
If \[a,b,c\] are vectors such that \[[abc] = 4\] then \[[a \times bb \times cc \times a] = \]
A. 16
B. 64
C. 4
D. 8
Answer
164.4k+ views
Hint: If a, b, and c are three vectors, the scalar triple product is defined as the dot product of ‘a’ and \[\left( {b{\rm{ }} \times {\rm{ }}c} \right)\]. It is commonly represented by \[a{\rm{ }}.{\rm{ }}\left( {b{\rm{ }} \times {\rm{ }}c} \right){\rm{ }}or{\rm{ }}\left[ {a{\rm{ }}b{\rm{ }}c} \right]\]. \[\left[ {\left( {a{\rm{ }} + {\rm{ }}b} \right){\rm{ }}c{\rm{ }}d} \right]{\rm{ }} = {\rm{ }}\left[ {a{\rm{ }}c{\rm{ }}d} \right]{\rm{ }} + {\rm{ }}\left[ {b{\rm{ }}c{\rm{ }}d} \right]\], if a, b, c, d are the given four vectors.
Formula Used:The positions of the dot and cross in a scalar triple product can be swapped as long as the vectors' cyclic order remains constant.
\[\left( {a{\rm{ }} \times {\rm{ }}b} \right).{\rm{ }}c{\rm{ }} = {\rm{ }}a.{\rm{ }}\left( {b{\rm{ }} \times {\rm{ }}c} \right)\]
Complete step by step solution:Before solving that, we should know the property of scalar triple product, that is
The value of the scalar triple product remains constant while a, b, c are cyclically permuted,
\[\left( {a{\rm{ }} \times {\rm{ }}b} \right).{\rm{ }}c{\rm{ }} = {\rm{ }}\left( {b{\rm{ }} \times {\rm{ }}c} \right).{\rm{ }}a{\rm{ }} = {\rm{ }}\left( {c{\rm{ }} \times {\rm{ }}a} \right).{\rm{ }}b{\rm{ }}or{\rm{ }}\left[ {a{\rm{ }}b{\rm{ }}c} \right]{\rm{ }} = {\rm{ }}\left[ {b{\rm{ }}c{\rm{ }}a} \right]{\rm{ }} = {\rm{ }}\left[ {c{\rm{ }}a{\rm{ }}b} \right]\]
The cyclic arrangement of vectors in a scalar triple product affects the sign but not the magnitude of the scalar triple product, \[\left[ {a{\rm{ }}b{\rm{ }}c} \right] = - \left[ {b{\rm{ }}a{\rm{ }}c} \right] = - \left[ {c{\rm{ }}b{\rm{ }}a} \right] = - \left[ {a{\rm{ }}c{\rm{ }}b} \right]\]
If any two of the vectors are equal, the scalar triple product of the three vectors is zero.
We have been given in the problem that, \[a,b,c\] are vectors.
Such that,
\[[abc] = 4\]
We are asked to determine the value of\[[\vec a \times \vec b\vec b \times \vec c\vec c \times \vec a]\]
\[[\vec a \times \vec b\vec b \times \vec c\vec c \times \vec a] = {[\vec a\vec b\vec c]^2}\]
Given that,
\[[\vec a\vec b\vec c] = 4\]
From the given information, we can write that as,
So, \[\left[ {{\rm{ }}a \times b{\rm{ }}b \times c{\rm{ }}c \times a} \right]\]
\[ = {4^2}\]
On solving the above expression by squaring it, it gives
\[ = 16\]
Therefore, if \[a,b,c\] are vectors such that \[[abc] = 4\] then \[[a \times bb \times cc \times a] = 16\]
Option ‘A’ is correct
Note:If two of the vectors are parallel or collinear, the scalar triple product is zero. The condition that three non-zero non-collinear vectors must satisfy in order to be coplanar is \[\left[ {a{\rm{ }}b{\rm{ }}c} \right] = {\rm{ }}0\] . Four locations with position vector a, b, c, d are coplanar. The volume of a parallelepiped with coterminous edges a, b, c is \[\left[ {a{\rm{ }}b{\rm{ }}c} \right]\] or \[a\left( {b{\rm{ }} \times {\rm{ }}c} \right)\].
Formula Used:The positions of the dot and cross in a scalar triple product can be swapped as long as the vectors' cyclic order remains constant.
\[\left( {a{\rm{ }} \times {\rm{ }}b} \right).{\rm{ }}c{\rm{ }} = {\rm{ }}a.{\rm{ }}\left( {b{\rm{ }} \times {\rm{ }}c} \right)\]
Complete step by step solution:Before solving that, we should know the property of scalar triple product, that is
The value of the scalar triple product remains constant while a, b, c are cyclically permuted,
\[\left( {a{\rm{ }} \times {\rm{ }}b} \right).{\rm{ }}c{\rm{ }} = {\rm{ }}\left( {b{\rm{ }} \times {\rm{ }}c} \right).{\rm{ }}a{\rm{ }} = {\rm{ }}\left( {c{\rm{ }} \times {\rm{ }}a} \right).{\rm{ }}b{\rm{ }}or{\rm{ }}\left[ {a{\rm{ }}b{\rm{ }}c} \right]{\rm{ }} = {\rm{ }}\left[ {b{\rm{ }}c{\rm{ }}a} \right]{\rm{ }} = {\rm{ }}\left[ {c{\rm{ }}a{\rm{ }}b} \right]\]
The cyclic arrangement of vectors in a scalar triple product affects the sign but not the magnitude of the scalar triple product, \[\left[ {a{\rm{ }}b{\rm{ }}c} \right] = - \left[ {b{\rm{ }}a{\rm{ }}c} \right] = - \left[ {c{\rm{ }}b{\rm{ }}a} \right] = - \left[ {a{\rm{ }}c{\rm{ }}b} \right]\]
If any two of the vectors are equal, the scalar triple product of the three vectors is zero.
We have been given in the problem that, \[a,b,c\] are vectors.
Such that,
\[[abc] = 4\]
We are asked to determine the value of\[[\vec a \times \vec b\vec b \times \vec c\vec c \times \vec a]\]
\[[\vec a \times \vec b\vec b \times \vec c\vec c \times \vec a] = {[\vec a\vec b\vec c]^2}\]
Given that,
\[[\vec a\vec b\vec c] = 4\]
From the given information, we can write that as,
So, \[\left[ {{\rm{ }}a \times b{\rm{ }}b \times c{\rm{ }}c \times a} \right]\]
\[ = {4^2}\]
On solving the above expression by squaring it, it gives
\[ = 16\]
Therefore, if \[a,b,c\] are vectors such that \[[abc] = 4\] then \[[a \times bb \times cc \times a] = 16\]
Option ‘A’ is correct
Note:If two of the vectors are parallel or collinear, the scalar triple product is zero. The condition that three non-zero non-collinear vectors must satisfy in order to be coplanar is \[\left[ {a{\rm{ }}b{\rm{ }}c} \right] = {\rm{ }}0\] . Four locations with position vector a, b, c, d are coplanar. The volume of a parallelepiped with coterminous edges a, b, c is \[\left[ {a{\rm{ }}b{\rm{ }}c} \right]\] or \[a\left( {b{\rm{ }} \times {\rm{ }}c} \right)\].
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