If a→ and b→ are non-zero vectors such that |a→+b→|=|a→−2b→| then

|a→+b→|=|a→−2b→| ⇒a→+b→2=|a→−2b→|2⇒a →⋅b→=|b→|22also a→⋅b→+1|b→|2+2=|b→|2+22+1|b→|2+2−1(using A.M. ≥G.M.)≥2−1

If a→ and b→ are non-zero vectors such that |a→+b→|=|a→−2b→| then

|a→+b→|=|a→−2b→| ⇒a→+b→2=|a→−2b→|2⇒a →⋅b→=|b→|22also a→⋅b→+1|b→|2+2=|b→|2+22+1|b→|2+2−1(using A.M. ≥G.M.)≥2−1

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If $\vec{a} and \vec{b}$ are non-zero vectors such that $| \vec{a} + \vec{b} | = | \vec{a} - 2 \vec{b} |$ then

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$2 \vec{a} \cdot \vec{b} = | \vec{b} |^{2}$

$\vec{a} \cdot \vec{b} = | \vec{b} |^{2}$

Least value of $| \vec{a} \cdot \vec{b |} + \frac{1}{| \vec{b} |^{2} + 2} is \sqrt{2}$

least value of $| \vec{a} \cdot \vec{b} | + \frac{1}{| \vec{b} |^{2} + 2} is \sqrt{2} -1$

answer is A, D.

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Detailed Solution

$\begin{array}{l} | \vec{a} + \vec{b} | = | \vec{a} - 2 \vec{b} | \Rightarrow {|\vec{a} + \vec{b}|}^{2} = | \vec{a} - 2 \vec{b} |^{2} \\ \vec{\Rightarrow a} \cdot \vec{b} = \frac{| \vec{b} |^{2}}{2} \\ also \vec{a} \cdot \vec{b} + \frac{1}{| \vec{b} |^{2} + 2} \\ = \frac{| \vec{b} |^{2} + 2}{2} + \frac{1}{| \vec{b} |^{2} + 2} - 1 \end{array}$