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If $\vec{a},\vec{b},\vec{c},\vec{d}$are unit vectors such that $(\vec{a}\times\vec{b}).(\vec{c}\times\vec{d})=1$ and $\vec{a}.\vec{c}=\frac{1}{2}$,then
$(A)\vec{a},\vec{b},\vec{c}$ are non-coplanar

$(B)\vec{a},\vec{b},\vec{d}$ are non-coplanar

$(C)\vec{b},\vec{d}$ are non-parallel.

$(D)\vec{a},\vec{d}$ are parallel and $\vec{b}$ and $\vec{c}$ are parallel.


My Attempt:
$(\vec{a}\times\vec{b}).(\vec{c}\times\vec{d})=1$
$(\vec{a}.\vec{c})(\vec{b}.\vec{d})-(\vec{a}.\vec{d})(\vec{b}.\vec{c})=1$
$\frac{1}{2}(\vec{b}.\vec{d})-(\vec{a}.\vec{d})(\vec{b}.\vec{c})=1$


Book's Solution which i did not understand:
$(\vec{a}\times\vec{b}).(\vec{c}\times\vec{d})=1$ is possible only when $|\vec{a}\times\vec{b}|=|\vec{c}\times\vec{d}|=0$ and $(\vec{a}\times\vec{b})$ is parallel to $(\vec{c}\times\vec{d})$ And the correct option given is $(C)$.

I could not solve this problem after some efforts.Please help me.Thanks.

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    $\begingroup$ Is there a typo? There isn't any information about $\vec{b}$ $\endgroup$ – Nicholas Nov 10 '15 at 16:11
  • $\begingroup$ Don't you mean $|\vec{a}\times\vec{b}|=|\vec{c}\times\vec{d}|=1$? $\endgroup$ – G-man Nov 13 '15 at 6:59
  • $\begingroup$ In the book,it is written as it is i have presented here.But i seriously doubt,there is some printing mistake in the book. $\endgroup$ – Vinod Kumar Punia Nov 13 '15 at 7:02
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While Michael's hint was a bit vague, it did answer the question nicely. This answer is just an elaboration on it.

Since the vectors given are all unit vectors, obviously the maximum value of $|\vec{a}\times\vec{b}|$ and $|\vec{c}\times\vec{d}|$ is unity. Then it follows that the maximum value of $(\vec{a}\times\vec{b}).(\vec{c}\times\vec{d})$ is also unity. This maximum clearly occurs when $\vec{a}\perp\vec{b}$ , $\vec{c}\perp\vec{d}$ and $\vec{a}\times\vec{b}\parallel\vec{c}\times\vec{d} $

Now you just have to use your imagination. Since the two planes that are spanned by the pairs of vectors $\vec{a},\vec{b}$ and $\vec{c},\vec{d}$ are parallel, and $\vec{a}$ is not collinear with $\vec{c}$, there is no way for $\vec{b}$ to be collinear with $\vec{d}$.

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There is some information to find from $$|u.v|=|u||v|\cos\theta$$ and $$|u\times v|=|u||v|\sin\theta$$

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  • $\begingroup$ While these are correct formulas, it isn't clear to me how you mean this to answer to question. This might work better as a comment, if you were confining yourself to a broad hint. $\endgroup$ – hardmath Nov 10 '15 at 16:50
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    $\begingroup$ What are the angles between a and b, between c and d, and between axb and cxd? $\endgroup$ – Empy2 Nov 10 '15 at 17:04

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