
Volume of parallelepiped determined by vectors \[\vec a\] , \[\vec b\] and \[\vec c\] is 2. Then the volume of the parallelepiped determined by vectors \[2\left( {\vec a \times \vec b} \right)\] , \[3\left( {\vec b \times \vec c} \right)\] and \[2\left( {\vec c \times \vec a} \right)\] .
Answer
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Hint: Vector triple product is given in the form of or as
In this question the value of the volume of vectors in determinant form is given so to find the value of volume for the vectors \[2\left( {\vec a \times \vec b} \right)\] , \[3\left( {\vec b \times \vec c} \right)\] and \[2\left( {\vec c \times \vec a} \right)\] we will first find the Vector triple product of those three vectors and then we will find their values.
Complete step-by-step answer:
Given the value of the volume of parallelepiped determined by vectors is \[\left[ {\begin{array}{*{20}{c}}
{\vec a}&{\vec b}&{\vec c}
\end{array}} \right] = 2\]
Now we have to find the value for the volume of parallelepiped determined by vectors is \[\left[ {\begin{array}{*{20}{c}}
{2\left( {\vec a \times \vec b} \right)}&{3\left( {\vec b \times \vec c} \right)}&{2\left( {\vec c \times \vec a} \right)}
\end{array}} \right] \]
Let us assume the vectors whose volume is to be found is to be
\[
2\left( {\vec a \times \vec b} \right) = \vec p \\
3\left( {\vec b \times \vec c} \right) = \vec q \\
2\left( {\vec c \times \vec a} \right) = \vec r \;
\]
Hence we can write these vectors as \[\left[ {\begin{array}{*{20}{c}}
{\vec p}&{\vec q}&{\vec r}
\end{array}} \right] \]
Now as we know the Vector triple product is given in the form of, hence we can write the vector of the volume of parallelepiped as
Now substitute the values of the vectors, we get
By solving this we get
Now we know that the cross product of the four vectors is given in the form of \[\left( {\vec a \times \vec b} \right) \times \left( {\vec c \times \vec d} \right) = \left[ {\begin{array}{*{20}{c}}
{\vec a}&{\vec c}&{\vec d}
\end{array}} \right] \vec b - \left[ {\begin{array}{*{20}{c}}
{\vec b}&{\vec c}&{\vec d}
\end{array}} \right] \vec a\] , hence we can further write this vector as
Now using the property of the Vector triple product, we can write
\[\left[ {\begin{array}{*{20}{c}}
{2\left( {\vec a \times \vec b} \right)}&{3\left( {\vec b \times \vec c} \right)}&{2\left( {\vec c \times \vec a} \right)}
\end{array}} \right] = 12\left( { - \left[ {\begin{array}{*{20}{c}}
{\vec a}&{\vec b}&{\vec c}
\end{array}} \right] \left[ {\begin{array}{*{20}{c}}
{\vec b}&{\vec c}&{\vec a}
\end{array}} \right] } \right)\]
Now by interchanging the positions, we get
\[
\left[ {\begin{array}{*{20}{c}}
{2\left( {\vec a \times \vec b} \right)}&{3\left( {\vec b \times \vec c} \right)}&{2\left( {\vec c \times \vec a} \right)}
\end{array}} \right] = 12\left( {\left[ {\begin{array}{*{20}{c}}
{\vec a}&{\vec b}&{\vec c}
\end{array}} \right] \left[ {\begin{array}{*{20}{c}}
{\vec a}&{\vec b}&{\vec c}
\end{array}} \right] } \right) \\
= 12{\left[ {\begin{array}{*{20}{c}}
{\vec a}&{\vec b}&{\vec c}
\end{array}} \right] ^2} \;
\]
Now since we already know \[\left[ {\begin{array}{*{20}{c}}
{\vec a}&{\vec b}&{\vec c}
\end{array}} \right] = 2\] , hence we can write the value vector as
\[
\left[ {\begin{array}{*{20}{c}}
{2\left( {\vec a \times \vec b} \right)}&{3\left( {\vec b \times \vec c} \right)}&{2\left( {\vec c \times \vec a} \right)}
\end{array}} \right] = 12{\left[ 2 \right] ^2} \\
= 12 \times 4 \\
= 48 \;
\]
Therefore the volume of the parallelepiped determined by vectors \[\left[ {\begin{array}{*{20}{c}}
{2\left( {\vec a \times \vec b} \right)}&{3\left( {\vec b \times \vec c} \right)}&{2\left( {\vec c \times \vec a} \right)}
\end{array}} \right] = 48\]
So, the correct answer is “48 ”.
Additional Information:
The cross product of four vectors is \[\left( {\vec a \times \vec b} \right) \times \left( {\vec c \times \vec d} \right) = \left[ {\begin{array}{*{20}{c}}
{\vec a}&{\vec c}&{\vec d}
\end{array}} \right] \vec b - \left[ {\begin{array}{*{20}{c}}
{\vec b}&{\vec c}&{\vec d}
\end{array}} \right] \vec a\]
Note: The cross product of two vectors tells us about how perpendicular the two vectors are given as \[\vec A \times \vec B = AB\sin \theta \]
Here \[\theta \] is the angle between the two vectors and if the two vectors are parallel then their cross product is zero and the cross product will be maximum if they are perpendicular.
