7.5 Q-6

Question Statement

Find the volume of the tetrahedron with the following vertices:

(i) $(0,1,2)$, $(3,2,1)$, $(1,2,1)$, and $(5,5,6)$

(ii) $(2,1,8)$, $(3,2,9)$, $(2,1,4)$, and $(3,3,10)$

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Background and Explanation

The volume of a tetrahedron can be calculated using the scalar triple product. The formula for the volume of a tetrahedron with vertices $A$, $B$, $C$, and $D$ is:

$$V = \frac{1}{6} \left| \vec{AB} \cdot (\vec{AC} \times \vec{AD}) \right|$$

Where:

• $\vec{AB}, \vec{AC}, \vec{AD}$ are vectors formed by subtracting the coordinates of the vertices.
• The cross product of two vectors gives a vector that is perpendicular to both, and the dot product gives the scalar value we need.

The determinant of the matrix formed by the components of these vectors also gives the same result.

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Solution

Part (i)

Given the vertices $A(0,1,2)$, $B(3,2,1)$, $C(1,2,1)$, and $D(5,5,6)$, we first compute the vectors:

1. Vector $\hat{a} = \vec{AB}$:

$$\hat{a} = (5-0) \hat{i} + (2-1) \hat{j} + (6-2) \hat{k} = 5\hat{i} + 4\hat{j} + 4\hat{k}$$

2. Vector $\hat{b} = \vec{AC}$:

$$\hat{b} = (1-0) \hat{i} + (2-1) \hat{j} + (1-2) \hat{k} = \hat{i} + \hat{j} - \hat{k}$$

3. Vector $\hat{c} = \vec{AD}$:

$$\hat{c} = (3-0) \hat{i} + (2-1) \hat{j} + (1-2) \hat{k} = 3\hat{i} + \hat{j} - \hat{k}$$

Next, the volume of the tetrahedron is given by the determinant of the matrix formed by these vectors:

$$V = \frac{1}{6} \left| \begin{array}{ccc} 5 & 4 & 4 1 & 1 & -1 3 & -1 & -1 \end{array} \right|$$

Expand the determinant along the first row:

$$V = \frac{1}{6} \left[ 5 \left( \left| \begin{array}{cc} 1 & -1 -1 & -1 \end{array} \right| \right) - 4 \left( \left| \begin{array}{cc} 1 & -1 3 & -1 \end{array} \right| \right) + 4 \left( \left| \begin{array}{cc} 1 & 1 3 & -1 \end{array} \right| \right) \right]$$

Simplifying the 2x2 determinants:

$$= \frac{1}{6} \left[ 5(0) - 4(2) + 4(-2) \right] = \frac{1}{6} (0 - 8 - 8) = \frac{-16}{6} = \frac{8}{3}$$

Thus, the volume of the tetrahedron for part (i) is $\frac{8}{3}$ cubic units.

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Part (ii)

Given the vertices $A(2,1,8)$, $B(3,2,9)$, $C(2,1,4)$, and $D(3,3,10)$, we compute the vectors:

1. Vector $\hat{a} = \vec{AB}$:

$$\hat{a} = (3-2) \hat{i} + (2-1) \hat{j} + (9-8) \hat{k} = 1\hat{i} + 1\hat{j} + 1\hat{k}$$

2. Vector $\hat{b} = \vec{AC}$:

$$\hat{b} = (2-2) \hat{i} + (1-1) \hat{j} + (4-8) \hat{k} = 0\hat{i} + 0\hat{j} - 4\hat{k}$$

3. Vector $\hat{c} = \vec{AD}$:

$$\hat{c} = (3-2) \hat{i} + (3-1) \hat{j} + (10-8) \hat{k} = 1\hat{i} + 2\hat{j} + 2\hat{k}$$

Next, the volume of the tetrahedron is given by the determinant of the matrix formed by these vectors:

$$V = \frac{1}{6} \left| \begin{array}{ccc} 1 & 1 & 1 0 & 0 & -4 1 & 2 & 2 \end{array} \right|$$

Expanding the determinant:

$$V = \frac{1}{6} \left[ 1 \left( \left| \begin{array}{cc} 0 & -4 2 & 2 \end{array} \right| \right) - 1 \left( \left| \begin{array}{cc} 0 & -4 1 & 2 \end{array} \right| \right) + 1 \left( \left| \begin{array}{cc} 0 & 0 1 & 2 \end{array} \right| \right) \right]$$

Simplifying the 2x2 determinants:

$$= \frac{1}{6} \left[ 1(8) - 1(4) + 1(0) \right] = \frac{1}{6} (8 - 4) = \frac{1}{6} (4) = \frac{2}{3}$$

Thus, the volume of the tetrahedron for part (ii) is $\frac{2}{3}$ cubic units.

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Key Formulas or Methods Used

• Volume of Tetrahedron:

$$V = \frac{1}{6} \left| \vec{AB} \cdot (\vec{AC} \times \vec{AD}) \right|$$

• Determinant of a Matrix: The volume is also calculated using the determinant of a matrix formed by the vectors $\hat{a}, \hat{b}, \hat{c}$.

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Summary of Steps

1. Compute vectors: Find vectors $\hat{a}$, $\hat{b}$, and $\hat{c}$ by subtracting the coordinates of the vertices.
2. Set up the determinant: Use the vectors in the determinant formula.
3. Expand the determinant: Perform the necessary row or column expansions to simplify the determinant.
4. Calculate the volume: Multiply the determinant by $\frac{1}{6}$ and take the absolute value.