Find the volume of the tetrahedron having coterminus edges represented by vectors `a=j+k, b=i+k" and "c=i+j`.

Answer

To find the volume of the tetrahedron with coterminus edges represented by the vectors **a**, **b**, and **c**, we can use the formula for the volume of a tetrahedron:

\[ \text{Volume} = \frac{1}{6} | \mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) | \]

Where:
- **a** = \( \mathbf{j} + \mathbf{k} \)
- **b** = \( \mathbf{i} + \mathbf{k} \)
- **c** = \( \mathbf{i} + \mathbf{j} \)

### Step 1: Write the vectors in component form
We can express the vectors as:
- \( \mathbf{a} = (0, 1, 1) \)
- \( \mathbf{b} = (1, 0, 1) \)
- \( \mathbf{c} = (1, 1, 0) \)

### Step 2: Calculate the cross product \( \mathbf{b} \times \mathbf{c} \)
To find \( \mathbf{b} \times \mathbf{c} \), we use the determinant of a matrix formed by the unit vectors and the components of **b** and **c**:

\[ \mathbf{b} \times \mathbf{c} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{vmatrix} \]

Calculating this determinant:

\[ = \mathbf{i} \begin{vmatrix} 0 & 1 \\ 1 & 0 \end{vmatrix} - \mathbf{j} \begin{vmatrix} 1 & 1 \\ 1 & 0 \end{vmatrix} + \mathbf{k} \begin{vmatrix} 1 & 0 \\ 1 & 1 \end{vmatrix} \]

Calculating the 2x2 determinants:

\[ = \mathbf{i} (0 \cdot 0 - 1 \cdot 1) - \mathbf{j} (1 \cdot 0 - 1 \cdot 1) + \mathbf{k} (1 \cdot 1 - 1 \cdot 0) \]

\[ = -\mathbf{i} - (-\mathbf{j}) + \mathbf{k} \]

\[ = -\mathbf{i} + \mathbf{j} + \mathbf{k} \]

So, \( \mathbf{b} \times \mathbf{c} = (-1, 1, 1) \).

### Step 3: Calculate the dot product \( \mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) \)
Now, we compute the dot product:

\[ \mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) = (0, 1, 1) \cdot (-1, 1, 1) \]

Calculating this:

\[ = 0 \cdot (-1) + 1 \cdot 1 + 1 \cdot 1 = 0 + 1 + 1 = 2 \]

### Step 4: Calculate the volume
Now we substitute this result into the volume formula:

\[ \text{Volume} = \frac{1}{6} |2| = \frac{2}{6} = \frac{1}{3} \]

### Final Answer
Thus, the volume of the tetrahedron is:

\[ \text{Volume} = \frac{1}{3} \text{ cubic units} \]