
Volume of Parallelepiped formed by vectors $\overrightarrow{a}$ $\times $ $\overrightarrow{b}$ , $\overrightarrow{b}$$\times $ $\overrightarrow{c}$ and $\overrightarrow{c}$ $\times $ $\overrightarrow{a}$ is 36 sq. units.
Prove that
\[A)\,\,\,\,\left[ \overrightarrow{a}\overrightarrow{b}\overrightarrow{c} \right]=6\]
B) Volume of tetrahedron formed by vectors $\overrightarrow{a}$, $\overrightarrow{b}$, $\overrightarrow{c}$ is 1
$C)\,\,\,\left[ \overrightarrow{a}+\overrightarrow{b}\overrightarrow{b}+\overrightarrow{c}\overrightarrow{c}+\overrightarrow{a} \right]=12$
D) Differences between vectors are coplanar.
Answer
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Hint: Formula for finding the volume of a parallelepiped is [$\overrightarrow{a}$ $\times $ $\overrightarrow{b}$ $\overrightarrow{b}$$\times $ $\overrightarrow{c}$ $\overrightarrow{c}$ $\times $ $\overrightarrow{a}$ ]. We can equate the given value of volume with the above formula and through that we can the value of [$\overrightarrow{a}$$\overrightarrow{b}$$\overrightarrow{c}$] as it is the square root of volume of parallelepiped.
Complete step-by-step solution:
Given Volume of Parallelepiped formed by vectors $\overrightarrow{a}$ $\times $ $\overrightarrow{b}$ , $\overrightarrow{b}$$\times $ $\overrightarrow{c}$ and $\overrightarrow{c}$ $\times $ $\overrightarrow{a}$ is 36 sq. units.
Therefore
[$\overrightarrow{a}$$\times $ $\overrightarrow{b}$ $\overrightarrow{b}$$\times $ $\overrightarrow{c}$ $\overrightarrow{c}$ $\times $ $\overrightarrow{a}$ ] = 36
We know that one of the property of box product is [$\overrightarrow{a}$ $\times $ $\overrightarrow{b}$ $\overrightarrow{b}$$\times $ $\overrightarrow{c}$ $\overrightarrow{c}$ $\times $ $\overrightarrow{a}$ ] = ${{\left[ \overrightarrow{a}\overrightarrow{b}\overrightarrow{c} \right]}^{2}}$
By this we can say that [$\overrightarrow{a}$$\overrightarrow{b}$$\overrightarrow{c}$] = 6
Hence the statement (a) is proved
Now let us move to the next statement
We know that Volume of tetrahedron formed by vectors $\overrightarrow{a}$, $\overrightarrow{b}$, $\overrightarrow{c}$ is $\dfrac{1}{6}$ [$\overrightarrow{a}$$\overrightarrow{b}$$\overrightarrow{c}$]
Therefore , The Volume of tetrahedron formed by vectors $\overrightarrow{a}$, $\overrightarrow{b}$, $\overrightarrow{c}$ = $\dfrac{1}{6}$ [$\overrightarrow{a}$$\overrightarrow{b}$$\overrightarrow{c}$] = $\dfrac{1}{6}$ ( 6 ) = 1
The Volume of tetrahedron formed by vectors $\overrightarrow{a}$, $\overrightarrow{b}$, $\overrightarrow{c}$ = 1
Hence the statement (b) is proved.
Now solving the third statement
c) [ $\overrightarrow{a}$ + $\overrightarrow{b}$ $\overrightarrow{b}$ + $\overrightarrow{c}$ $\overrightarrow{c}$ + $\overrightarrow{a}$ ]
We know that [ $\overrightarrow{a}$ + $\overrightarrow{b}$ $\overrightarrow{b}$ + $\overrightarrow{c}$ $\overrightarrow{c}$ + $\overrightarrow{a}$ ] = 2 [$\overrightarrow{a}$$\overrightarrow{b}$$\overrightarrow{c}$ ] = 2 ( 6 ) = 12
Hence the statement (c) is proved.
d) $\overrightarrow{a} -\overrightarrow{b}$, $\overrightarrow{b} -\overrightarrow{c}$ and $\overrightarrow{c} - \overrightarrow{a}$ can be said that they are coplanar only when
[ $\overrightarrow{a}-\overrightarrow{b}$ $\overrightarrow{b}-\overrightarrow{c}$ $\overrightarrow{c} - \overrightarrow{a}$ ] is equal to 0.
We also know that if the lines are $\overrightarrow{a} - \overrightarrow{b}$, $\overrightarrow{b} - \overrightarrow{c}$ and $\overrightarrow{c} - \overrightarrow{a}$ the determinant of those is 0, which implies they are coplanar.
