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$\vec{a}, \vec{b} and \vec{c}$ are three non-zero & non-coplanar vectors with fixed magnitudes. Angles between any of the vector with the normal of the plane containing other two is $α$ . Volume of parallelepiped formed is V and its surface area is S. Which of the following is/are true?

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if angle between $\vec{a}, \vec{b}$ and $θ$ , then angle between $\vec{c}$ and plane of $\vec{a} and \vec{b}$ is $(90^{0} - θ)$

$\frac{V}{S} = \frac{| cosα |}{2} {(\frac{1}{| \vec{a} |} + \frac{1}{| \vec{b} |} + \frac{1}{| \vec{c} |})}^{- 1}$

$\vec{a}, \vec{b} and \vec{c}$ are equally inclined to each other

$V_{\max} = | \vec{a} | | \vec{b} | | \vec{c} |$

answer is A, B.

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Detailed Solution

$V = | \vec{a} \cdot (\vec{b} \times \vec{c}) | = | \vec{a} | | \vec{b} \times \vec{c} | | \cos α |$

Similarly, $V = | \vec{b} | | \vec{c} \times \vec{a} | | \cos α |$

& $V = | c | | \vec{a} \times \vec{b} | | \cos α |$

While $S = 2 (| \vec{a} \times \vec{b} | + | \vec{b} \times \vec{c} | + | \vec{c} \times \vec{a} |)$

$= 2 (\frac{V}{| \vec{c} | | \cos α |} + \frac{V}{| \vec{a} | | \cos α |} + \frac{V}{| \vec{b} | | \cos α |})$

$\Rightarrow \frac{V}{S} = \frac{| \cos α |}{2} {[\frac{1}{| \vec{a} |} + \frac{1}{| \vec{b} |} + \frac{1}{| \vec{c} |}]}^{- 1}$

Now, $V = | \vec{a} \cdot (\vec{b} \times \vec{c}) | = | \vec{a} | | \vec{b} \times \vec{c} | | \cos α |$

$= | \vec{a} | | \vec{b} | | \vec{c} | \sin (\vec{b}, \vec{c}) | \cos α | \leq \vec{a} | | \vec{b} | | \vec{c} ∣$

Also, $\sin (\vec{a}, \vec{b}) = \sin (\vec{b}, \vec{c}) = \sin (\vec{c}, \vec{a})$

So, it not necessary that they are equally inclined.

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