The angle $θ$ between two non-zero vectors $a$ and $b$ satisfies the relation $\cos θ = (a \times \hat{i}) \cdot (b \times \hat{i}) + (a \times \hat{j}) \cdot (b \times \hat{j}) + (a^{'} \times \hat{k}) \cdot (b \times \hat{k})$ then the least value of $| a | + | b |$ is equal to (where $θ \neq 90 °$ )

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$2$

$\frac{1}{2}$

$\sqrt{2}$

$4$

answer is C.

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Detailed Solution

Given, $\cos θ = (a \times \hat{i}) \cdot (b \times \hat{i}) + (a + \hat{j}) \cdot (b \times \hat{j})$ $+ (a \times \hat{k}) \cdot (b \times \hat{k})$ …(i)
Consider, $(a \times \hat{i}) \cdot (b \times \hat{i}) = [(a \times \hat{i}) b \hat{i}] = ((a \times \hat{i}) \times b) \cdot \hat{i}$

$((a \cdot b) \hat{i}) - (\hat{i} \cdot b) a) \hat{i} = (a \cdot b) (\hat{i} \cdot \hat{i}) - (\hat{i} \cdot \hat{b}) (a \cdot \hat{i})$
$= a \cdot b - a_{1} b_{1}$
$Similarly, (a \times \hat{j}) \cdot (b \times \hat{j}) = a \cdot b - a_{2} b_{2}$
$and (a \times \hat{k}) (b \times \hat{k}) = a \cdot b - a_{3} b_{3}$

$∴$ From Eq. (i), we get
$\cos θ = 3 a \cdot b - (a_{1} b_{1} + a_{2} b_{2} + a_{3} b_{3})$
$= 3 a \cdot b - a \cdot b$
$\Rightarrow \frac{a \cdot b}{| a | | b |} = 2 a \cdot b \Rightarrow | a | | b | = \frac{1}{2}$
$Now, use AM \geq GM on | a |, | b |$
$∴ \frac{| a | + | b |}{2} \geq (| a | \cdot | b |)^{\frac{1}{2}} \Rightarrow | a | + | b | \geq \frac{2}{\sqrt{2}}$
$\Rightarrow | a | + | b | \geq \sqrt{2}$