Find `|vec(a)xx vec(b)|`, if `vec(a)=2hat(i)+hat(j)+3hat(k)` and `vec(b)=3hat(i)+5hat(j)-2hat(k)`.

Answer

To find the modulus of the cross product of vectors \(\vec{a}\) and \(\vec{b}\), we will follow these steps:

### Step 1: Define the vectors
Given:
\[ \vec{a} = 2\hat{i} + \hat{j} + 3\hat{k} \]
\[ \vec{b} = 3\hat{i} + 5\hat{j} - 2\hat{k} \]

### Step 2: Set up the determinant for the cross product
The cross product \(\vec{a} \times \vec{b}\) can be calculated using the determinant of a matrix formed by the unit vectors \(\hat{i}\), \(\hat{j}\), and \(\hat{k}\) and the components of \(\vec{a}\) and \(\vec{b}\):

\[ \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & 3 \\ 3 & 5 & -2 \end{vmatrix} \]

### Step 3: Calculate the determinant
To find the determinant, we can expand it as follows:

\[ \vec{a} \times \vec{b} = \hat{i} \begin{vmatrix} 1 & 3 \\ 5 & -2 \end{vmatrix} - \hat{j} \begin{vmatrix} 2 & 3 \\ 3 & -2 \end{vmatrix} + \hat{k} \begin{vmatrix} 2 & 1 \\ 3 & 5 \end{vmatrix} \]

Calculating each of these 2x2 determinants:

1. For \(\hat{i}\):
\[ \begin{vmatrix} 1 & 3 \\ 5 & -2 \end{vmatrix} = (1 \cdot -2) - (3 \cdot 5) = -2 - 15 = -17 \]

2. For \(\hat{j}\):
\[ \begin{vmatrix} 2 & 3 \\ 3 & -2 \end{vmatrix} = (2 \cdot -2) - (3 \cdot 3) = -4 - 9 = -13 \]

3. For \(\hat{k}\):
\[ \begin{vmatrix} 2 & 1 \\ 3 & 5 \end{vmatrix} = (2 \cdot 5) - (1 \cdot 3) = 10 - 3 = 7 \]

Putting it all together:
\[ \vec{a} \times \vec{b} = -17\hat{i} + 13\hat{j} + 7\hat{k} \]

### Step 4: Find the modulus of the cross product
The modulus of the vector \(\vec{a} \times \vec{b}\) is given by:
\[ |\vec{a} \times \vec{b}| = \sqrt{(-17)^2 + (13)^2 + (7)^2} \]

Calculating each term:
\[ (-17)^2 = 289, \quad (13)^2 = 169, \quad (7)^2 = 49 \]

Adding these values:
\[ |\vec{a} \times \vec{b}| = \sqrt{289 + 169 + 49} = \sqrt{507} \]

### Step 5: Simplify the modulus
We can simplify \(\sqrt{507}\):
\[ 507 = 3 \times 169 = 3 \times 13^2 \]
Thus,
\[ |\vec{a} \times \vec{b}| = 13\sqrt{3} \]

### Final Answer
The modulus of the cross product \(|\vec{a} \times \vec{b}|\) is:
\[ \boxed{13\sqrt{3}} \]