Question

If $ \vec a $ , $ \vec b $ , $ \vec c $ are three vectors such that $ \vec a \times \vec b = \vec c $ and $ \vec b \times \vec c = \vec a $ , then
1. $ \vec a \ne \vec b \ne \vec c $
2. $ \vec a = \vec b = \vec c $
3. $ \vec a \ne \vec b \ne \vec c \ne 1 $
4. $ \vec a $ , $ \vec b $ , $ \vec c $ are orthogonal in pairs

Answer 1

Hint: In the question it is given that $ \vec a \times \vec b = \vec c $ and $ \vec b \times \vec c = \vec a $ . First we have to substitute the value of vector $ \vec a $ in the equation $ \vec a \times \vec b = \vec c $
Now the equation becomes $ \left( {\vec b \times \vec c} \right) \times \vec b = \vec c $
Next we have to substitute value of vector $ \vec c $ in the equation $ \vec b \times \vec c = \vec a $
Now the equation becomes $ \vec b \times \left( {\vec a \times \vec b} \right) = \vec a $
We know that the formula for vector triple product of three vectors is given by,
 $ \vec a \times \left( {\vec b \times \vec c} \right) = \left( {\vec a \cdot \vec c} \right)\vec b - \left( {\vec a \cdot \vec b} \right)\vec c $ , and
 $ \left( {\vec a \times \vec b} \right) \times \vec c = \left( {\vec a \cdot \vec c} \right)\vec b - \left( {\vec b \cdot \vec c} \right)\vec a $
Applying the formula of a vector triple product of three vectors we have to reduce the equations and solve them.

Complete step-by-step answer:
Let us consider the equations given in the question.
 $ \vec a \times \vec b = \vec c - - - - - - - \left( 1 \right) $
 $ \vec b \times \vec c = \vec a - - - - - - - \left( 2 \right) $
Substituting the equation $ \left( 2 \right) $ in equation $ \left( 1 \right) $ we get,
 $ \left( {\vec b \times \vec c} \right) \times \vec b = \vec c $
Applying the formula of vector triple product we get,
 $ \left( {\vec b \cdot \vec b} \right)\vec c - \left( {\vec c \cdot \vec b} \right)\vec b = \vec c $
We know that $ \vec b \cdot \vec b = {\left| b \right|^2} $ , hence the equation becomes
 $ {\left| b \right|^2}\vec c - \left( {\vec c \cdot \vec b} \right)\vec b = \vec c $
Comparing both the sides we get
 $ {\left| b \right|^2} = 1 $ and $ \left( {\vec c \cdot \vec b} \right) = 0 $
We know that if the dot product of two vectors is zero then the vectors are orthogonal to each other or perpendicular to each other.
Therefore vectors $ \vec c $ and $ \vec b $ are perpendicular to each other.
Now, substituting the equation $ \left( 1 \right) $ in equation $ \left( 2 \right) $ we get,
 $ \vec b \times \left( {\vec a \times \vec b} \right) = \vec a $
Applying the vector triple product formula of three vectors we get,
 $ \left( {\vec b \cdot \vec b} \right)\vec a - \left( {\vec b \cdot \vec a} \right)\vec b = \vec a $
We know that $ \vec b \cdot \vec b = {\left| b \right|^2} $ , hence the equation becomes
 $ {\left| b \right|^2}\vec a - \left( {\vec b \cdot \vec a} \right)\vec b = \vec a $
Comparing both the sides we get
 $ {\left| b \right|^2} = 1 $ and $ \left( {\vec b \cdot \vec a} \right) = 0 $
Therefore vectors $ \vec b $ and $ \vec a $ are orthogonal to each other.
This shows that vectors $ \vec a $ , $ \vec b $ , $ \vec c $ are orthogonal in pairs.
So, the correct answer is “Option 4”.

Note: This can also be solved using the vector triple product properties.
According to the property,
If $ \vec r = \vec a \times \left( {\vec b \times \vec c} \right) $
Then vector $ \vec r $ is perpendicular to vector $ \vec a $ and remains in the plane of vector $ \vec b $ and $ \vec c $ .
Therefore the equation
 $ \vec b \times \left( {\vec a \times \vec b} \right) = \vec a $ implies that vector $ \vec a $ is perpendicular to vector $ \vec b $ and
 $ \left( {\vec b \times \vec c} \right) \times \vec b = \vec c $ implies that vector $ \vec c $ is perpendicular to vector $ \vec b $ .