If a¯ and b¯ are non zero vectors such that a¯+b¯=a¯−2b¯ then

a¯+b¯=a¯−2b¯S.O.B.Sa¯2+b¯2+2a¯.b¯=a¯2+4b¯2−4a¯.b¯6a¯.b¯=3b¯2a¯.b¯=b¯22Also a¯.b¯+1b¯2+2=b¯2+22+1b¯2+2−1≥2−1

If a¯ and b¯ are non zero vectors such that a¯+b¯=a¯−2b¯ then

a¯+b¯=a¯−2b¯S.O.B.Sa¯2+b¯2+2a¯.b¯=a¯2+4b¯2−4a¯.b¯6a¯.b¯=3b¯2a¯.b¯=b¯22Also a¯.b¯+1b¯2+2=b¯2+22+1b¯2+2−1≥2−1

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

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Least value of $\bar{a .} \bar{b} + \frac{1}{{|\bar{b}|}^{2} + 2}$ is $\sqrt{2} - 1$

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

$2 \bar{a .} \bar{b} = {|\bar{b}|}^{2}$

$\bar{a .} \bar{b} = {|\bar{b}|}^{2}$

answer is A, D.

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Detailed Solution

$|\bar{a} + \bar{b}| = |\bar{a} - 2 \bar{b}|$

S.O.B.S

${|\bar{a}|}^{2} + {|\bar{b}|}^{2} + 2 (\bar{a} . \bar{b}) = {|\bar{a}|}^{2} + 4 {|\bar{b}|}^{2} - 4 (\bar{a} . \bar{b})$

$\begin{array}{l} 6 (\bar{a} . \bar{b}) = 3 {|\bar{b}|}^{2} \\ \bar{a} . \bar{b} = \frac{{|\bar{b}|}^{2}}{2} \end{array}$

Also $\bar{a} . \bar{b} + \frac{1}{{|\bar{b}|}^{2} + 2}$

$= \frac{{|\bar{b}|}^{2} + 2}{2} + \frac{1}{{|\bar{b}|}^{2} + 2} - 1$

$\geq \sqrt{2} - 1$

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