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# If $\vec a,\,\vec b,\,\vec c$ are three non-coplanar non-zero vectors and $\vec r$ is any vector in space, then$\left( {\vec a \times \vec b} \right) \times \left( {\vec r \times \vec c} \right) + \left( {\vec b \times \vec c} \right) \times \left( {\vec r \times \vec a} \right) + \left( {\vec c \times \vec a} \right) \times \left( {\vec r \times \vec b} \right)$ is equal toA.$2\left[ {\vec a\,\vec b\,\vec c} \right]\vec r$B.$3\left[ {\vec a\,\vec b\,\vec c} \right]\vec r$C.$\left[ {\vec a\,\vec b\,\vec c} \right]\vec r$D.$4\left[ {\vec a\,\vec b\,\vec c} \right]\vec r$

Last updated date: 20th Jun 2024
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Hint: Here we will apply the identity of the cross multiplication of the vectors to expand the given equation in terms of the matrices of the given vectors. Then we will simplify that equation to get the required answer.

Given equation is $\left( {\vec a \times \vec b} \right) \times \left( {\vec r \times \vec c} \right) + \left( {\vec b \times \vec c} \right) \times \left( {\vec r \times \vec a} \right) + \left( {\vec c \times \vec a} \right) \times \left( {\vec r \times \vec b} \right)$………………..$\left( 1 \right)$
We will use the identity of the vectors, $\vec a \times \left( {\vec b \times \vec c} \right) = \left( {\vec a\, \cdot \vec c} \right)\vec b - \left( {\vec a\, \cdot \vec b} \right)\vec c$ to simplify this equation..
So, firstly we will take the first term from the equation $\left( 1 \right)$ i.e. $\left( {\vec a \times \vec b} \right) \times \left( {\vec r \times \vec c} \right)$ and we will apply the identity on it.
Therefore, we get
$\left( {\vec a \times \vec b} \right) \times \left( {\vec r \times \vec c} \right) = \left( {\left( {\vec a \times \vec b} \right)\vec c} \right)\vec r - \left( {\left( {\vec a \times \vec b} \right)\vec r} \right)\vec c$
We know that $\left( {\vec a \times \vec b} \right)\vec c$can be written in terms of the matrix form as$\left[ {\vec a\,\vec b\,\vec c} \right]$i.e. $\left( {\vec a \times \vec b} \right)\vec c = \left[ {\vec a\,\vec b\,\vec c} \right]$. Therefore, we get
$\Rightarrow \left( {\vec a \times \vec b} \right) \times \left( {\vec r \times \vec c} \right) = \left( {\left( {\vec a \times \vec b} \right)\vec c} \right)\vec r - \left( {\left( {\vec a \times \vec b} \right)\vec r} \right)\vec c = \left[ {\vec a\,\vec b\,\vec c} \right]\vec r - \left[ {\vec a\,\vec b\,\vec r} \right]\vec c$
Similarly, we will find this for the second terms of the equation $1$ i.e. $\left( {\vec b \times \vec c} \right) \times \left( {\vec r \times \vec a} \right)$.
$\left( {\vec b \times \vec c} \right) \times \left( {\vec r \times \vec a} \right) = \left( {\left( {\vec b \times \vec c} \right)\vec a} \right)\vec r - \left( {\left( {\vec b \times \vec c} \right)\vec r} \right)\vec a = \left[ {\vec a\,\vec b\,\vec c} \right]\vec r - \left[ {\,\vec b\vec c\,\vec r} \right]\vec a$
Again, we will find this for the third terms of the equation $1$ i.e. $\left( {\vec c \times \vec a} \right) \times \left( {\vec r \times \vec b} \right)$.
$\left( {\vec c \times \vec a} \right) \times \left( {\vec r \times \vec b} \right) = \left( {\left( {\vec c \times \vec a} \right)\vec b} \right)\vec r - \left( {\left( {\vec c \times \vec a} \right)\vec r} \right)\vec b = \left[ {\vec a\,\vec b\,\vec c} \right]\vec r - \left[ {\,\vec a\vec c\,\vec r} \right]\vec b$
