Question

If \[\vec a,\vec b,\vec c\] and \[\vec d\] are unit vectors such that \[\left( {\vec a \times \vec b} \right) \cdot \left( {\vec c \times \vec d} \right) = 1\] and \[\vec a \cdot \vec c = \dfrac{1}{2}\], then
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

Answer 1

Hint: First of consider the angle between the vectors \[\left( {\vec a \times \vec b} \right)\] and \[\left( {\vec c \times \vec d} \right)\] as a variable and find its value to show that the two vectors are in parallel to each other. Likewise find the angle between the vectors \[\vec a\] and \[\vec c\]. Then draw a figure for the vectors \[\vec a,\vec b,\vec c\] and \[\vec d\] to get the required answer.

Complete step-by-step answer:
Given that \[\vec a,\vec b,\vec c\] and \[\vec d\] are unit vectors. So, we have \[\left| {\vec a} \right| = \left| {\vec b} \right| = \left| {\vec c} \right| = \left| {\vec d} \right| = 1\].
Also given that \[\left( {\vec a \times \vec b} \right) \cdot \left( {\vec c \times \vec d} \right) = 1\]
Let the angle between \[\left( {\vec a \times \vec b} \right)\] and \[\left( {\vec c \times \vec d} \right)\] be \[\theta \].
We know that for the two vectors \[\vec x\] and \[\vec y\], the dot product is given by \[\vec x \cdot \vec y = \left| {\vec x} \right|\left| {\vec y} \right|\cos \theta \] where \[\theta \] is the angle between the two vectors \[\vec x\] and \[\vec y\].
By using this formula, we have
\[
   \Rightarrow \left( {\vec a \times \vec b} \right) \cdot \left( {\vec c \times \vec d} \right) = \left| {\vec a \times \vec b} \right|\left| {\vec c \times \vec d} \right|\cos \theta = 1 \\
   \Rightarrow \left| {\vec a \times \vec b} \right|\left| {\vec c \times \vec d} \right|\cos \theta = 1 \\
\]
This is only possible when \[\left| {\vec a \times \vec b} \right| = \left| {\vec c \times \vec d} \right| = \cos \theta = 1\]
So, we have \[\cos \theta = 1\]. Thus, \[\theta = {90^0}\].
Since the angle between the vectors \[\left( {\vec a \times \vec b} \right)\] and \[\left( {\vec c \times \vec d} \right)\] is \[{90^0}\] the two vectors \[\left( {\vec a \times \vec b} \right)\] and \[\left( {\vec c \times \vec d} \right)\] are parallel to each other i.e., \[\left( {\vec a \times \vec b} \right)\parallel \left( {\vec c \times \vec d} \right)..................................\left( 1 \right)\]
Also given that \[\vec a \cdot \vec c = \dfrac{1}{2}\]. Let the angle between the two vectors \[\vec a,\vec c\] be \[\alpha \]. So, we have
\[
   \Rightarrow \left| {\vec a} \right|\left| {\vec c} \right|\cos \alpha = \dfrac{1}{2} \\
   \Rightarrow 1 \times 1 \times \cos \alpha = \dfrac{1}{2}{\text{ }}\left[ {\because \left| {\vec a} \right| = \left| {\vec c} \right| = 1} \right] \\
   \Rightarrow \cos \alpha = \dfrac{1}{2} \\
  \therefore \alpha = {60^0}{\text{ }}\left[ {\because \cos {{60}^0} = \dfrac{1}{2}} \right]{\text{ }} \\
\]
We can draw the vectors as

Since the two vectors \[\left( {\vec a \times \vec b} \right)\] and \[\left( {\vec c \times \vec d} \right)\] are parallel to each other and the angle between the vectors \[\vec a\] and \[\vec c\]is \[{60^0}\] we can say that \[\vec b\] and \[\vec d\] are not parallel.
Thus, the correct option is C. \[\vec b,\vec d\] are non-parallel

So, the correct answer is “Option C”.

Note: For the two vectors \[\vec x\] and \[\vec y\], the dot product is given by \[\vec x \cdot \vec y = \left| {\vec x} \right|\left| {\vec y} \right|\cos \theta \] where \[\theta \] is the angle between the two vectors \[\vec x\] and \[\vec y\]. And for the two vectors \[\vec x\] and \[\vec y\], the cross product is given by \[\vec x \times \vec y = \left| {\vec x} \right|\left| {\vec y} \right|\sin \theta \] where \[\theta \] is the angle between the two vectors \[\vec x\] and \[\vec y\].