A unit vector in the plane of `hat(i)+2hat(j)+hat(k) and hat(i)+hat(j)+2hat(k)` and perpendicular to `2hat(i)+hat(j)+hat(k)`, is

Options

A. `hat(j)-hat(k)`
B. `(hat(i)+hat(j))/(sqrt(2))`
C. `(hat(j)+hat(k))/(sqrt(2))`
D. `(hat(j)-hat(k))/(sqrt(2))`

Explanation

To solve the problem of finding a unit vector in the plane of the vectors \( \hat{i} + 2\hat{j} + \hat{k} \) and \( \hat{i} + \hat{j} + 2\hat{k} \), and perpendicular to the vector \( 2\hat{i} + \hat{j} + \hat{k} \), we can follow these steps:

### Step 1: Identify the Given Vectors
We have the following vectors:
- \( \mathbf{a} = \hat{i} + 2\hat{j} + \hat{k} \)
- \( \mathbf{b} = \hat{i} + \hat{j} + 2\hat{k} \)
- \( \mathbf{c} = 2\hat{i} + \hat{j} + \hat{k} \)

### Step 2: Find a Vector in the Plane of \( \mathbf{a} \) and \( \mathbf{b} \)
A vector \( \mathbf{v} \) in the plane of \( \mathbf{a} \) and \( \mathbf{b} \) can be expressed as a linear combination:
\[ \mathbf{v} = x(\hat{i} + 2\hat{j} + \hat{k}) + y(\hat{i} + \hat{j} + 2\hat{k}) \]
Expanding this, we get:
\[ \mathbf{v} = (x + y)\hat{i} + (2x + y)\hat{j} + (x + 2y)\hat{k} \]

### Step 3: Set Up the Perpendicular Condition
Since \( \mathbf{v} \) must be perpendicular to \( \mathbf{c} \), we use the dot product:
\[ \mathbf{v} \cdot \mathbf{c} = 0 \]
Calculating the dot product:
\[ (x + y)(2) + (2x + y)(1) + (x + 2y)(1) = 0 \]
This simplifies to:
\[ 2(x + y) + (2x + y) + (x + 2y) = 0 \]
Combining like terms:
\[ 2x + 2y + 2x + y + x + 2y = 0 \implies 5x + 5y = 0 \]
Thus, we have:
\[ x + y = 0 \implies y = -x \]

### Step 4: Substitute Back to Find \( \mathbf{v} \)
Substituting \( y = -x \) into the expression for \( \mathbf{v} \):
\[ \mathbf{v} = x(\hat{i} + 2\hat{j} + \hat{k}) - x(\hat{i} + \hat{j} + 2\hat{k}) \]
This simplifies to:
\[ \mathbf{v} = x\left[(\hat{i} - \hat{i}) + (2\hat{j} - \hat{j}) + (\hat{k} - 2\hat{k})\right] = x(0 + \hat{j} - \hat{k}) = x(\hat{j} - \hat{k}) \]

### Step 5: Find the Unit Vector
To find the unit vector, we need to normalize \( \mathbf{v} \):
\[ \text{Magnitude of } \mathbf{v} = |x| \sqrt{1^2 + (-1)^2} = |x| \sqrt{2} \]
Thus, the unit vector \( \mathbf{u} \) is given by:
\[ \mathbf{u} = \frac{\mathbf{v}}{|\mathbf{v}|} = \frac{x(\hat{j} - \hat{k})}{|x|\sqrt{2}} = \frac{\hat{j} - \hat{k}}{\sqrt{2}} \]

### Final Answer
The unit vector is:
\[ \mathbf{u} = \frac{\hat{j} - \hat{k}}{\sqrt{2}} \]