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Multiplication with Vectors

If a⃗b⃗, an...

Question

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

A

4

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B

9

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C

8

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D

6

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Solution

The correct option is B $9$
$\begin{matrix} {∣ ∣ ∣ \to a - \to b ∣ ∣ ∣}^{2} = {∣ ∣ \to a ∣ ∣}^{2} + {∣ ∣ ∣ \to b ∣ ∣ ∣}^{2} - 2 \to a . \to b {∣ ∣ ∣ \to b - \to c ∣ ∣ ∣}^{2} = {∣ ∣ ∣ \to b ∣ ∣ ∣}^{2} + {∣ ∣ \to c ∣ ∣}^{2} - 2 \to b . \to c {∣ ∣ \to c - \to a ∣ ∣}^{2} = {∣ ∣ \to c ∣ ∣}^{2} + {∣ ∣ \to a ∣ ∣}^{2} - 2 \to c . \to a {∣ ∣ ∣ \to a - \to b ∣ ∣ ∣}^{2} + {∣ ∣ ∣ \to b - \to c ∣ ∣ ∣}^{2} + {∣ ∣ \to c - \to a ∣ ∣}^{2} = 2 ({∣ ∣ \to a ∣ ∣}^{2} + {∣ ∣ ∣ \to b ∣ ∣ ∣}^{2} + {∣ ∣ \to c ∣ ∣}^{2} - \to a . \to b - \to b . \to c - \to c . \to a) = 2 (3 - \to a . \to b - \to b . \to c - \to c . \to a) = 6 - 2 (\to a . \to b + \to b . \to c + \to c . \to a) (\to a + \to b + \to c) \geq 0 {∣ ∣ \to a ∣ ∣}^{2} + {∣ ∣ ∣ \to b ∣ ∣ ∣}^{2} + {∣ ∣ \to c ∣ ∣}^{2} + 2 (\to a . \to b + \to b . \to c + \to c . \to a) \geq 0 \to a . \to b + \to b . \to c + \to c . \to a \geq - \frac{3}{2} \Rightarrow - 2 (\to a . \to b + \to b . \to c + \to c . \to a) \leq 3 {∣ ∣ ∣ \to a - \to b ∣ ∣ ∣}^{2} + {∣ ∣ ∣ \to b - \to c ∣ ∣ ∣}^{2} + {∣ ∣ \to c - \to a ∣ ∣}^{2} \leq 6 + 3 \leq 9 O p t i o n B i s c o r r e c t . \end{matrix}$

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