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 $\vec b$ .