Find the altitude of a parallelepiped determined by the vectors `veca, vecb and vecc` if the base is taken as parallelogram detemined by `veca and vecb and if veca = hati+ hatj + hatk, vecb = 2 hati + 4 hatj- hatk and vec c = hati + hatj + 3 hatk.`

Answer

To find the altitude of a parallelepiped determined by the vectors \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\), we will follow these steps:

### Step 1: Identify the vectors
Given:
- \(\vec{a} = \hat{i} + \hat{j} + \hat{k}\)
- \(\vec{b} = 2\hat{i} + 4\hat{j} - \hat{k}\)
- \(\vec{c} = \hat{i} + \hat{j} + 3\hat{k}\)

### Step 2: Calculate the volume of the parallelepiped
The volume \(V\) of the parallelepiped can be calculated using the scalar triple product:
\[ V = \vec{a} \cdot (\vec{b} \times \vec{c}) \]

### Step 3: Calculate \(\vec{b} \times \vec{c}\)
To find \(\vec{b} \times \vec{c}\), we set up the determinant:
\[ \vec{b} \times \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 4 & -1 \\ 1 & 1 & 3 \end{vmatrix} \]

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

### Step 4: Calculate \(\vec{a} \cdot (\vec{b} \times \vec{c})\)
Now we compute the dot product:
\[ V = \vec{a} \cdot (13\hat{i} - 7\hat{j} - 2\hat{k}) = (1\hat{i} + 1\hat{j} + 1\hat{k}) \cdot (13\hat{i} - 7\hat{j} - 2\hat{k}) \]
\[ = 1 \cdot 13 + 1 \cdot (-7) + 1 \cdot (-2) = 13 - 7 - 2 = 4 \]

### Step 5: Calculate the area of the base
The area \(A\) of the base (parallelogram) formed by \(\vec{a}\) and \(\vec{b}\) is given by:
\[ A = |\vec{a} \times \vec{b}| \]

Calculating \(\vec{a} \times \vec{b}\):
\[ \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 2 & 4 & -1 \end{vmatrix} \]
Calculating the determinant:
\[ = \hat{i} \begin{vmatrix} 1 & 1 \\ 4 & -1 \end{vmatrix} - \hat{j} \begin{vmatrix} 1 & 1 \\ 2 & -1 \end{vmatrix} + \hat{k} \begin{vmatrix} 1 & 1 \\ 2 & 4 \end{vmatrix} \]
\[ = \hat{i} (1 \cdot -1 - 1 \cdot 4) - \hat{j} (1 \cdot -1 - 1 \cdot 2) + \hat{k} (1 \cdot 4 - 1 \cdot 2) \]
\[ = \hat{i} (-1 - 4) - \hat{j} (-1 - 2) + \hat{k} (4 - 2) \]
\[ = -5\hat{i} + 3\hat{j} + 2\hat{k} \]

### Step 6: Calculate the magnitude of \(\vec{a} \times \vec{b}\)
\[ A = \sqrt{(-5)^2 + 3^2 + 2^2} = \sqrt{25 + 9 + 4} = \sqrt{38} \]

### Step 7: Calculate the altitude
The altitude \(h\) of the parallelepiped is given by:
\[ h = \frac{V}{A} = \frac{4}{\sqrt{38}} \]

### Final Answer
The altitude of the parallelepiped is:
\[ h = \frac{4}{\sqrt{38}} \text{ units} \]