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

|a→+b→+c→|≥0 ⇒|a→+b→+c→|2=1+1+1+2(a→⋅b→+b→⋅c→+c→⋅a→)≥0∴ a→⋅b→+b→⋅c→+c→⋅a→≥−32 …..(i)|a→−b→|2+|b→−c→|2+|c→−a|2=2|a→|2+2|b→|2+2|c→|2−2(a→⋅b→+b→⋅c→+c→⋅a→)=6−2(a→⋅b→+b→⋅c→+c→⋅a→)≤6+3=9, [using (i)]

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

|a→+b→+c→|≥0 ⇒|a→+b→+c→|2=1+1+1+2(a→⋅b→+b→⋅c→+c→⋅a→)≥0∴ a→⋅b→+b→⋅c→+c→⋅a→≥−32 …..(i)|a→−b→|2+|b→−c→|2+|c→−a|2=2|a→|2+2|b→|2+2|c→|2−2(a→⋅b→+b→⋅c→+c→⋅a→)=6−2(a→⋅b→+b→⋅c→+c→⋅a→)≤6+3=9, [using (i)]

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

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$4$

$9$

$8$

$6$

answer is B.

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Detailed Solution

$| \vec{a} + \vec{b} + \vec{c} | \geq 0$

$\begin{array}{l} \Rightarrow | \vec{a} + \vec{b} + \vec{c} |^{2} = 1 + 1 + 1 + 2 (\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) \geq 0 \\ ∴ \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} \geq - \frac{3}{2} \end{array}$ …..(i)

$\begin{array}{l} | \vec{a} - \vec{b} |^{2} + | \vec{b} - \vec{c} |^{2} + | \vec{c} - a |^{2} \\ = 2 | \vec{a} |^{2} + 2 | \vec{b} |^{2} + 2 | \vec{c} |^{2} - 2 (\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) \\ = 6 - 2 (\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) \leq 6 + 3 = 9, [using (i)] \end{array}$