Now we will find the sum of all the terms of the equation $\left( 1 \right)$. Therefore, we get
$\left( {\vec a \times \vec b} \right) \times \left( {\vec r \times \vec c} \right) + \left( {\vec b \times \vec c} \right) \times \left( {\vec r \times \vec a} \right) + \left( {\vec c \times \vec a} \right) \times \left( {\vec r \times \vec b} \right) = \left( {\left[ {\vec a\,\vec b\,\vec c} \right]\vec r - \left[ {\vec a\,\vec b\,\vec r} \right]\vec c} \right) + \left( {\left[ {\vec a\,\vec b\,\vec c} \right]\vec r - \left[ {\,\vec b\vec c\,\vec r} \right]\vec a} \right) + \left( {\left[ {\vec a\,\vec b\,\vec c} \right]\vec r - \left[ {\,\vec a\vec c\,\vec r} \right]\vec b} \right)$
Now by solving this, we get
$\Rightarrow \left( {\vec a \times \vec b} \right) \times \left( {\vec r \times \vec c} \right) + \left( {\vec b \times \vec c} \right) \times \left( {\vec r \times \vec a} \right) + \left( {\vec c \times \vec a} \right) \times \left( {\vec r \times \vec b} \right) = 3\left[ {\vec a\,\vec b\,\vec c} \right]\vec r - \left[ {\vec a\,\vec b\,\vec r} \right]\vec c - \left[ {\,\vec b\vec c\,\vec r} \right]\vec a - \left[ {\,\vec a\vec c\,\vec r} \right]\vec b$
$\Rightarrow \left( {\vec a \times \vec b} \right) \times \left( {\vec r \times \vec c} \right) + \left( {\vec b \times \vec c} \right) \times \left( {\vec r \times \vec a} \right) + \left( {\vec c \times \vec a} \right) \times \left( {\vec r \times \vec b} \right) = 3\left[ {\vec a\,\vec b\,\vec c} \right]\vec r - \left( {\left[ {\vec a\,\vec b\,\vec r} \right]\vec c + \left[ {\,\vec b\vec c\,\vec r} \right]\vec a + \left[ {\,\vec a\vec c\,\vec r} \right]\vec b} \right)$
We know that the value of $\left[ {\vec a\,\vec b\,\vec r} \right]\vec c + \left[ {\,\vec b\vec c\,\vec r} \right]\vec a + \left[ {\,\vec a\vec c\,\vec r} \right]\vec b = \left[ {\vec a\,\vec b\,\vec c} \right]\vec r$.
Substituting $\left[ {\vec a\,\vec b\,\vec r} \right]\vec c + \left[ {\,\vec b\vec c\,\vec r} \right]\vec a + \left[ {\,\vec a\vec c\,\vec r} \right]\vec b = \left[ {\vec a\,\vec b\,\vec c} \right]\vec r$ in the above equation, we get
$\Rightarrow \left( {\vec a \times \vec b} \right) \times \left( {\vec r \times \vec c} \right) + \left( {\vec b \times \vec c} \right) \times \left( {\vec r \times \vec a} \right) + \left( {\vec c \times \vec a} \right) \times \left( {\vec r \times \vec b} \right) = 3\left[ {\vec a\,\vec b\,\vec c} \right]\vec r - \left[ {\vec a\,\vec b\,\vec c} \right]\vec r = 2\left[ {\vec a\,\vec b\,\vec c} \right]\vec r$
Hence, $2\left[ {\vec a\,\vec b\,\vec c} \right]\vec r$ is the value of the given equation.
So, option A is the correct option.

Note: Vector is the geometric object that has both the magnitude and the direction of an object. So while calculating the equation of a line vector we should know that it is equal to the difference between the final point vector and the starting point vector of that line. Vectors have three components i.e. $x$ component, $y$ component and $z$ component and all the three components of the vectors are perpendicular to each other. Unit vector is a vector which has a magnitude of 1 unit and zero vector is a vector which has a magnitude of 0 unit.