IF `bara=hati+hatj+hatk,barb=2hatj-hatk and barr times bara=barb times bara,barr . barb`=0 then `barr/|barr|` is equal to

Options

A. `1/sqrt11(hati+3hatj-hatk)`
B. `1/sqrt11(hati+3hatj+hatk)`
C. `1/sqrt3(hati+hatj+hatk)`
D. None of these

Answer

To solve the problem, we need to find the unit vector \( \frac{\mathbf{r}}{|\mathbf{r}|} \) given the conditions involving vectors \( \mathbf{a} \), \( \mathbf{b} \), and \( \mathbf{r} \).

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

### Step 2: Understand the conditions
We have two conditions:
1. \( \mathbf{r} \times \mathbf{a} = \mathbf{b} \times \mathbf{a} \)
2. \( \mathbf{r} \cdot \mathbf{b} = 0 \)

### Step 3: Calculate \( \mathbf{b} \times \mathbf{a} \)
To find \( \mathbf{b} \times \mathbf{a} \):
\[ \mathbf{b} \times \mathbf{a} = (2\hat{j} - \hat{k}) \times (\hat{i} + \hat{j} + \hat{k}) \]
Using the determinant method for cross products:
\[ \mathbf{b} \times \mathbf{a} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 2 & -1 \\ 1 & 1 & 1 \end{vmatrix} \]
Calculating this determinant:
\[ = \hat{i} \begin{vmatrix} 2 & -1 \\ 1 & 1 \end{vmatrix} - \hat{j} \begin{vmatrix} 0 & -1 \\ 1 & 1 \end{vmatrix} + \hat{k} \begin{vmatrix} 0 & 2 \\ 1 & 1 \end{vmatrix} \]
Calculating each of the 2x2 determinants:
\[ = \hat{i} (2 \cdot 1 - (-1) \cdot 1) - \hat{j} (0 \cdot 1 - (-1) \cdot 1) + \hat{k} (0 \cdot 1 - 2 \cdot 1) \]
\[ = \hat{i} (2 + 1) - \hat{j} (0 + 1) + \hat{k} (0 - 2) \]
\[ = 3\hat{i} - \hat{j} - 2\hat{k} \]

### Step 4: Set \( \mathbf{r} \times \mathbf{a} = \mathbf{b} \times \mathbf{a} \)
This gives us:
\[ \mathbf{r} \times \mathbf{a} = 3\hat{i} - \hat{j} - 2\hat{k} \]

### Step 5: Use the second condition \( \mathbf{r} \cdot \mathbf{b} = 0 \)
Let \( \mathbf{r} = x\hat{i} + y\hat{j} + z\hat{k} \). Then:
\[ \mathbf{r} \cdot \mathbf{b} = (x\hat{i} + y\hat{j} + z\hat{k}) \cdot (2\hat{j} - \hat{k}) = 2y - z = 0 \]
Thus, we have:
\[ z = 2y \]

### Step 6: Substitute \( z \) in \( \mathbf{r} \)
Now, substituting \( z = 2y \) into \( \mathbf{r} \):
\[ \mathbf{r} = x\hat{i} + y\hat{j} + 2y\hat{k} = x\hat{i} + y\hat{j} + 2y\hat{k} \]

### Step 7: Find the magnitude of \( \mathbf{r} \)
The magnitude \( |\mathbf{r}| \) is:
\[ |\mathbf{r}| = \sqrt{x^2 + y^2 + (2y)^2} = \sqrt{x^2 + y^2 + 4y^2} = \sqrt{x^2 + 5y^2} \]

### Step 8: Find the unit vector \( \frac{\mathbf{r}}{|\mathbf{r}|} \)
The unit vector is:
\[ \frac{\mathbf{r}}{|\mathbf{r}|} = \frac{x\hat{i} + y\hat{j} + 2y\hat{k}}{\sqrt{x^2 + 5y^2}} \]

### Step 9: Simplify the unit vector
To express it in terms of \( \hat{i}, \hat{j}, \hat{k} \):
\[ \frac{\mathbf{r}}{|\mathbf{r}|} = \frac{x}{\sqrt{x^2 + 5y^2}}\hat{i} + \frac{y}{\sqrt{x^2 + 5y^2}}\hat{j} + \frac{2y}{\sqrt{x^2 + 5y^2}}\hat{k} \]

### Conclusion
The final expression for the unit vector \( \frac{\mathbf{r}}{|\mathbf{r}|} \) is:
\[ \frac{\mathbf{r}}{|\mathbf{r}|} = \frac{x}{\sqrt{x^2 + 5y^2}}\hat{i} + \frac{y}{\sqrt{x^2 + 5y^2}}\hat{j} + \frac{2y}{\sqrt{x^2 + 5y^2}}\hat{k} \]