MHT CET · Maths · Vector Algebra
If the position vectors of the vertices \(A, B\) and \(C\) are 6i, \(6 \mathbf{j}\) and \(\mathbf{k}\) respectively w.r.t. origin \(O\), then the volume of the tetrahedron \(O A B C\) is
- A 6
- B 3
- C \(\frac{1}{6}\)
- D \(\frac{1}{3}\)
Answer & Solution
Correct Answer
(A) 6
Step-by-step Solution
Detailed explanation
Given that, the position vectors of the vertices \(A, B\) and \(C\) are
\(
\begin{array}{l}
\mathbf{O A}=6 \mathbf{i}=6 \mathbf{i}+0 \mathbf{j}+0 \mathbf{k} \\
\mathbf{O B}=6 \mathbf{j}=0 \mathbf{i}+6 \mathbf{j}+0 \mathbf{k} \\
\mathbf{O C}=\mathbf{k}=0 \mathbf{i}+0 \mathbf{j}+\mathbf{k}
\end{array}
\)
Now, volume of the tetrahedron
\(
\begin{array}{l}
=\frac{1}{6}[\mathbf{O A} \mathbf{O B} \mathbf{O C}] \\
=\frac{1}{6}\left|\begin{array}{lll}
6 & 0 & 0 \\
0 & 6 & 0 \\
0 & 0 & 1
\end{array}\right| \\
=\frac{1}{6}(6 \times 6 \times 1)=6
\end{array}
\)
\(
\begin{array}{l}
\mathbf{O A}=6 \mathbf{i}=6 \mathbf{i}+0 \mathbf{j}+0 \mathbf{k} \\
\mathbf{O B}=6 \mathbf{j}=0 \mathbf{i}+6 \mathbf{j}+0 \mathbf{k} \\
\mathbf{O C}=\mathbf{k}=0 \mathbf{i}+0 \mathbf{j}+\mathbf{k}
\end{array}
\)
Now, volume of the tetrahedron
\(
\begin{array}{l}
=\frac{1}{6}[\mathbf{O A} \mathbf{O B} \mathbf{O C}] \\
=\frac{1}{6}\left|\begin{array}{lll}
6 & 0 & 0 \\
0 & 6 & 0 \\
0 & 0 & 1
\end{array}\right| \\
=\frac{1}{6}(6 \times 6 \times 1)=6
\end{array}
\)
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