Note: Learn all the formulae and properties of vectors. The main properties that are used are
1) [$\overrightarrow{a}$ $\times $ $\overrightarrow{b}$ $\overrightarrow{b}$$\times $ $\overrightarrow{c}$ $\overrightarrow{c}$ $\times $ $\overrightarrow{a}$ ] = ${{\left[ \overrightarrow{a}\overrightarrow{b}\overrightarrow{c} \right]}^{2}}$
2) [ $\overrightarrow{a}$ + $\overrightarrow{b}$ $\overrightarrow{b}$ + $\overrightarrow{c}$ $\overrightarrow{c}$ + $\overrightarrow{a}$ ] = 2 [$\overrightarrow{a}$$\overrightarrow{b}$$\overrightarrow{c}$ ] and
3) Volume of tetrahedron formed by vectors $\overrightarrow{a}$, $\overrightarrow{b}$, $\overrightarrow{c}$ is $\dfrac{1}{6}$ [$\overrightarrow{a}$$\overrightarrow{b}$$\overrightarrow{c}$]
Find the values of determinants without making calculation mistakes. Also learn the formulae of planes formed by the vectors and lines formed by the vectors.
Complete step-by-step solution:
Given Volume of Parallelepiped formed by vectors $\overrightarrow{a}$ $\times $ $\overrightarrow{b}$ , $\overrightarrow{b}$$\times $ $\overrightarrow{c}$ and $\overrightarrow{c}$ $\times $ $\overrightarrow{a}$ is 36 sq. units.
Therefore
[$\overrightarrow{a}$$\times $ $\overrightarrow{b}$ $\overrightarrow{b}$$\times $ $\overrightarrow{c}$ $\overrightarrow{c}$ $\times $ $\overrightarrow{a}$ ] = 36
We know that one of the property of box product is [$\overrightarrow{a}$ $\times $ $\overrightarrow{b}$ $\overrightarrow{b}$$\times $ $\overrightarrow{c}$ $\overrightarrow{c}$ $\times $ $\overrightarrow{a}$ ] = ${{\left[ \overrightarrow{a}\overrightarrow{b}\overrightarrow{c} \right]}^{2}}$
By this we can say that [$\overrightarrow{a}$$\overrightarrow{b}$$\overrightarrow{c}$] = 6
Hence the statement (a) is proved
Now let us move to the next statement
We know that Volume of tetrahedron formed by vectors $\overrightarrow{a}$, $\overrightarrow{b}$, $\overrightarrow{c}$ is $\dfrac{1}{6}$ [$\overrightarrow{a}$$\overrightarrow{b}$$\overrightarrow{c}$]
Therefore , The Volume of tetrahedron formed by vectors $\overrightarrow{a}$, $\overrightarrow{b}$, $\overrightarrow{c}$ = $\dfrac{1}{6}$ [$\overrightarrow{a}$$\overrightarrow{b}$$\overrightarrow{c}$] = $\dfrac{1}{6}$ ( 6 ) = 1
The Volume of tetrahedron formed by vectors $\overrightarrow{a}$, $\overrightarrow{b}$, $\overrightarrow{c}$ = 1
Hence the statement (b) is proved.
Now solving the third statement
c) [ $\overrightarrow{a}$ + $\overrightarrow{b}$ $\overrightarrow{b}$ + $\overrightarrow{c}$ $\overrightarrow{c}$ + $\overrightarrow{a}$ ]
We know that [ $\overrightarrow{a}$ + $\overrightarrow{b}$ $\overrightarrow{b}$ + $\overrightarrow{c}$ $\overrightarrow{c}$ + $\overrightarrow{a}$ ] = 2 [$\overrightarrow{a}$$\overrightarrow{b}$$\overrightarrow{c}$ ] = 2 ( 6 ) = 12
Hence the statement (c) is proved.
d) $\overrightarrow{a} -\overrightarrow{b}$, $\overrightarrow{b} -\overrightarrow{c}$ and $\overrightarrow{c} - \overrightarrow{a}$ can be said that they are coplanar only when
[ $\overrightarrow{a}-\overrightarrow{b}$ $\overrightarrow{b}-\overrightarrow{c}$ $\overrightarrow{c} - \overrightarrow{a}$ ] is equal to 0.
We also know that if the lines are $\overrightarrow{a} - \overrightarrow{b}$, $\overrightarrow{b} - \overrightarrow{c}$ and $\overrightarrow{c} - \overrightarrow{a}$ the determinant of those is 0, which implies they are coplanar.
Note: Learn all the formulae and properties of vectors. The main properties that are used are
1) [$\overrightarrow{a}$ $\times $ $\overrightarrow{b}$ $\overrightarrow{b}$$\times $ $\overrightarrow{c}$ $\overrightarrow{c}$ $\times $ $\overrightarrow{a}$ ] = ${{\left[ \overrightarrow{a}\overrightarrow{b}\overrightarrow{c} \right]}^{2}}$
2) [ $\overrightarrow{a}$ + $\overrightarrow{b}$ $\overrightarrow{b}$ + $\overrightarrow{c}$ $\overrightarrow{c}$ + $\overrightarrow{a}$ ] = 2 [$\overrightarrow{a}$$\overrightarrow{b}$$\overrightarrow{c}$ ] and
3) Volume of tetrahedron formed by vectors $\overrightarrow{a}$, $\overrightarrow{b}$, $\overrightarrow{c}$ is $\dfrac{1}{6}$ [$\overrightarrow{a}$$\overrightarrow{b}$$\overrightarrow{c}$]
Find the values of determinants without making calculation mistakes. Also learn the formulae of planes formed by the vectors and lines formed by the vectors.
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