Find the altitude of a parallelopiped whose three conterminous edges are verctors `A=hat(i)+hat(j)+hat(k), B=2hat(i)+4hat(j)-hat(k) and C=hat(i)+hat(j)+3hat(k)` with A and B as the sides of the base of the parallelopiped.

Answer

To find the altitude of a parallelepiped whose three conterminous edges are given by the vectors \( \mathbf{A} = \hat{i} + \hat{j} + \hat{k} \), \( \mathbf{B} = 2\hat{i} + 4\hat{j} - \hat{k} \), and \( \mathbf{C} = \hat{i} + \hat{j} + 3\hat{k} \), we will follow these steps:

### Step 1: Calculate the Volume of the Parallelepiped
The volume \( V \) of the parallelepiped can be calculated using the scalar triple product of the vectors \( \mathbf{A} \), \( \mathbf{B} \), and \( \mathbf{C} \):
\[ V = \mathbf{A} \cdot (\mathbf{B} \times \mathbf{C}) \]

First, we need to calculate the cross product \( \mathbf{B} \times \mathbf{C} \).

### Step 2: Compute the Cross Product \( \mathbf{B} \times \mathbf{C} \)
Using the determinant form:
\[ \mathbf{B} \times \mathbf{C} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 4 & -1 \\ 1 & 1 & 3 \end{vmatrix} \]

Calculating the determinant:
\[ = \hat{i} \begin{vmatrix} 4 & -1 \\ 1 & 3 \end{vmatrix} - \hat{j} \begin{vmatrix} 2 & -1 \\ 1 & 3 \end{vmatrix} + \hat{k} \begin{vmatrix} 2 & 4 \\ 1 & 1 \end{vmatrix} \]
Calculating each of the 2x2 determinants:
1. \( \begin{vmatrix} 4 & -1 \\ 1 & 3 \end{vmatrix} = (4 \cdot 3) - (-1 \cdot 1) = 12 + 1 = 13 \)
2. \( \begin{vmatrix} 2 & -1 \\ 1 & 3 \end{vmatrix} = (2 \cdot 3) - (-1 \cdot 1) = 6 + 1 = 7 \)
3. \( \begin{vmatrix} 2 & 4 \\ 1 & 1 \end{vmatrix} = (2 \cdot 1) - (4 \cdot 1) = 2 - 4 = -2 \)

Putting it all together:
\[ \mathbf{B} \times \mathbf{C} = 13\hat{i} - 7\hat{j} - 2\hat{k} \]

### Step 3: Calculate the Dot Product \( \mathbf{A} \cdot (\mathbf{B} \times \mathbf{C}) \)
Now, we compute the dot product:
\[ V = \mathbf{A} \cdot (13\hat{i} - 7\hat{j} - 2\hat{k}) = (1)(13) + (1)(-7) + (1)(-2) = 13 - 7 - 2 = 4 \]

### Step 4: Calculate the Area of the Base
The area \( A \) of the base of the parallelepiped, which is formed by vectors \( \mathbf{A} \) and \( \mathbf{B} \), is given by the magnitude of the cross product \( \mathbf{A} \times \mathbf{B} \):
\[ \mathbf{A} \times \mathbf{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 2 & 4 & -1 \end{vmatrix} \]

Calculating this determinant:
\[ = \hat{i} \begin{vmatrix} 1 & 1 \\ 4 & -1 \end{vmatrix} - \hat{j} \begin{vmatrix} 1 & 1 \\ 2 & -1 \end{vmatrix} + \hat{k} \begin{vmatrix} 1 & 1 \\ 2 & 4 \end{vmatrix} \]
Calculating each of the 2x2 determinants:
1. \( \begin{vmatrix} 1 & 1 \\ 4 & -1 \end{vmatrix} = (1)(-1) - (1)(4) = -1 - 4 = -5 \)
2. \( \begin{vmatrix} 1 & 1 \\ 2 & -1 \end{vmatrix} = (1)(-1) - (1)(2) = -1 - 2 = -3 \)
3. \( \begin{vmatrix} 1 & 1 \\ 2 & 4 \end{vmatrix} = (1)(4) - (1)(2) = 4 - 2 = 2 \)

Putting it all together:
\[ \mathbf{A} \times \mathbf{B} = -5\hat{i} + 3\hat{j} + 2\hat{k} \]

### Step 5: Calculate the Magnitude of the Cross Product
Now, we find the magnitude:
\[ |\mathbf{A} \times \mathbf{B}| = \sqrt{(-5)^2 + 3^2 + 2^2} = \sqrt{25 + 9 + 4} = \sqrt{38} \]

### Step 6: Calculate the Altitude
The altitude \( h \) of the parallelepiped can be found using the formula:
\[ h = \frac{V}{A} \]
Substituting the values we found:
\[ h = \frac{4}{\sqrt{38}} \]

To rationalize the denominator:
\[ h = \frac{4\sqrt{38}}{38} = \frac{2\sqrt{38}}{19} \]

### Final Answer
The altitude of the parallelepiped is:
\[ h = \frac{2\sqrt{38}}{19} \]