MHT CET · Maths · Vector Algebra
If \(\mathrm{A}(-1,2,3), \mathrm{B}(3,-2,1), \mathrm{C}(2,1,3)\) and \(\mathrm{D}(-1,-2,4)\) are the vertices of a tetrahedron, then its volume is
- A \(\frac{16}{3}\) Cu. units
- B \(\frac{13}{6}\) cu. units
- C \(\frac{16}{31}\) CU. units
- D \(\frac{31}{6}\) cu. units
Answer & Solution
Correct Answer
(A) \(\frac{16}{3}\) Cu. units
Step-by-step Solution
Detailed explanation
We have \(\overline{A B}=4 \hat{i}-4 \hat{j}-2 \hat{k}, \overline{A C}=3 \hat{i}-\hat{j}\) and \(\overline{A D}=-4 \bar{j}+\bar{k}\)
Volume of tetrahedron \(=\frac{1}{6}[\overline{\mathrm{AB}} \overline{\mathrm{AC}} \overline{\mathrm{AD}}]\)
\(
\begin{array}{l}
=\frac{1}{6}\left|\begin{array}{ccc}
4 & -4 & -2 \\
3 & -1 & 0 \\
0 & -4 & 1
\end{array}\right| \end{array}\)
\(=\frac{1}{6}[4(-1)+4(3)-2(-12)]=\frac{1}{6}(-4+12+24)=\) \(\frac{32}{6}=\frac{16}{3}\)
Volume of tetrahedron \(=\frac{1}{6}[\overline{\mathrm{AB}} \overline{\mathrm{AC}} \overline{\mathrm{AD}}]\)
\(
\begin{array}{l}
=\frac{1}{6}\left|\begin{array}{ccc}
4 & -4 & -2 \\
3 & -1 & 0 \\
0 & -4 & 1
\end{array}\right| \end{array}\)
\(=\frac{1}{6}[4(-1)+4(3)-2(-12)]=\frac{1}{6}(-4+12+24)=\) \(\frac{32}{6}=\frac{16}{3}\)
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