The magnitude of `vec(a)xx vec(b)` if `vec(a)=2hat(i)+hat(k)` and `vec(b)=hat(i)+hat(j)+hat(k)` is ___________.

Answer

To find the magnitude of the cross product \(\vec{a} \times \vec{b}\) where \(\vec{a} = 2\hat{i} + \hat{k}\) and \(\vec{b} = \hat{i} + \hat{j} + \hat{k}\), we can follow these steps:

### Step 1: Write the vectors in component form
\[ \vec{a} = 2\hat{i} + 0\hat{j} + 1\hat{k} \]
\[ \vec{b} = 1\hat{i} + 1\hat{j} + 1\hat{k} \]

### 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 and the components of \(\vec{a}\) and \(\vec{b}\):

\[ \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 0 & 1 \\ 1 & 1 & 1 \end{vmatrix} \]

### Step 3: Calculate the determinant
Using the determinant formula, we expand it as follows:

\[ \vec{a} \times \vec{b} = \hat{i} \begin{vmatrix} 0 & 1 \\ 1 & 1 \end{vmatrix} - \hat{j} \begin{vmatrix} 2 & 1 \\ 1 & 1 \end{vmatrix} + \hat{k} \begin{vmatrix} 2 & 0 \\ 1 & 1 \end{vmatrix} \]

Calculating each of the 2x2 determinants:

1. \(\begin{vmatrix} 0 & 1 \\ 1 & 1 \end{vmatrix} = (0 \cdot 1 - 1 \cdot 1) = -1\)

2. \(\begin{vmatrix} 2 & 1 \\ 1 & 1 \end{vmatrix} = (2 \cdot 1 - 1 \cdot 1) = 1\)

3. \(\begin{vmatrix} 2 & 0 \\ 1 & 1 \end{vmatrix} = (2 \cdot 1 - 0 \cdot 1) = 2\)

Putting it all together:

\[ \vec{a} \times \vec{b} = \hat{i}(-1) - \hat{j}(1) + \hat{k}(2) = -\hat{i} - \hat{j} + 2\hat{k} \]

### Step 4: Write the result in vector form
Thus, we have:
\[ \vec{a} \times \vec{b} = -\hat{i} - \hat{j} + 2\hat{k} \]

### Step 5: Find the magnitude of the cross product
The magnitude of a vector \(\vec{v} = a\hat{i} + b\hat{j} + c\hat{k}\) is given by:
\[ |\vec{v}| = \sqrt{a^2 + b^2 + c^2} \]

For our vector \(-\hat{i} - \hat{j} + 2\hat{k}\):
\[ |\vec{a} \times \vec{b}| = \sqrt{(-1)^2 + (-1)^2 + (2)^2} = \sqrt{1 + 1 + 4} = \sqrt{6} \]

### Final Answer
The magnitude of \(\vec{a} \times \vec{b}\) is \(\sqrt{6}\).

---