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Let $\hat{a}, \hat{b} a n d \hat{c}$ are three unit vectors mutually inclined to each other at an angle $θ$ . $\hat{d}$ is also a unit vector such that $\hat{d} . \hat{a} = \hat{d} . \hat{b} = \hat{d} . \hat{c} = \cos α$ . Match the entries for $θ$ given in column-I with their corresponding values for $α$ given in column-II
Column–I
Column–II
A)
$\frac{π}{2}$
P)
$\cos^{- 1} (\frac{\sqrt{5}}{3})$
B)
$\frac{π}{3}$
Q)
$\cos^{- 1} (\sqrt{\frac{1 + \sqrt{2}}{3}})$
C)
$\cos^{- 1} (\frac{1}{3})$
R)
$\cos^{- 1} \sqrt{\frac{2}{3}}$
D)
$\cos^{- 1} (\frac{5}{8})$
S)
$\frac{π}{6}$

T)
$\frac{π}{4}$

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A–Q, B–R, C–P, D–S

A–R, B–Q, C–S, D–P

A–Q, B–S, C–R, D–P

A–R, B–Q, C–P, D–S

answer is A.

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Detailed Solution

$\bar{d} = λ (\bar{a} + \bar{b} + \bar{c})$

Any four vectors are linearly dependent

$\bar{d} . \bar{a} = \bar{d} . \bar{b} = \bar{d} . \bar{c} = \cos α 1 = λ \sqrt{3 + 6 \cos θ} \bar{d} . \bar{a} = \cos α = λ (1 + \cos θ + \cos θ) \Rightarrow λ = \frac{\cos α}{1 + 2 \cos θ} 1 = \frac{\cos α}{(1 + 2 \cos θ)} \sqrt{3 + 6 \cos θ} \Rightarrow \cos α = \sqrt{\frac{1 + 2 \cos θ}{3}}$

A) $θ = \frac{π}{4} \Rightarrow α = \cos^{- 1} (\sqrt{\frac{1 + \sqrt{2}}{3}})$
B) $θ = \frac{π}{3} \Rightarrow α = \cos^{- 1} (\sqrt{\frac{2}{3}})$
C) $θ = \cos^{- 1} (\frac{1}{3}) \Rightarrow α = \cos^{- 1} (\sqrt{\frac{1 + 2 (\frac{1}{3})}{3}}) = \cos^{- 1} (\sqrt{\frac{5}{9}})$
D) $θ = \cos^{- 1} (\frac{5}{8}) \Rightarrow α = \cos^{- 1} (\sqrt{\frac{1 + 2 (\frac{5}{8})}{3}}) = \cos^{- 1} (\sqrt{\frac{3}{4}}) = \cos^{- 1} (\frac{\sqrt{3}}{2}) = \frac{π}{6}$

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