Question

Find the volume of parallelepiped whose coterminous edges are given by vectors \[2\vec{i}+3\vec{j}-4\vec{k}\], \[5\vec{i}+7\vec{j}+5\vec{k}\] and \[4\vec{i}+4\vec{j}-2\vec{k}\].

Answer 1

Hint: Now let us compare \[a\vec{i}+b\vec{j}+c\vec{k}\] with \[2\vec{i}+3\vec{j}-4\vec{k}\], then we get the values a, b and c. Now let us compare \[d\vec{i}+e\vec{j}+f\vec{k}\] with \[5\vec{i}+7\vec{j}+5\vec{k}\], then we get the values of d, e and f. Now let us compare \[g\vec{i}+h\vec{j}+i\vec{k}\] with \[4\vec{i}+4\vec{j}-2\vec{k}\], then we get the values of g, h and i. Let us assume the volume of parallelepiped is equal to V.

Complete step by step answer:
We know that the volume of parallelepiped whose coterminous edges are \[a\vec{i}+b\vec{j}+c\vec{k}\], \[d\vec{i}+e\vec{j}+f\vec{k}\] and \[g\vec{i}+h\vec{j}+i\vec{k}\] is given by \[\left| \begin{matrix}
   a & b & c \\
   d & e & f \\
   g & h & h \\
\end{matrix} \right|\]. Now we should substitute the values of a, b, c, d, e, f, g, h and I in \[\left| \begin{matrix}
   a & b & c \\
   d & e & f \\
   g & h & h \\
\end{matrix} \right|\]. This gives us the volume of parallelepiped whose coterminous edges are given by vectors \[2\vec{i}+3\vec{j}-4\vec{k}\], \[5\vec{i}+7\vec{j}+5\vec{k}\] and \[4\vec{i}+4\vec{j}-2\vec{k}\].



Before solving the question, we should know that the volume of parallelepiped whose coterminous edges are \[a\vec{i}+b\vec{j}+c\vec{k}\], \[d\vec{i}+e\vec{j}+f\vec{k}\] and \[g\vec{i}+h\vec{j}+i\vec{k}\] is given by \[\left| \begin{matrix}
   a & b & c \\
   d & e & f \\
   g & h & h \\
\end{matrix} \right|\].
From the question, it is clear that we have to find the volume of parallelepiped whose coterminous edges are given by vectors \[2\vec{i}+3\vec{j}-4\vec{k}\], \[5\vec{i}+7\vec{j}+5\vec{k}\] and \[4\vec{i}+4\vec{j}-2\vec{k}\].
Now let us compare \[a\vec{i}+b\vec{j}+c\vec{k}\] with \[2\vec{i}+3\vec{j}-4\vec{k}\], then we get
\[\begin{align}
  & a=2....(1) \\
 & b=3.....(2) \\
 & c=-4....(3) \\
\end{align}\]
Now let us compare \[d\vec{i}+e\vec{j}+f\vec{k}\] with \[5\vec{i}+7\vec{j}+5\vec{k}\], then we get
\[\begin{align}
  & d=5....(4) \\
 & e=7.....(5) \\
 & f=5....(6) \\
\end{align}\]
Now let us compare \[g\vec{i}+h\vec{j}+i\vec{k}\] with \[4\vec{i}+4\vec{j}-2\vec{k}\], then we get
\[\begin{align}
  & g=4....(7) \\
 & h=4.....(8) \\
 & i=-2....(9) \\
\end{align}\]

Let us assume the volume of the parallelepiped is equal to V.
We should know that the volume of parallelepiped whose coterminous edges are \[a\vec{i}+b\vec{j}+c\vec{k}\], \[d\vec{i}+e\vec{j}+f\vec{k}\] and \[g\vec{i}+h\vec{j}+i\vec{k}\] is given by \[\left| \begin{matrix}
   a & b & c \\
   d & e & f \\
   g & h & h \\
\end{matrix} \right|\].
Then, we can write
\[V=\left| \begin{matrix}
   a & b & c \\
   d & e & f \\
   g & h & h \\
\end{matrix} \right|....(10)\]
Now we have to calculate the value of V.
Now we will substitute equation (1), equation (2), equation (3), equation (4), equation (5), equation (6), equation (7), equation (8) and equation (9) in equation (10).
\[\begin{align}
  & \Rightarrow V=\left| \begin{matrix}
   2 & 3 & -4 \\
   5 & 7 & 5 \\
   4 & 4 & -2 \\
\end{matrix} \right| \\
 & \Rightarrow V=2((7)(-2)-(5)(4))-3((5)(-2)-(5)(4))-4((5)(4)-(7)(4)) \\
 & \Rightarrow V=2(-14-20)-3(-10-20)-4(20-28) \\
 & \Rightarrow V=2(-34)-3(-30)-4(-8) \\
 & \Rightarrow V=-68+60+32 \\
 & \Rightarrow V=24.....(11) \\
\end{align}\]
So, from equation (11) we get the value of V is equal to 24.
So, the volume of parallelepiped whose coterminous edges are given by vectors \[2\vec{i}+3\vec{j}-4\vec{k}\], \[5\vec{i}+7\vec{j}+5\vec{k}\] and \[4\vec{i}+4\vec{j}-2\vec{k}\] is equal to 24.

Note:
 Students may go wrong while calculating the determinant of \[\left| \begin{matrix}
   a & b & c \\
   d & e & f \\
   g & h & h \\
\end{matrix} \right|\].
We know that
\[\left| \begin{matrix}
   a & b & c \\
   d & e & f \\
   g & h & h \\
\end{matrix} \right|=a(ei-fh)-b(di-fg)+c(dh-gh)\].
But some students have a misconception that
\[\left| \begin{matrix}
   a & b & c \\
   d & e & f \\
   g & h & h \\
\end{matrix} \right|=a(ei-fh)+b(di-fg)+c(dh-gh)\].
This will give us a wrong result. So, this type of misconception should be avoided.