Question

To find: The volume of the parallelepiped determined by the vectors a, b and c.

Short answer

The volume of the parallelepiped determined by the vectors a, b and \(c\) is \(9\) cubic units.

Step-by-step solution

Concept of parallelepiped with the help of Formula’s

Consider two three-dimensional vectors such as,

\(\begin{array}{l}a = \left\langle {{a_1},{a_2},{a_3}} \right\rangle \\{\rm{b}} = \left\langle {{b_1},{b_2},{b_3}} \right\rangle \end{array}\)

Write the expression for cross product between\(a\)and\({\rm{b}}\)vectors.

\(a \times b = \left| {\begin{array}{*{20}{c}}{\rm{i}}&{\rm{j}}&{\rm{k}}\\{{a_1}}&{{a_2}}&{{a_3}}\\{{b_1}}&{{b_2}}&{{b_3}}\end{array}} \right|(1)\)

Write the expression for dot product between\(a\)and\({\rm{b}}\)vectors.

\(a \cdot b = {a_1}{b_1} + {a_2}{b_2} + {a_3}{b_3}(2)\)

Write the expression to calculate the parallelepiped's volume.

\(V = |{\rm{a}} \cdot ({\rm{b}} \times {\rm{c}})|(3)\)

Volume of parallelepiped by calculating through Formula.

\(a = (1,2,3),b( - 1,1,2){\rm{ and }}c = (2,1,4)\)

Modify equation (\(1\)).

\(b \times c = \left| {\begin{array}{*{20}{c}}{\rm{i}}&{\rm{j}}&{\rm{k}}\\{{b_1}}&{{b_2}}&{{b_3}}\\{{c_1}}&{{c_2}}&{{c_3}}\end{array}} \right|\)

We can substitute \( - 1\) for \({b_1},1\) for \({b_2},2\) for \({b_3},2\) for \({c_1},1\) for \({c_2}\) and \(4\) for \({c_3}\),

\(\begin{array}{l}b \times c = \left| {\begin{array}{*{20}{c}}{\rm{i}}&{\rm{j}}&{\rm{k}}\\{ - 1}&1&2\\2&1&4\end{array}} \right|\\b \times c = \left| {\begin{array}{*{20}{l}}1&2\\1&4\end{array}} \right|{\rm{i}} - \left| {\begin{array}{*{20}{l}}{ - 1}&2\\2&4\end{array}} \right|{\rm{j}} + \left| {\begin{array}{*{20}{l}}{ - 1}&1\\2&1\end{array}} \right|{\rm{k}}\\b \times c = (4 - 2){\rm{i}} - ( - 4 - 4){\rm{j}} + ( - 1 - 2){\rm{k}}\\b \times c = 2{\rm{i}} + 8{\rm{j}} - 3{\rm{k}}\end{array}\)

Modify equation \(\left( 2 \right)\)

\(a \cdot (b \times c) = {a_1}\left( {{b_1} \times {c_1}} \right) + {a_2}\left( {{b_2} \times {c_2}} \right) + {a_3}\left( {{b_3} \times {c_3}} \right)\)

We are substituting \(1\) for \({a_1},2\) for \({a_2},3\) for \({a_3},2\) for \(\left( {{b_1} \times {c_1}} \right),8\) for \(\left( {{b_2} \times {c_2}} \right)\) and \( - 3\) for \(\left( {{b_3} \times {c_3}} \right)\),

\(\begin{array}{l}a \cdot (b \times c) = 1(2) + 2(8) + 3( - 3)\\a \cdot (b \times c) = 2 + 16 - 9\\a \cdot (b \times c) = 9\end{array}\)

We are substituting \(9\) for \(a \cdot (b \times c)\) in equation \((3)\)

\(\begin{array}{l}V = |9|\\V = 9\end{array}\)

Thus, the volume of the parallelepiped determined by the vectors \(a\), \(b\) and \(c\) is \(9\) cubic units.