Let $\vec a,\vec b,\vec c$ and $\vec d$ be four unit vectors such that $(\vec a \times \vec b).(\vec c \times \vec d)=1$ and $\vec a . \vec c = \frac {1} {2}$. Then which of the following is true$?$

$(a)$ $\vec a,\vec b,\vec c$ are non-coplanar$.$

$(b)$ $\vec b,\vec c,\vec d$ are non-coplanar$.$

$(c)$ $\vec b,\vec d$ are non-parallel$.$

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

I tried to solve it by simplifying $(\vec a \times \vec b).(\vec c \times \vec d)=1$ which yields $(\vec a . \vec c)(\vec b . \vec d)-(\vec b . \vec c)(\vec a . \vec d)=1.$ Putting the value of $\vec a.\vec c$ we have $\frac {1} {2}(\vec b . \vec d)-(\vec b . \vec c)(\vec a . \vec d)=1.$

Clearly $(d)$ is not correct.Because if the angle between $\vec b$ and $\vec d$ is taken to be $\theta$ therefore if $(d)$ holds then we would have $\cos \theta = 4$ which is absurd.

Now I also observe that since all the vectors are unit modulus so the maximum value of $(\vec a \times \vec b).(\vec c \times \vec d)$ is $1$ and it has been attained iff $|\vec a \times \vec b|=1$ and $|\vec c \times \vec d|=1$. So from these three conditions we have $\vec a$ is perpendicular to $ \vec b$, $\vec c$ is perpendicular to $\vec d$ and $(\vec a \times \vec b)$ is parallel to $(\vec c \times \vec d)$. Now it may so happen all the vectors are coplanar because suppose $\vec a$ is in the positive $x$-direction,$\vec b$ is in the positive $y$-direction.Now if I make rotation of the coordinate axes through an angle $60^o$ then if I take $\vec c$ and $\vec d$ along the new $x$ and $y$ direction respectively then the given condition holds. So I think $(a)$ and $(c)$ are in general false. Now if $\vec b$ and $\vec d$ are parallel then since $\vec c$ is perpendicular to $\vec d$ and since $\vec a$ is perpendicular to $\vec b$ so we would have either $\vec a$ is perpendicular to $\vec c$ or $\vec a$ is parallel to $\vec c$.Neither of them holds since $\vec a . \vec c=\frac {1} {2}$ and $\vec a$ and $\vec c$ are unit vectors. So $\vec b$ and $\vec d$ are non-parallel. Thus $(c)$ is the only correct option.

Is the above reasoning correct at all? Please verify it.

Thank you in advance.

  • $\begingroup$ if you had used an enumeration with roman numerals, you could have simply called the vectors $a$, $b$, $c$, $d$ (without arrows) and saved yourself 5 minutes of typing. Also, use \cdot for scalar products $\endgroup$ – Bananach Jun 17 '17 at 17:20
  • $\begingroup$ I agree that $c$ is correct, but I don't know how to efficiently prove it at the moment. Just one more note: $a \parallel b$ commonly means that $a$ and $b$ are parallel. You can use $a \bot b$ to denote orthogonal vectors $\endgroup$ – Bananach Jun 17 '17 at 17:23

$(\vec{a} \times \vec{b}) \cdot (\vec{c} \times \vec{d}) = 1 \implies ||\vec{a} \times \vec{b}|| \; ||\vec{c} \times \vec{d}|| \; \cos{\phi} = 1 \implies ab \; |\sin{\alpha}| \; cd\; |\sin{\beta}| \; \cos{\phi} = 1 \implies |\sin{\alpha}| \; |\sin{\beta}| \; \cos{\phi} = 1$. ($\alpha = $ angle between $\vec{a}$ and $\vec{b}$, $\beta = $ angle between $\vec{c}$ and $\vec{d}$, $\phi = $ angle between $\vec{a} \times \vec{b}$ and $\vec{c} \times \vec{d}$ and $a,b,c,d = $ magnitudes of respective vectors = $1$ since they are all unit vectors).

