A unit vector `veca` in the plane of `vecb=2hati+hatj` and `vecc=hati-hatj+hatk` is such that angle between `veca` and `vecd` is same as angle between `veca` and `vecb` where `vecd=vecj+2veck`. Then `veca` is

Options

A. `(veci+vecj+veck)/sqrt(3)`
B. `(veci-vecj+veck)/sqrt(3)`
C. `(2veci+vecj)/sqrt(5)`
D. `(2veci-vecj)/sqrt(5)`

Explanation

To solve the problem, we need to find the unit vector \(\vec{a}\) in the plane of the vectors \(\vec{b}\) and \(\vec{c}\), given the conditions regarding the angles between the vectors. Let's break down the solution step by step.

### Step 1: Identify the vectors
We are given:
- \(\vec{b} = 2\hat{i} + \hat{j}\)
- \(\vec{c} = \hat{i} - \hat{j} + \hat{k}\)
- \(\vec{d} = \hat{j} + 2\hat{k}\)

### Step 2: Establish the relationship of the unit vector \(\vec{a}\)
Since \(\vec{a}\) lies in the plane of \(\vec{b}\) and \(\vec{c}\), we can express \(\vec{a}\) as a linear combination of \(\vec{b}\) and \(\vec{c}\):
\[ \vec{a} = \lambda \vec{b} + \mu \vec{c} \]
where \(\lambda\) and \(\mu\) are scalars.

### Step 3: Use the angle condition
We know that the angle between \(\vec{a}\) and \(\vec{d}\) is the same as the angle between \(\vec{a}\) and \(\vec{b}\). This gives us the equation:
\[ \frac{\vec{a} \cdot \vec{d}}{|\vec{a}| |\vec{d}|} = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|} \]
Since \(|\vec{a}|\) is a unit vector, this simplifies to:
\[ \vec{a} \cdot \vec{d} = \frac{|\vec{d}|}{|\vec{b}|} \vec{a} \cdot \vec{b} \]

### Step 4: Calculate the dot products
First, we need to calculate \(|\vec{b}|\) and \(|\vec{d}|\):
\[ |\vec{b}| = \sqrt{(2^2 + 1^2)} = \sqrt{5} \]
\[ |\vec{d}| = \sqrt{(0^2 + 1^2 + 2^2)} = \sqrt{5} \]

Now, we compute the dot products:
1. \(\vec{b} \cdot \vec{d} = (2)(0) + (1)(1) + (0)(2) = 1\)
2. \(\vec{c} \cdot \vec{d} = (1)(0) + (-1)(1) + (1)(2) = -1 + 2 = 1\)
3. \(\vec{b} \cdot \vec{b} = 5\)

### Step 5: Substitute into the angle equation
Substituting into the angle equation gives:
\[ \vec{a} \cdot \vec{d} = \frac{\sqrt{5}}{\sqrt{5}} \vec{a} \cdot \vec{b} \]
This simplifies to:
\[ \vec{a} \cdot \vec{d} = \vec{a} \cdot \vec{b} \]

### Step 6: Substitute \(\vec{a}\) into the dot products
Substituting \(\vec{a} = \lambda \vec{b} + \mu \vec{c}\) into the dot products:
\[ (\lambda \vec{b} + \mu \vec{c}) \cdot \vec{d} = (\lambda \vec{b} + \mu \vec{c}) \cdot \vec{b} \]
This expands to:
\[ \lambda (\vec{b} \cdot \vec{d}) + \mu (\vec{c} \cdot \vec{d}) = \lambda (\vec{b} \cdot \vec{b}) + \mu (\vec{c} \cdot \vec{b}) \]
Substituting the values we calculated:
\[ \lambda(1) + \mu(1) = \lambda(5) + \mu(1) \]

### Step 7: Solve for \(\lambda\) and \(\mu\)
This gives us:
\[ \lambda + \mu = 5\lambda + \mu \]
Subtracting \(\mu\) from both sides:
\[ \lambda = 5\lambda \]
This implies:
\[ 4\lambda = 0 \implies \lambda = 0 \]

### Step 8: Substitute \(\lambda\) back
Substituting \(\lambda = 0\) into the expression for \(\vec{a}\):
\[ \vec{a} = 0 \cdot \vec{b} + \mu \vec{c} = \mu \vec{c} \]

### Step 9: Normalize \(\vec{a}\)
Since \(\vec{a}\) is a unit vector:
\[ |\vec{a}| = |\mu \vec{c}| = |\mu| |\vec{c}| = 1 \]
Calculating \(|\vec{c}|\):
\[ |\vec{c}| = \sqrt{1^2 + (-1)^2 + 1^2} = \sqrt{3} \]
Thus:
\[ |\mu| \sqrt{3} = 1 \implies |\mu| = \frac{1}{\sqrt{3}} \]

### Step 10: Final expression for \(\vec{a}\)
Choosing \(\mu = \frac{1}{\sqrt{3}}\):
\[ \vec{a} = \frac{1}{\sqrt{3}} (\hat{i} - \hat{j} + \hat{k}) \]

### Conclusion
Thus, the unit vector \(\vec{a}\) is:
\[ \vec{a} = \frac{1}{\sqrt{3}} \hat{i} - \frac{1}{\sqrt{3}} \hat{j} + \frac{1}{\sqrt{3}} \hat{k} \]