Question

If a, b, c are three non-zero vector such that no two of them are collinear and $\left( {a \times b} \right) \times c = \dfrac{1}{3}\left| b \right|\left| c \right|\overrightarrow a $ . Find $\sin \theta $ if $\theta $ Is angle between $\left( {b \times c} \right)$.

Answer 1

Hint:
We can expand the LHS of the given relation using the property of vector triple product \[a \times \left( {b \times c} \right) = b\left( {a.c} \right) - c\left( {a.b} \right)\]. Then we can find the dot product using the relation \[a.b = \left| a \right|\left| b \right|\cos \theta \]. Then we can equate the coefficients of the vectors. By solving we get the value of \[\cos \theta \]. Then we can find the value of $\sin \theta $ using the relation ${\sin ^2}\theta = 1 - {\cos ^2}\theta $.

Complete step by step solution:
We have the vector triple product of 3 vectors \[\vec a\] , \[\vec b\] and \[\vec c\] where no two of them are collinear given as,
 $\left( {a \times b} \right) \times c = \dfrac{1}{3}\left| b \right|\left| c \right|\overrightarrow a $
By property of cross product of vectors, \[\vec a \times \vec b = - \vec b \times \vec a\], on applying this, we get,
 $ \Rightarrow - c \times \left( {a \times b} \right) = \dfrac{1}{3}\left| b \right|\left| c \right|\overrightarrow a $
We know that the vector triple product is defined as \[a \times \left( {b \times c} \right) = b\left( {a.c} \right) - c\left( {a.b} \right)\] . On expanding the above equation, we get,
  \[ \Rightarrow - \left( {c.b} \right)\vec a + \left( {c.a} \right)\vec b = \dfrac{1}{3}\left| b \right|\left| c \right|\overrightarrow a \]
As no two vectors are collinear, the coefficient of \[\vec b\] will be zero.
 \[ \Rightarrow - \left( {c.b} \right)\vec a = \dfrac{1}{3}\left| b \right|\left| c \right|\overrightarrow a \]
We know that dot product is given by, \[a.b = \left| a \right|\left| b \right|\cos \theta \] , where $\theta $ is the angle between the vectors.
 \[ \Rightarrow - \left( {\left| b \right|\left| c \right|\cos \theta } \right)\vec a = \dfrac{1}{3}\left| b \right|\left| c \right|\overrightarrow a \]
On equating the coefficient, we get,
 \[ \Rightarrow - \left| b \right|\left| c \right|\cos \theta = \dfrac{1}{3}\left| b \right|\left| c \right|\]
On further simplification, we get,
 \[ \Rightarrow \cos \theta = - \dfrac{1}{3}\]
We need to find the sin of the angle. We know that ${\sin ^2}\theta = 1 - {\cos ^2}\theta $ . On substituting the value, we get,
 $ \Rightarrow {\sin ^2}\theta = 1 - {\left( { - \dfrac{1}{3}} \right)^2}$
On simplification we get,
 $ \Rightarrow {\sin ^2}\theta = 1 - \dfrac{1}{9}$
On taking LCM we get,
 $ \Rightarrow {\sin ^2}\theta = \dfrac{{9 - 1}}{9}$
On further simplification we get,
 $ \Rightarrow {\sin ^2}\theta = \dfrac{8}{9}$
On taking the square root, we get,
 \[ \Rightarrow \sin \theta = \pm \sqrt {\dfrac{8}{9}} \]
 \[ \Rightarrow \sin \theta = \pm \dfrac{{2\sqrt 2 }}{3}\]

Note:
Vector triple product of 3 vectors is defined as the cross product of one vector with the cross product of the other two vectors. Vector triple products will always give a vector. For the vector triple product \[a \times \left( {b \times c} \right)\] , the resultant vector will be coplanar with b and c and will be perpendicular to a. We can write the vector product as the linear combinations of vectors b and c using the relation \[a \times \left( {b \times c} \right) = b\left( {a.c} \right) - c\left( {a.b} \right)\] . While taking the square root, we must take both positive and negative values as the quadrant is not mentioned.