A parallelepiped is formed using three non-collinear vectors, $\vec{a}, \vec{b}$ and $\vec{c}$ with fixed magnitudes. Angles between any of the vector with normal of the plane determined by the other two is $α$ and volume of the parallelepiped is T and its surface area is Y. If $\frac{Y}{T} = 4 (\frac{1}{\vec{| a |}} + \frac{1}{\vec{| b |}} + \frac{1}{\vec{| c |}})$ then:

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$\sin^{2} α + \sin^{4} α = \frac{21}{16}$

$\cos^{2} α + \cos α = \frac{3}{4}$

$\cos^{2} α + \cos α = \frac{3 + 2 \sqrt{3}}{4}$

$\sin^{2} α + \sin^{4} α = \frac{5}{16}$

answer is A, B.

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Detailed Solution

Y = surface area = $2 (\frac{| \vec{b} \times \vec{c} |}{[\vec{a} \vec{b} \vec{c}]} + \frac{| \vec{c} \times \vec{a} |}{[\vec{a} \vec{b} \vec{c}]} + \frac{| \vec{a} \times \vec{b} |}{[\vec{a} \vec{b} \vec{c}]})$
$2 (\frac{1}{| \vec{a} | \cos α} + \frac{1}{| \vec{b} | \cos α} + \frac{1}{| \vec{c} | \cos α}) ∴ \frac{2}{\cos α} = 4 \Rightarrow \cos α = \frac{1}{2}$