Let p¯=i^+2j^−k^, q¯=2i^−j^+k^. If a¯ and b¯ are two vectors such that p¯=a¯−2b¯ q¯=2a¯+b¯, then the angle between a¯ and b¯ is

p=a−2b and q=2a+bso, 5a=p+2q−(i−2j−k)+2(2i−j+k)=5i+k⇒a=i+k5Similarly b=q−2a−(2i−j+k)−2i−2k5=−j+3k5So, angle between and b isθ=cos−1a.b|a||b|=cos−1325263×346=cos−1323×7=cos−132221

Let p¯=i^+2j^−k^, q¯=2i^−j^+k^. If a¯ and b¯ are two vectors such that p¯=a¯−2b¯ q¯=2a¯+b¯, then the angle between a¯ and b¯ is

p=a−2b and q=2a+bso, 5a=p+2q−(i−2j−k)+2(2i−j+k)=5i+k⇒a=i+k5Similarly b=q−2a−(2i−j+k)−2i−2k5=−j+3k5So, angle between and b isθ=cos−1a.b|a||b|=cos−1325263×346=cos−1323×7=cos−132221

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Let $\bar{p} = \hat{i} + 2 \hat{j} - \hat{k}, \bar{q} = 2 \hat{i} - \hat{j} + \hat{k} .$ If $\bar{a}$ and $\bar{b}$ are two vectors such that $\bar{p} = \bar{a} - 2 \bar{b} \bar{q} = 2 \bar{a} + \bar{b}$ , then the angle between $\bar{a}$ and $\bar{b}$ is

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$\cos^{- 1} (\frac{3}{2 \sqrt{221}})$

$\frac{π}{2}$

$\cos^{- 1} \frac{7}{\sqrt{143}}$

$\frac{π}{3}$

answer is A.

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Detailed Solution

$p = a - 2 b a n d q = 2 a + b$
$\begin{array}{l} s o, 5 a = p + 2 q - (i - 2 j - k) + 2 (2 i - j + k) \\ = 5 i + k \\ \Rightarrow a = i + \frac{k}{5} \end{array}$
Similarly $b = q - 2 a - (2 i - j + k) - 2 i - \frac{2 k}{5}$
$= - j + \frac{3 k}{5}$
So, angle between and b is
$\begin{array}{l} θ = \cos^{- 1} |\frac{a . b}{| a | | b |}| = \cos^{- 1} [\frac{\frac{3}{25}}{\frac{\sqrt{26}}{3} \times \frac{\sqrt{34}}{6}}] \\ = \cos^{- 1} (\frac{3}{2 \sqrt{3} \times 7}) = \cos^{- 1} (\frac{3}{2 \sqrt{221}}) \end{array}$