This is possible iff $|\sin{\alpha}| = |\sin{\beta}| = \cos{\phi} = 1$.

$(\cos{\phi} = 1 \implies \phi = 0)$ and $(||\vec{a} \times \vec{b}|| = ||\vec{c} \times \vec{d}|| = 1)$ together imply that $\vec{a} \times \vec{b} = \vec{c} \times \vec{d}$.

Also, $\vec{a}$, $\vec{b}$ lie in a plane perpendicular to $\vec{a} \times \vec{b}$ and $\vec{c}$, $\vec{d}$ lie in a plane perpendicular to $\vec{c} \times \vec{d}$. Since the two cross products produce the same vector, all of $\vec{a}, \vec{b}, \vec{c}, \vec{d}$ must be co-planar.

Since $|\sin{\alpha}| = 1 = |\sin{\beta}|$, it can be concluded that $\vec{a} \perp \vec{b}$ and $\vec{c} \perp \vec{d}$.

$\vec{a} \cdot \vec{c} = 0.5 \implies$ angle between $\vec{a}$ and $\vec{c}$ is 60 degress (since $a = c = 1$).

With all this information, draw a picture of all 4 vectors on a plane. (We have already concluded that they are all co-planar). It's not hard to see that choice (c) is the only correct answer.


The fact that all the vectors are unitary implies that the module of the cross product between any two of them will not be greater than $1$. $$ \left| a \right| = \left| b \right| = \left| c \right| = \left| d \right| = 1\quad \Rightarrow \quad \left\{ \matrix{ 0 \le \left| {a \times b} \right| \le 1 \hfill \cr 0 \le \left| {c \times d} \right| \le 1 \hfill \cr} \right. $$ Then adding the condition of the dot product between $a \times b$ and $c \times d$ be $1$, implies that each one of the cross products have module $1$, and that they are parallel parallel. $$ \eqalign{ & \left\{ \matrix{ \left| a \right| = \left| b \right| = \left| c \right| = \left| d \right| = 1 \hfill \cr 0 \le \left| {a \times b} \right| \le 1 \hfill \cr 0 \le \left| {c \times d} \right| \le 1 \hfill \cr \left( {a \times b} \right) \cdot \left( {c \times d} \right) = 1 \hfill \cr} \right.\quad \Rightarrow \quad \left\{ \matrix{ \left| a \right| = \left| b \right| = \left| c \right| = \left| d \right| = 1 \hfill \cr \left| {a \times b} \right| = 1 \hfill \cr \left| {c \times d} \right| = 1 \hfill \cr \left( {a \times b} \right)||\left( {c \times d} \right) \hfill \cr} \right.\quad \Rightarrow \cr & \Rightarrow \quad \left\{ \matrix{ \left| a \right| = \left| b \right| = \left| c \right| = \left| d \right| = 1 \hfill \cr a \bot b \hfill \cr c \bot d \hfill \cr a,b,c,d\;{\rm coplanar} \hfill \cr} \right. \cr} $$ which consequently implies that in each cross product the terms be normal to each other, and since they give cross products that are parallel, all four vectors shall be coplanar.

Finally adding the condition that $a$ makes an angle of $\pi/3$ with $c$, which implies the same between $b$ and $d$, we get the following final situation $$ \left\{ \matrix{ \left| a \right| = \left| b \right| = \left| c \right| = \left| d \right| = 1 \hfill \cr a \bot b \hfill \cr c \bot d \hfill \cr a,b,c,d\;{\rm coplanar} \hfill \cr a \cdot c = 1/2 \hfill \cr} \right.\quad \Rightarrow \quad \left\{ \matrix{ \left| a \right| = \left| b \right| = \left| c \right| = \left| d \right| = 1 \hfill \cr a,b,c,d\;{\rm coplanar} \hfill \cr a \bot b,\;c \bot d \hfill \cr \,a\angle c,\;b\angle d \hfill \cr} \right. $$ ($a\angle c$ meaning that $a$ is "angled" with $c$)
which does not fit with any of the proposed answers.


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