First slide

Dot product or scalar product of vectors

Question

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

Moderate

A

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

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

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

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

Solution

$\begin{array}{l} | \vec{a} + \vec{b} | = | \vec{a} - 2 \vec{b} | \Rightarrow {|\vec{a} + \vec{b}|}^{2} = | \vec{a} - 2 \vec{b} |^{2} \\ \vec{\Rightarrow a} \cdot \vec{b} = \frac{| \vec{b} |^{2}}{2} \\ also \vec{a} \cdot \vec{b} + \frac{1}{| \vec{b} |^{2} + 2} \\ = \frac{| \vec{b} |^{2} + 2}{2} + \frac{1}{| \vec{b} |^{2} + 2} - 1 \end{array}$
(using A.M. $\geq$ G.M.)
$\geq \sqrt{2} - 1$

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