Let `a=i+j+k, c=j-k` and a vector b is such that `a times b=c" and "a*b=3`. Then `abs(b)` equals:

Options

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

Answer

To solve the problem, we need to find the magnitude of vector **b** given the conditions:

1. **a × b = c**
2. **a · b = 3**

Where:
- **a = i + j + k**
- **c = j - k**

Let's denote vector **b** as:
**b = x i + y j + z k**

### Step 1: Set up the cross product equation
We need to calculate the cross product **a × b**.

**a × b = (i + j + k) × (x i + y j + z k)**

Using the determinant method for the cross product, we can write:

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

Calculating this determinant, we get:

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

This simplifies to:

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

### Step 2: Set the cross product equal to vector c
Now we set this equal to vector **c**:

\[ (z - y) \hat{i} - (z - x) \hat{j} + (y - x) \hat{k} = 0 \hat{i} + 1 \hat{j} - 1 \hat{k} \]

From this, we can equate the components:

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

### Step 3: Solve the equations
From equation (1):
\[ z = y \]

Substituting \( z \) in equation (2):
\[ -(y - x) = 1 \implies y - x = -1 \implies y = x - 1 \]

Now substituting \( y \) in equation (3):
\[ (x - 1) - x = -1 \implies -1 = -1 \text{ (True)} \]

Now we can express everything in terms of \( x \):
- \( y = x - 1 \)
- \( z = y = x - 1 \)

### Step 4: Use the dot product condition
Now we use the second condition \( a · b = 3 \):

\[ (i + j + k) · (x i + (x - 1) j + (x - 1) k) = 3 \]

Calculating the dot product, we have:

\[ x + (x - 1) + (x - 1) = 3 \]

This simplifies to:

\[ 3x - 2 = 3 \implies 3x = 5 \implies x = \frac{5}{3} \]

Now substituting back to find \( y \) and \( z \):

\[ y = x - 1 = \frac{5}{3} - 1 = \frac{2}{3} \]
\[ z = y = \frac{2}{3} \]

### Step 5: Write vector b
Now we have:

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

### Step 6: Find the magnitude of vector b
The magnitude of vector **b** is given by:

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

Calculating this:

\[ = \sqrt{\frac{25}{9} + \frac{4}{9} + \frac{4}{9}} = \sqrt{\frac{33}{9}} = \sqrt{\frac{11}{3}} \]

### Final Answer
Thus, the magnitude of vector **b** is:

\[ |b| = \sqrt{\frac{11}{3}} \]