Question

If the four plane faces of a tetrahedron are represented by the equation $\overrightarrow{r} \cdot (l\hat{i}+m\hat{j}) = 0$, $\overrightarrow{r} \cdot (m\hat{j}+n\hat{k}) = 0$, $\overrightarrow{r} \cdot (n\hat{k}+l\hat{i}) = 0$, $\overrightarrow{r} \cdot (l\hat{i}+m\hat{j}+n\hat{k}) = p$, then the volume of the tetrahedron, is :

• A. $\frac{p^3}{6lmn}$
• B. $\frac{2p^3}{3lmn}$
• C. $\frac{3p^3}{lmn}$
• D. $\frac{6p^3}{lmn}$

Answer 1

Answer: (B)

$\frac{2p^3}{3lmn}$

Answer 2

The four given planes are:

Plane 1: $lx + my = 0$
Plane 2: $my + nz = 0$
Plane 3: $lx + nz = 0$
Plane 4: $lx + my + nz = p$

Step 1: Find the Vertices

• Vertex A: Intersection of planes 1, 2, and 3. Solving the system of equations $lx + my = 0$, $my + nz = 0$, and $lx + nz = 0$ yields $x = y = z = 0$. Thus, $A = (0, 0, 0)$.

• Vertex B: Intersection of planes 1, 2, and 4. Solving the system $lx + my = 0$, $my + nz = 0$, and $lx + my + nz = p$ gives $B = (\frac{p}{l}, -\frac{p}{m}, \frac{p}{n})$.

• Vertex C: Intersection of planes 1, 3, and 4. Solving the system $lx + my = 0$, $lx + nz = 0$, and $lx + my + nz = p$ gives $C = (-\frac{p}{l}, \frac{p}{m}, \frac{p}{n})$.

• Vertex D: Intersection of planes 2, 3, and 4. Solving the system $my + nz = 0$, $lx + nz = 0$, and $lx + my + nz = p$ gives $D = (\frac{p}{l}, \frac{p}{m}, -\frac{p}{n})$.

Step 2: Compute the Volume

The volume $V$ of a tetrahedron with vertices $A$, $B$, $C$, and $D$ (with $A$ at the origin) is:

$V = \frac{1}{6} \left| \det(\vec{B},\vec{C},\vec{D}) \right|$

Where:

$\vec{B} = \left(\frac{p}{l}, -\frac{p}{m}, \frac{p}{n}\right)$
$\vec{C} = \left(-\frac{p}{l}, \frac{p}{m}, \frac{p}{n}\right)$
$\vec{D} = \left(\frac{p}{l}, \frac{p}{m}, -\frac{p}{n}\right)$

Factoring out $p$ from each vector:

$\det(\vec{B},\vec{C},\vec{D}) = p^3 \det \begin{pmatrix} \frac{1}{l} & -\frac{1}{m} & \frac{1}{n} \\ -\frac{1}{l} & \frac{1}{m} & \frac{1}{n} \\ \frac{1}{l} & \frac{1}{m} & -\frac{1}{n} \end{pmatrix}$

Let $M$ denote the matrix:

$M = \begin{pmatrix} \frac{1}{l} & -\frac{1}{m} & \frac{1}{n} \\ -\frac{1}{l} & \frac{1}{m} & \frac{1}{n} \\ \frac{1}{l} & \frac{1}{m} & -\frac{1}{n} \end{pmatrix}$

Calculate $\det(M)$ by expansion:

$\det(M) = \frac{1}{l}\,\det\begin{pmatrix}\frac{1}{m} & \frac{1}{n} \\ \frac{1}{m} & -\frac{1}{n}\end{pmatrix} - \left(-\frac{1}{m}\right)\,\det\begin{pmatrix} -\frac{1}{l} & \frac{1}{n} \\ \frac{1}{l} & -\frac{1}{n}\end{pmatrix} + \frac{1}{n}\,\det\begin{pmatrix} -\frac{1}{l} & \frac{1}{m} \\ \frac{1}{l} & \frac{1}{m}\end{pmatrix}$

Evaluate the 2x2 determinants:

1. $\det\begin{pmatrix}\frac{1}{m} & \frac{1}{n} \\ \frac{1}{m} & -\frac{1}{n}\end{pmatrix} = \frac{1}{m}\left(-\frac{1}{n}\right) - \frac{1}{n}\left(\frac{1}{m}\right) = -\frac{2}{m n}.$

2. $\det\begin{pmatrix}-\frac{1}{l} & \frac{1}{n} \\ \frac{1}{l} & -\frac{1}{n}\end{pmatrix} = \left(-\frac{1}{l}\right)\left(-\frac{1}{n}\right) - \left(\frac{1}{n}\right)\left(\frac{1}{l}\right) = \frac{1}{l n} - \frac{1}{l n} = 0.$

3. $\det\begin{pmatrix}-\frac{1}{l} & \frac{1}{m} \\ \frac{1}{l} & \frac{1}{m}\end{pmatrix} = \left(-\frac{1}{l}\right)\left(\frac{1}{m}\right) - \left(\frac{1}{l}\right)\left(\frac{1}{m}\right) = -\frac{2}{l m}.$

Substitute back:

$\det(M) = \frac{1}{l}\left(-\frac{2}{m n}\right) + \frac{1}{m}\cdot 0 + \frac{1}{n}\left(-\frac{2}{l m}\right) = -\frac{2}{l m n} - \frac{2}{l m n} = -\frac{4}{l m n}.$

Taking the absolute value gives:

$\left|\det(M)\right| = \frac{4}{l m n}.$

Thus, the volume is:

$V = \frac{1}{6} \, p^3 \cdot \frac{4}{l m n} = \frac{4p^3}{6l m n} = \frac{2p^3}{3l m n}.$