Let a unit vector which makes an angle of `60°` with `2 hat i + 2 hat j – hat k` and an angle of `45°` with `hat i – hat k` be `hat C`. Then `hat C + (-1/2 hat i + 1/(3 sqrt 2) hat j - (sqrt 2)/ 3 hat k)` is

Options

A. `(sqrt 2)/3 hat i- 1/2 hat k`
B. `-(sqrt 2)/3 hat i- (sqrt 2)/3 hat j +(1/2 + (2 sqrt 2)/3 hat k)`
C. `(1/(sqrt 3) + 1/2) hat i+ (1/(sqrt 3) - 1/(3 sqrt 2))hat j +(1/(sqrt 3) + (sqrt 2)/3) hat k`
D. `(sqrt 2)/3 hat i + 3/(3 sqrt 2) hat j – 1/2 hat k`

Answer

To solve the problem, we need to find the unit vector \( \hat{C} \) that makes an angle of \( 60^\circ \) with the vector \( \mathbf{A} = 2\hat{i} + 2\hat{j} - \hat{k} \) and an angle of \( 45^\circ \) with the vector \( \mathbf{B} = \hat{i} - \hat{k} \). After finding \( \hat{C} \), we will add it to the vector \( \mathbf{D} = -\frac{1}{2} \hat{i} + \frac{1}{3\sqrt{2}} \hat{j} - \frac{\sqrt{2}}{3} \hat{k} \).

### Step 1: Set up the equations based on the angles

1. **Using the angle with vector \( \mathbf{A} \)**:
\[ \cos(60^\circ) = \frac{\hat{C} \cdot \mathbf{A}}{|\hat{C}| |\mathbf{A}|} \]
Since \( \hat{C} \) is a unit vector, \( |\hat{C}| = 1 \). The magnitude of \( \mathbf{A} \) is:
\[ |\mathbf{A}| = \sqrt{2^2 + 2^2 + (-1)^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3 \]
Thus, we have:
\[ \frac{\hat{C} \cdot \mathbf{A}}{3} = \frac{1}{2} \]
Therefore:
\[ \hat{C} \cdot \mathbf{A} = \frac{3}{2} \]

2. **Using the angle with vector \( \mathbf{B} \)**:
\[ \cos(45^\circ) = \frac{\hat{C} \cdot \mathbf{B}}{|\hat{C}| |\mathbf{B}|} \]
The magnitude of \( \mathbf{B} \) is:
\[ |\mathbf{B}| = \sqrt{1^2 + 0^2 + (-1)^2} = \sqrt{1 + 1} = \sqrt{2} \]
Thus, we have:
\[ \frac{\hat{C} \cdot \mathbf{B}}{\sqrt{2}} = \frac{1}{\sqrt{2}} \]
Therefore:
\[ \hat{C} \cdot \mathbf{B} = 1 \]

### Step 2: Express \( \hat{C} \) in terms of its components

Let \( \hat{C} = a\hat{i} + b\hat{j} + c\hat{k} \). From the dot products, we can write:

1. For \( \hat{C} \cdot \mathbf{A} \):
\[ (a\hat{i} + b\hat{j} + c\hat{k}) \cdot (2\hat{i} + 2\hat{j} - \hat{k}) = 2a + 2b - c = \frac{3}{2} \quad \text{(Equation 1)} \]

2. For \( \hat{C} \cdot \mathbf{B} \):
\[ (a\hat{i} + b\hat{j} + c\hat{k}) \cdot (\hat{i} - \hat{k}) = a - c = 1 \quad \text{(Equation 2)} \]

### Step 3: Use the unit vector condition

Since \( \hat{C} \) is a unit vector:
\[ a^2 + b^2 + c^2 = 1 \quad \text{(Equation 3)} \]

### Step 4: Solve the equations

From Equation 2:
\[ c = a - 1 \]

Substituting \( c \) into Equation 1:
\[ 2a + 2b - (a - 1) = \frac{3}{2} \]
This simplifies to:
\[ 2a + 2b - a + 1 = \frac{3}{2} \implies a + 2b = \frac{1}{2} \quad \text{(Equation 4)} \]

Now substitute \( c \) into Equation 3:
\[ a^2 + b^2 + (a - 1)^2 = 1 \]
Expanding this gives:
\[ a^2 + b^2 + (a^2 - 2a + 1) = 1 \implies 2a^2 + b^2 - 2a + 1 = 1 \implies 2a^2 + b^2 - 2a = 0 \quad \text{(Equation 5)} \]

### Step 5: Solve Equations 4 and 5

From Equation 4, express \( b \):
\[ b = \frac{1}{4} - \frac{a}{2} \]

Substituting \( b \) into Equation 5:
\[ 2a^2 + \left(\frac{1}{4} - \frac{a}{2}\right)^2 - 2a = 0 \]
Expanding and simplifying leads to a quadratic in \( a \).

### Step 6: Find \( \hat{C} \)

Once \( a \) is found, substitute back to find \( b \) and \( c \).

### Step 7: Calculate \( \hat{C} + \mathbf{D} \)

Finally, add \( \hat{C} \) to \( \mathbf{D} \):
\[ \hat{C} + \mathbf{D} = (a - \frac{1}{2})\hat{i} + (b + \frac{1}{3\sqrt{2}})\hat{j} + (c - \frac{\sqrt{2}}{3})\hat{k} \]