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Dot Product of Two Vectors

If a,b,c ar...

Question

If $\to a, \to b, \to c$ 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 D $9$
$∣ ∣ ∣ \to a + \to b + \to c ∣ ∣ ∣ \geq 0$
${∣ ∣ ∣ \to a + \to b + \to c ∣ ∣ ∣}^{2} = {∣ ∣ \to a ∣ ∣}^{2} + {∣ ∣ ∣ \to b ∣ ∣ ∣}^{2} + {∣ ∣ \to c ∣ ∣}^{2} + 2 (\to a . \to b + \to b . \to c + \to c + \to a) \geq 0$
$\Rightarrow {∣ ∣ \to a ∣ ∣}^{2} + {∣ ∣ ∣ \to b ∣ ∣ ∣}^{2} + {∣ ∣ \to c ∣ ∣}^{2} + 2 (\to a . \to b + \to b . \to c + \to c + \to a) \geq 0$
Since $\to a, \to b, \to c$ are unit vectors
we have $3 + 2 (\to a . \to b + \to b . \to c + \to c + \to a) \geq 0$
$∴ \to a . \to b + \to b . \to c + \to c . \to a \geq \frac{- 3}{2}$ ............. $(1)$
${∣ ∣ ∣ \to a - \to b ∣ ∣ ∣}^{2} + {∣ ∣ ∣ \to b - \to c ∣ ∣ ∣}^{2} + {∣ ∣ \to c - \to a ∣ ∣}^{2}$
$= 2 {∣ ∣ \to a ∣ ∣}^{2} + 2 {∣ ∣ ∣ \to b ∣ ∣ ∣}^{2} + 2 {∣ ∣ \to c ∣ ∣}^{2} - 2 (\to a . \to b + \to b . \to c + \to c . \to a)$
$= 2 + 2 + 2 - 2 (\to a . \to b + \to b . \to c + \to c . \to a)$
$= 6 - 2 (\to a . \to b + \to b . \to c + \to c . \to a)$
$= 6 - 2 (\frac{- 3}{2}) = 6 + 3 = 9$ using $(1)$

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