Question

How do you find the volume of the parallelepiped determined by the vectors: $ < 1,3,7 > $ , $ < 2,1,5 > $ and $ < 3,1,1 > $ ?

Answer 1

Hint: Whenever we are given the three vectors to find the volume then we need to remember one thing that the triple product of three vectors gives us the volume which we require. Therefore, we can make use of the formula which is given by: $Volume = |(a \times b).c|$ , where $a = \{ 1,3,7\} $ , $b = \{ 2,1,5\} $ and $c = \{ 3,1,1\} $ . With these we can find the volume of parallelepiped.

Complete step by step answer:
Here in this question, they have asked to find the volume of the parallelepiped. They have given three vectors to find the volume. So we know that the triple product of the three vectors gives us the volume of a parallelepiped. We have the formula to find the volume which is given by: $Volume = \left| {\left( {\overrightarrow a \times \overrightarrow b } \right).\overrightarrow c } \right|$
From the given question $a = \{ 1,3,7\} $
 $b = \{ 2,1,5\} $
 $c = \{ 3,1,1\} $
The formula $Volume = \left| {\left( {\overrightarrow a \times \overrightarrow b } \right).\overrightarrow c } \right|$ is written as below for simplification purposes.
\[Volume = \left| {\overrightarrow a .\left( {\overrightarrow b \times \overrightarrow c } \right)} \right| = \left| {\begin{array}{*{20}{c}}
  {{a_1}}&{{a_2}}&{{a_3}} \\
  {{b_1}}&{{b_2}}&{{b_3}} \\
  {{c_1}}&{{c_2}}&{{c_3}}
\end{array}} \right|\]
The above determinant form can be further written in simplified version for the easy simplification.
\[\left| {\begin{array}{*{20}{c}}
  {{a_1}}&{{a_2}}&{{a_3}} \\
  {{b_1}}&{{b_2}}&{{b_3}} \\
  {{c_1}}&{{c_2}}&{{c_3}}
\end{array}} \right| = {a_1}\left[ {\left( {{b_2}{c_3}} \right) - \left( {{c_2}{b_3}} \right)} \right] - {a_2}\left[ {\left( {{b_1}{c_3}} \right) - \left( {{c_1}{b_3}} \right)} \right] + {a_3}\left[ {\left( {{b_1}{c_2}} \right) - \left( {{c_1}{b_2}} \right)} \right]\]
Now, substitute the vectors given in the question that is $a = \{ 1,3,7\} $ , $b = \{ 2,1,5\} $ and $c = \{ 3,1,1\} $.
Therefore, we get
$Volume = \left| {\overrightarrow a .\left( {\overrightarrow b \times \overrightarrow c } \right)} \right| = \left| {\begin{array}{*{20}{c}}
  1&3&7 \\
  2&1&5 \\
  3&1&1
\end{array}} \right|$
Now the above step can be simplified as we discussed in the above steps. So we get
$\left| {\begin{array}{*{20}{c}}
  1&3&7 \\
  2&1&5 \\
  3&1&1
\end{array}} \right| = 1\left[ {\left( {1 \times 1} \right) - \left( {1 \times 5} \right)} \right] - 3\left[ {\left( {2 \times 1} \right) - \left( {3 \times 5} \right)} \right] + 7\left[ {\left( {2 \times 1} \right) - \left( {3 \times 1} \right)} \right]$
Now simplify the above expression to get the required volume. Therefore, we get
$ \Rightarrow \left| {\begin{array}{*{20}{c}}
  1&3&7 \\
  2&1&5 \\
  3&1&1
\end{array}} \right| = 1\left[ {1 - 5} \right] - 3\left[ {2 - 15} \right] + 7\left[ {2 - 3} \right]$
$ \Rightarrow \left| {\begin{array}{*{20}{c}}
  1&3&7 \\
  2&1&5 \\
  3&1&1
\end{array}} \right| = 1\left[ { - 4} \right] - 3\left[ { - 13} \right] + 7\left[ { - 1} \right]$
$ \Rightarrow \left| {\begin{array}{*{20}{c}}
  1&3&7 \\
  2&1&5 \\
  3&1&1
\end{array}} \right| = - 4 + 39 - 7$
$ \Rightarrow \left| {\begin{array}{*{20}{c}}
  1&3&7 \\
  2&1&5 \\
  3&1&1
\end{array}} \right| = 28uni{t^3}$

Therefore the volume of parallelepiped is $28uni{t^3}$.

Note: Whenever they ask to find out the volume if we are given with three vectors then only we can find by using the above procedure, if not then we cannot make use of this formula. While simplifying the vectors be careful because we may make some minor mistakes that can lead to wrong answers.