Vectors `hata` in the plane of `vecb = 2 hati +hatj and vecc = hati-hatj + hatk` is such that it is equally inclined to `vecb and vecd " where " vecd= hatj + 2hatk ` the value of `hata ` is (a) `(hati+hatj+hatk)/sqrt3` (b) `(hati-hatj+hatk)/sqrt3` (c) `( 2 hati + hatj)/sqrt5` (d) `( 2 hati + hatj)/sqrt5`

Options

A. `(hati+hatj+hatk)/sqrt3`
B. `(hati-hatj+hatk)/sqrt3`
C. `( 2 hati + hatj)/sqrt5`
D. `( 2 hati + hatj)/sqrt5`

Explanation

To solve the problem step by step, we need to find the vector \( \hat{a} \) that is in the plane of vectors \( \vec{b} \) and \( \vec{c} \), and is equally inclined to vectors \( \vec{b} \) and \( \vec{d} \).

### Step 1: Define the Vectors
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: Express \( \hat{a} \) in terms of \( \vec{b} \) and \( \vec{c} \)
Since \( \hat{a} \) lies in the plane of \( \vec{b} \) and \( \vec{c} \), we can express \( \hat{a} \) as a linear combination of \( \vec{b} \) and \( \vec{c} \):
\[ \hat{a} = \lambda \vec{b} + \mu \vec{c} \]
Substituting the values of \( \vec{b} \) and \( \vec{c} \):
\[ \hat{a} = \lambda (2 \hat{i} + \hat{j}) + \mu (\hat{i} - \hat{j} + \hat{k}) \]
\[ = (2\lambda + \mu) \hat{i} + (\lambda - \mu) \hat{j} + \mu \hat{k} \]

### Step 3: Condition for Equal Inclination
The condition for \( \hat{a} \) to be equally inclined to \( \vec{b} \) and \( \vec{d} \) can be expressed using the dot product:
\[ \frac{\hat{a} \cdot \vec{b}}{|\hat{a}| |\vec{b}|} = \frac{\hat{a} \cdot \vec{d}}{|\hat{a}| |\vec{d}|} \]
This implies:
\[ \hat{a} \cdot \vec{b} = k \hat{a} \cdot \vec{d} \]
for some scalar \( k \).

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

Now calculate \( \hat{a} \cdot \vec{b} \):
\[ \hat{a} \cdot \vec{b} = (2\lambda + \mu)(2) + (\lambda - \mu)(1) = 4\lambda + 2\mu + \lambda - \mu = 5\lambda + \mu \]

Now calculate \( \hat{a} \cdot \vec{d} \):
\[ \hat{a} \cdot \vec{d} = (2\lambda + \mu)(0) + (\lambda - \mu)(1) + \mu(2) = \lambda - \mu + 2\mu = \lambda + \mu \]

### Step 5: Set Up the Equation
Setting the two dot products equal gives:
\[ 5\lambda + \mu = k(\lambda + \mu) \]
Since \( |\hat{a}| \) cancels out, we can simplify this equation.

### Step 6: Solve for \( \lambda \) and \( \mu \)
We can rearrange the equation:
\[ 5\lambda + \mu = k\lambda + k\mu \]
Rearranging gives:
\[ (5 - k)\lambda + (1 - k)\mu = 0 \]
This implies that either \( \lambda = 0 \) or \( \mu \) can be expressed in terms of \( \lambda \).

### Step 7: Find the Unit Vector
Assuming \( \lambda = 0 \), we have:
\[ \hat{a} = \mu(\hat{i} - \hat{j} + \hat{k}) \]
To find the unit vector, we calculate the magnitude:
\[ |\hat{a}| = \sqrt{1^2 + (-1)^2 + 1^2} = \sqrt{3} \]
Thus,
\[ \hat{a} = \frac{\mu}{\sqrt{3}}(\hat{i} - \hat{j} + \hat{k}) \]
Choosing \( \mu = 1 \), we have:
\[ \hat{a} = \frac{1}{\sqrt{3}}(\hat{i} - \hat{j} + \hat{k}) \]

### Final Step: Check Options
Comparing with the options given:
- (a) \( \frac{\hat{i} + \hat{j} + \hat{k}}{\sqrt{3}} \)
- (b) \( \frac{\hat{i} - \hat{j} + \hat{k}}{\sqrt{3}} \)
- (c) \( \frac{2 \hat{i} + \hat{j}}{\sqrt{5}} \)
- (d) \( \frac{2 \hat{i} + \hat{j}}{\sqrt{5}} \)

The correct answer is:
**(b) \( \frac{\hat{i} - \hat{j} + \hat{k}}{\sqrt{3}} \)**