
The volume of the parallelepiped whose conterminous edges \[i - j + k,2i - 4j + 5k\] and \[3i - 5j + 2k\] is
A. 4
B. 3
C. 2
D. 8
Answer
162.6k+ views
Hint: Utilize the fact that A=a b gives the area of the parallelogram whose adjacent sides are determined by the vectors \[\overrightarrow a \] and \[\overrightarrow b \]. Find the parallelepiped's base area as a result. Utilize the knowledge that \[V = AH\] determines the volume of a parallelepiped with a base area of ‘A’ and a height of ‘H’. Consequently, ascertain the parallelepiped's volume.
Formula Used:The volume of parallelepiped' can be calculated using the formula
\[V = (a \times b) \cdot c\]
\[[\vec a,\vec b,\vec c] = \left| {\begin{array}{*{20}{c}}{{a_1}}&{{a_2}}&{{a_3}}\\{{b_1}}&{{b_2}}&{{b_3}}\\{{c_1}}&{{c_2}}&{{c_3}}\end{array}} \right|\]
Complete step by step solution:Here, we have been given that the vectors are
\[i - j + k,2i - 4j + 5k\] \[3i - 5j + 2k\]
We are aware that \[A = \left| {\overrightarrow a \times \overrightarrow b } \right|\]gives the area of the parallelogram whose neighboring sides are given by the vectors \[\overrightarrow a \] and \[\overrightarrow b \].
Consequently, the volume of the parallelepiped's base is given by
\[{\rm{V}} = [\overrightarrow {\rm{a}} \overrightarrow {\rm{b}} \overrightarrow {\rm{c}} ]\]
We know that, the matrix format is,
\[[\vec a,\vec b,\vec c] = \left| {\begin{array}{*{20}{c}}{{a_1}}&{{a_2}}&{{a_3}}\\{{b_1}}&{{b_2}}&{{b_3}}\\{{c_1}}&{{c_2}}&{{c_3}}\end{array}} \right|\]
Now, we have to write the given vectors in the question as matrix, we get
\[{\rm{V}} = \left| {\begin{array}{*{20}{l}}1&{ - 1}&1\\2&{ - 4}&5\\3&{ - 5}&2\end{array}} \right|\]
Now, let’s solve the matrix along row, we obtain
\[{\rm{V}} = 1( - 8 + 25) + 1(4 - 15) + 1( - 10 + 12)\]
On adding the terms inside the parentheses from the above equation, we get\[ = 1(17) + 1\left( { - 11} \right) + 1(2)\]
On multiplying each term in the above equation, we get
\[ = 17 - 11 + 2\]
On Adding/ subtracting the terms in above equation, we get
\[{\rm{v = 8}}\]
Therefore, the volume of the parallelepiped whose conterminous edges \[i - j + k,2i - 4j + 5k\] and \[3i - 5j + 2k\] is \[{\rm{v = 8cubic units}}\]
Option ‘D’ is correct
Note: Six parallelogram faces combine to produce a three-dimensional shape known as a parallelepiped. It contains three sets of four parallel, equal edges. It has 8 vertex points. Any tetrahedron that has three coterminous edges with a parallelepiped has a volume that is one-sixth that of the parallelepiped. The nature of the scalar triple product is cyclic. The three vectors are thought to be coplanar if it equals zero. The cross product is distributive over addition and anti-commutative.
Formula Used:The volume of parallelepiped' can be calculated using the formula
\[V = (a \times b) \cdot c\]
\[[\vec a,\vec b,\vec c] = \left| {\begin{array}{*{20}{c}}{{a_1}}&{{a_2}}&{{a_3}}\\{{b_1}}&{{b_2}}&{{b_3}}\\{{c_1}}&{{c_2}}&{{c_3}}\end{array}} \right|\]
Complete step by step solution:Here, we have been given that the vectors are
\[i - j + k,2i - 4j + 5k\] \[3i - 5j + 2k\]
We are aware that \[A = \left| {\overrightarrow a \times \overrightarrow b } \right|\]gives the area of the parallelogram whose neighboring sides are given by the vectors \[\overrightarrow a \] and \[\overrightarrow b \].
Consequently, the volume of the parallelepiped's base is given by
\[{\rm{V}} = [\overrightarrow {\rm{a}} \overrightarrow {\rm{b}} \overrightarrow {\rm{c}} ]\]
We know that, the matrix format is,
\[[\vec a,\vec b,\vec c] = \left| {\begin{array}{*{20}{c}}{{a_1}}&{{a_2}}&{{a_3}}\\{{b_1}}&{{b_2}}&{{b_3}}\\{{c_1}}&{{c_2}}&{{c_3}}\end{array}} \right|\]
Now, we have to write the given vectors in the question as matrix, we get
\[{\rm{V}} = \left| {\begin{array}{*{20}{l}}1&{ - 1}&1\\2&{ - 4}&5\\3&{ - 5}&2\end{array}} \right|\]
Now, let’s solve the matrix along row, we obtain
\[{\rm{V}} = 1( - 8 + 25) + 1(4 - 15) + 1( - 10 + 12)\]
On adding the terms inside the parentheses from the above equation, we get\[ = 1(17) + 1\left( { - 11} \right) + 1(2)\]
On multiplying each term in the above equation, we get
\[ = 17 - 11 + 2\]
On Adding/ subtracting the terms in above equation, we get
\[{\rm{v = 8}}\]
Therefore, the volume of the parallelepiped whose conterminous edges \[i - j + k,2i - 4j + 5k\] and \[3i - 5j + 2k\] is \[{\rm{v = 8cubic units}}\]
Option ‘D’ is correct
Note: Six parallelogram faces combine to produce a three-dimensional shape known as a parallelepiped. It contains three sets of four parallel, equal edges. It has 8 vertex points. Any tetrahedron that has three coterminous edges with a parallelepiped has a volume that is one-sixth that of the parallelepiped. The nature of the scalar triple product is cyclic. The three vectors are thought to be coplanar if it equals zero. The cross product is distributive over addition and anti-commutative.
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