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

We know that, |a + b+ nbsp;c |^2 ≥ 0 ⇒ |a |^2+|b |^2+|c |^2 + 2(a·b +b·c+ c·a)≥ 0 ⇒ 1+1+1+2(a·b +b·c+ c·a)≥ 0 ∵ |a |= |b |= |c |=1 ⇒ 2(a·b +b·c+ c·a) ≥ 3 We ...

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Standard XI

Mathematics

Cross Product of Two Vectors

If a, b, c ar...

Question

If $\to a, \to b, \to c$ 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

The correct option is C $9$
We know that,
$| \to a + \to b + \to c |^{2} \geq 0$
$\Rightarrow | \to a |^{2} + | \to b |^{2} + | \to c |^{2} + 2 (\to a . \to b + \to b . \to c + \to c . \to a) \geq 0$
$\Rightarrow 1 + 1 + 1 + 2 (\to a . \to b + \to b . \to c + \to c . \to a) \geq 0$
$∵ | \to a | = | \to b | = | \to c | = 1$
$\Rightarrow 2 (\to a . \to b + \to b . \to c + \to c . \to a) \geq - 3$
We know,
$| \to a - \to b |^{2} + | \to b - \to c |^{2} + | \to c - \to a |^{2} = 2 (| \to a |^{2} + | \to b |^{2} + | \to c |^{2} - \to a . \to b - \to b . \to c - \to c . \to a)$
$\Rightarrow | \to a - \to b |^{2} + | \to b - \to c |^{2} + | \to c - \to a |^{2} \leq 2 (1 + 1 + 1) - (- 3)$
$\Rightarrow | \to a - \to b |^{2} + | \to b - \to c |^{2} + | \to c - \to a |^{2} \leq 9$

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Cross Product of Two Vectors

Standard XI Mathematics

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

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