In this question the value of the volume of vectors in determinant form is given so to find the value of volume for the vectors \[2\left( {\vec a \times \vec b} \right)\] , \[3\left( {\vec b \times \vec c} \right)\] and \[2\left( {\vec c \times \vec a} \right)\] we will first find the Vector triple product of those three vectors and then we will find their values.
Complete step-by-step answer:
Given the value of the volume of parallelepiped determined by vectors is \[\left[ {\begin{array}{*{20}{c}}
{\vec a}&{\vec b}&{\vec c}
\end{array}} \right] = 2\]
Now we have to find the value for the volume of parallelepiped determined by vectors is \[\left[ {\begin{array}{*{20}{c}}
{2\left( {\vec a \times \vec b} \right)}&{3\left( {\vec b \times \vec c} \right)}&{2\left( {\vec c \times \vec a} \right)}
\end{array}} \right] \]
Let us assume the vectors whose volume is to be found is to be
\[
2\left( {\vec a \times \vec b} \right) = \vec p \\
3\left( {\vec b \times \vec c} \right) = \vec q \\
2\left( {\vec c \times \vec a} \right) = \vec r \;
\]
Hence we can write these vectors as \[\left[ {\begin{array}{*{20}{c}}
{\vec p}&{\vec q}&{\vec r}
\end{array}} \right] \]
Now as we know the Vector triple product is given in the form of, hence we can write the vector of the volume of parallelepiped as
Now substitute the values of the vectors, we get
By solving this we get
Now we know that the cross product of the four vectors is given in the form of \[\left( {\vec a \times \vec b} \right) \times \left( {\vec c \times \vec d} \right) = \left[ {\begin{array}{*{20}{c}}
{\vec a}&{\vec c}&{\vec d}
\end{array}} \right] \vec b - \left[ {\begin{array}{*{20}{c}}
{\vec b}&{\vec c}&{\vec d}
\end{array}} \right] \vec a\] , hence we can further write this vector as
Now using the property of the Vector triple product, we can write
\[\left[ {\begin{array}{*{20}{c}}
{2\left( {\vec a \times \vec b} \right)}&{3\left( {\vec b \times \vec c} \right)}&{2\left( {\vec c \times \vec a} \right)}
\end{array}} \right] = 12\left( { - \left[ {\begin{array}{*{20}{c}}
{\vec a}&{\vec b}&{\vec c}
\end{array}} \right] \left[ {\begin{array}{*{20}{c}}
{\vec b}&{\vec c}&{\vec a}
\end{array}} \right] } \right)\]
Now by interchanging the positions, we get
\[
\left[ {\begin{array}{*{20}{c}}
{2\left( {\vec a \times \vec b} \right)}&{3\left( {\vec b \times \vec c} \right)}&{2\left( {\vec c \times \vec a} \right)}
\end{array}} \right] = 12\left( {\left[ {\begin{array}{*{20}{c}}
{\vec a}&{\vec b}&{\vec c}
\end{array}} \right] \left[ {\begin{array}{*{20}{c}}
{\vec a}&{\vec b}&{\vec c}
\end{array}} \right] } \right) \\
= 12{\left[ {\begin{array}{*{20}{c}}
{\vec a}&{\vec b}&{\vec c}
\end{array}} \right] ^2} \;
\]
Now since we already know \[\left[ {\begin{array}{*{20}{c}}
{\vec a}&{\vec b}&{\vec c}
\end{array}} \right] = 2\] , hence we can write the value vector as
\[
\left[ {\begin{array}{*{20}{c}}
{2\left( {\vec a \times \vec b} \right)}&{3\left( {\vec b \times \vec c} \right)}&{2\left( {\vec c \times \vec a} \right)}
\end{array}} \right] = 12{\left[ 2 \right] ^2} \\
= 12 \times 4 \\
= 48 \;
\]
Therefore the volume of the parallelepiped determined by vectors \[\left[ {\begin{array}{*{20}{c}}
{2\left( {\vec a \times \vec b} \right)}&{3\left( {\vec b \times \vec c} \right)}&{2\left( {\vec c \times \vec a} \right)}
\end{array}} \right] = 48\]
So, the correct answer is “48 ”.
Additional Information:
The cross product of four vectors is \[\left( {\vec a \times \vec b} \right) \times \left( {\vec c \times \vec d} \right) = \left[ {\begin{array}{*{20}{c}}
{\vec a}&{\vec c}&{\vec d}
\end{array}} \right] \vec b - \left[ {\begin{array}{*{20}{c}}
{\vec b}&{\vec c}&{\vec d}
\end{array}} \right] \vec a\]
Note: The cross product of two vectors tells us about how perpendicular the two vectors are given as \[\vec A \times \vec B = AB\sin \theta \]
Here \[\theta \] is the angle between the two vectors and if the two vectors are parallel then their cross product is zero and the cross product will be maximum if they are perpendicular.
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