Let `veca = hati + hatj +hatjk, vecc =hatj - hatk` and a vector `vecb` be such that `veca xx vecb = vecc` and `veca.vecb=3`. Then `|vecb|` equals:

Options

A. `sqrt(11/3)`
B. `11/sqrt(3)`
C. `sqrt(11)/3`
D. `11/3`

Answer

To solve the problem, we will follow these steps:

### Step 1: Define the vectors
Let \(\vec{a} = \hat{i} + \hat{j} + \hat{k}\) and \(\vec{c} = \hat{j} - \hat{k}\). We need to find \(\vec{b}\) such that \(\vec{a} \times \vec{b} = \vec{c}\) and \(\vec{a} \cdot \vec{b} = 3\).

### Step 2: Assume the form of vector \(\vec{b}\)
Let \(\vec{b} = x \hat{i} + y \hat{j} + z \hat{k}\).

### Step 3: Compute the cross product \(\vec{a} \times \vec{b}\)
Using the determinant form for the cross product, we have:

\[ \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ x & y & z \end{vmatrix} \]

Calculating this determinant:

\[ = \hat{i} \begin{vmatrix} 1 & 1 \\ y & z \end{vmatrix} - \hat{j} \begin{vmatrix} 1 & 1 \\ x & z \end{vmatrix} + \hat{k} \begin{vmatrix} 1 & 1 \\ x & y \end{vmatrix} \]

Calculating the minors:

\[ = \hat{i} (1z - 1y) - \hat{j} (1z - 1x) + \hat{k} (1y - 1x) \]

This simplifies to:

\[ = (z - y) \hat{i} - (z - x) \hat{j} + (y - x) \hat{k} \]

### Step 4: Set the cross product equal to \(\vec{c}\)
We have:

\[ (z - y) \hat{i} - (z - x) \hat{j} + (y - x) \hat{k} = \hat{j} - \hat{k} \]

From this, we can equate the coefficients:

1. \(z - y = 0\) (1)
2. \(- (z - x) = 1\) (2)
3. \(y - x = -1\) (3)

### Step 5: Solve the equations
From equation (1), we have:

\[ z = y \]

Substituting \(z = y\) into equation (2):

\[ - (y - x) = 1 \implies y - x = -1 \implies x = y + 1 \]

Now substituting \(x = y + 1\) into equation (3):

\[ y - (y + 1) = -1 \implies -1 = -1 \quad \text{(which is true)} \]

### Step 6: Find values of \(x\), \(y\), and \(z\)
Now we can express everything in terms of \(y\):

1. \(z = y\)
2. \(x = y + 1\)

### Step 7: Use the dot product condition
Using the dot product condition \(\vec{a} \cdot \vec{b} = 3\):

\[ (1)(x) + (1)(y) + (1)(z) = 3 \]

Substituting \(x\) and \(z\):

\[ (y + 1) + y + y = 3 \implies 3y + 1 = 3 \implies 3y = 2 \implies y = \frac{2}{3} \]

Thus:

\[ z = \frac{2}{3}, \quad x = \frac{2}{3} + 1 = \frac{5}{3} \]

### Step 8: Write vector \(\vec{b}\)
Now we have:

\[ \vec{b} = \frac{5}{3} \hat{i} + \frac{2}{3} \hat{j} + \frac{2}{3} \hat{k} \]

### Step 9: Calculate the modulus of \(\vec{b}\)
The modulus of \(\vec{b}\) is given by:

\[ |\vec{b}| = \sqrt{\left(\frac{5}{3}\right)^2 + \left(\frac{2}{3}\right)^2 + \left(\frac{2}{3}\right)^2} \]

Calculating this:

\[ = \sqrt{\frac{25}{9} + \frac{4}{9} + \frac{4}{9}} = \sqrt{\frac{33}{9}} = \frac{\sqrt{33}}{3} \]

### Final Answer
Thus, the modulus of \(\vec{b}\) is:

\[ |\vec{b}| = \frac{\sqrt{33}}{3} \]