OABC is a tetrahedron. The position vectors of A, B and C are `i, i+j and j+k`, respectively. O is origin. The height of the tetrahedron (taking ABC as base) is

Options

A. `(1)/(2)`
B. `(1)/(sqrt(2))`
C. `(1)/(2sqrt(2))`
D. None of these

Answer

To find the height of the tetrahedron OABC, we will follow these steps:

### Step 1: Identify the position vectors
The position vectors of points A, B, and C are given as:
- \( \vec{A} = \hat{i} \)
- \( \vec{B} = \hat{i} + \hat{j} \)
- \( \vec{C} = \hat{j} + \hat{k} \)
- \( \vec{O} = \vec{0} \) (the origin)

### Step 2: Calculate the volume of the tetrahedron
The volume \( V \) of tetrahedron OABC can be calculated using the formula:
\[ V = \frac{1}{6} | \vec{OA} \cdot (\vec{OB} \times \vec{OC}) | \]
where:
- \( \vec{OA} = \vec{A} - \vec{O} = \hat{i} \)
- \( \vec{OB} = \vec{B} - \vec{O} = \hat{i} + \hat{j} \)
- \( \vec{OC} = \vec{C} - \vec{O} = \hat{j} + \hat{k} \)

### Step 3: Calculate the cross product \( \vec{OB} \times \vec{OC} \)
First, we need to compute the cross product \( \vec{OB} \times \vec{OC} \):
\[ \vec{OB} = \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}, \quad \vec{OC} = \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix} \]
Using the determinant to find the cross product:
\[ \vec{OB} \times \vec{OC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 0 \\ 0 & 1 & 1 \end{vmatrix} \]
Calculating this determinant:
\[ = \hat{i} \begin{vmatrix} 1 & 0 \\ 1 & 1 \end{vmatrix} - \hat{j} \begin{vmatrix} 1 & 0 \\ 0 & 1 \end{vmatrix} + \hat{k} \begin{vmatrix} 1 & 1 \\ 0 & 1 \end{vmatrix} \]
\[ = \hat{i}(1 - 0) - \hat{j}(1 - 0) + \hat{k}(1 - 0) = \hat{i} - \hat{j} + \hat{k} \]

### Step 4: Calculate the dot product \( \vec{OA} \cdot (\vec{OB} \times \vec{OC}) \)
Now we compute the dot product:
\[ \vec{OA} \cdot (\vec{OB} \times \vec{OC}) = \hat{i} \cdot (\hat{i} - \hat{j} + \hat{k}) = 1 \]

### Step 5: Calculate the volume \( V \)
Substituting back into the volume formula:
\[ V = \frac{1}{6} |1| = \frac{1}{6} \]

### Step 6: Calculate the area of triangle ABC
The area \( A \) of triangle ABC can be calculated using the formula:
\[ A = \frac{1}{2} | \vec{AB} \times \vec{AC} | \]
where:
- \( \vec{AB} = \vec{B} - \vec{A} = (\hat{i} + \hat{j}) - \hat{i} = \hat{j} \)
- \( \vec{AC} = \vec{C} - \vec{A} = (\hat{j} + \hat{k}) - \hat{i} = -\hat{i} + \hat{j} + \hat{k} \)

Now, calculate the cross product \( \vec{AB} \times \vec{AC} \):
\[ \vec{AB} \times \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 1 & 0 \\ -1 & 1 & 1 \end{vmatrix} \]
Calculating this determinant:
\[ = \hat{i} \begin{vmatrix} 1 & 0 \\ 1 & 1 \end{vmatrix} - \hat{j} \begin{vmatrix} 0 & 0 \\ -1 & 1 \end{vmatrix} + \hat{k} \begin{vmatrix} 0 & 1 \\ -1 & 1 \end{vmatrix} \]
\[ = \hat{i}(1) - \hat{j}(0) + \hat{k}(1) = \hat{i} + \hat{k} \]
The magnitude of this vector is:
\[ |\hat{i} + \hat{k}| = \sqrt{1^2 + 1^2} = \sqrt{2} \]
Thus, the area \( A \) is:
\[ A = \frac{1}{2} \sqrt{2} \]

### Step 7: Calculate the height \( h \) of the tetrahedron
Using the volume and area, we can find the height \( h \):
\[ V = \frac{1}{3} \times A \times h \]
Substituting the known values:
\[ \frac{1}{6} = \frac{1}{3} \times \frac{1}{2} \sqrt{2} \times h \]
Solving for \( h \):
\[ \frac{1}{6} = \frac{\sqrt{2}}{6} h \implies h = 1 \]

### Final Answer
The height of the tetrahedron OABC is \( h = 1 \).