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

Given expression, |a 2b|^2+|b 2c|^2+|c 2a|^2 =5(|a|^2+|b|^2+|c|^2) 4(a·b+b·c+c·a) =7(|a|^2+|b|^2+|c|^2) 2(|a|^2+|b|^2+|c|^2+2a·b+2b·c+2c·a) =7×3 2[(a+b+c)^2] ...

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Byju's Answer

Standard XII

Physics

Vector Notation

If a, b, c ar...

Question

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

18

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Solution

The correct option is B $21$
Given expression,
$| \to a - 2 \to b |^{2} + | \to b - 2 \to c |^{2} + | \to c - 2 \to a |^{2} = 5 (| \to a |^{2} + | \to b |^{2} + | \to c |^{2}) - 4 (\to a \cdot \to b + \to b \cdot \to c + \to c \cdot \to a) = 7 (| \to a |^{2} + | \to b |^{2} + | \to c |^{2}) - 2 (| \to a |^{2} + | \to b |^{2} + | \to c |^{2} + 2 \to a \cdot \to b + 2 \to b \cdot \to c + 2 \to c \cdot \to a) = 7 \times 3 - 2 [(\to a + \to b + \to c)^{2}] = 21 - 2 X$
where, $X = | \to a + \to b + \to c |^{2}$
So, for maximum value of the expression : $X = 0$
So, $| \to a - 2 \to b |^{2} + | \to b - 2 \to c |^{2} + | \to c - 2 \to a |^{2} \leq 21 - 2 \times 0 \leq 21$

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If →a,→b,→c are unit vectors, then |→a−2→b|2+|→b−2→c|2+|→c−2→a|2 does not exceed ?

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