c |^{2} + | \to c - \to a |^{2}$ does not exceed

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Solution

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Similar questions

\to a, \to b,

\to c

\to a + 2 \to b + 2 \to c = \to 0,

| \to a \times \to c |

\to a, \to b, \to c

\to a \neq 0

\to a \times \to b = 2 \to a \times \to c, | \to a | = | \to c | = 1, | \to b | = 4

| \to b \times \to c | = \sqrt{15}

\to b - 2 \to c = λ \to a

λ

\to a, \to b

\to c

\to p, \to q

\to r

\to p = \frac{\to b \times \to c}{[\to a \to b \to c]}, \to q = \frac{\to c \times \to a}{[\to a \to b \to c]}, \to r = \frac{\to a \times \to b}{[\to a \to b \to c]}

(\to a + \to b) \cdot \to p + (\to b + \to c) \cdot \to q + (\to c + \to a) \cdot \to r =

\to a, \to b

\to c

[\to a + \to b + \to c \to a + \to b \to a + \to c] = - [\to a \to b \to c]

\to A, \to B

\to C

(\to A + \to B + \to C)

[(\to A + \to B) \times (\to A + \to C)]

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