If `veca=2hati+3hatj+4hatk,veca.vecb=2 and vecaxxvecb=2hati-hatk`, then `vecb` is

Options

A. `(hati-2hatj+hatk)`
B. `(4hati-4hatj+2hatk)`
C. `1/2(3hati+7hatj+9hatk)`
D. `1/29(7hati-4hatj+14hatk)`

Answer

To find the vector \(\vec{b}\) given the conditions, we can follow these steps:

### Step 1: Define the vectors
We are given:
\[ \vec{a} = 2\hat{i} + 3\hat{j} + 4\hat{k} \]
Let us assume:
\[ \vec{b} = x\hat{i} + y\hat{j} + z\hat{k} \]

### Step 2: Use the dot product condition
The dot product \(\vec{a} \cdot \vec{b} = 2\):
\[ \vec{a} \cdot \vec{b} = (2\hat{i} + 3\hat{j} + 4\hat{k}) \cdot (x\hat{i} + y\hat{j} + z\hat{k}) = 2x + 3y + 4z = 2 \]
This gives us our first equation:
\[ 2x + 3y + 4z = 2 \quad \text{(1)} \]

### Step 3: Use the cross product condition
The cross product \(\vec{a} \times \vec{b} = 2\hat{j} - \hat{k}\):
\[ \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 4 \\ x & y & z \end{vmatrix} \]
Calculating the determinant:
\[ \vec{a} \times \vec{b} = \hat{i}(3z - 4y) - \hat{j}(2z - 4x) + \hat{k}(2y - 3x) \]
Setting this equal to \(2\hat{j} - \hat{k}\), we equate coefficients:
1. Coefficient of \(\hat{i}\): \(3z - 4y = 0\) \quad \text{(2)}
2. Coefficient of \(\hat{j}\): \(- (2z - 4x) = 2 \Rightarrow 2z - 4x = -2\) \quad \text{(3)}
3. Coefficient of \(\hat{k}\): \(2y - 3x = -1\) \quad \text{(4)}

### Step 4: Solve the equations
From equation (2):
\[ 3z = 4y \Rightarrow z = \frac{4y}{3} \quad \text{(5)} \]

Substituting (5) into equation (3):
\[ 2\left(\frac{4y}{3}\right) - 4x = -2 \]
Multiplying through by 3 to eliminate the fraction:
\[ 8y - 12x = -6 \Rightarrow 12x - 8y = 6 \quad \text{(6)} \]

Now substituting (5) into equation (1):
\[ 2x + 3y + 4\left(\frac{4y}{3}\right) = 2 \]
This simplifies to:
\[ 2x + 3y + \frac{16y}{3} = 2 \]
Multiplying through by 3:
\[ 6x + 9y + 16y = 6 \Rightarrow 6x + 25y = 6 \quad \text{(7)} \]

### Step 5: Solve equations (6) and (7)
We have:
1. \(12x - 8y = 6\) (from equation (6))
2. \(6x + 25y = 6\) (from equation (7))

From equation (6), express \(x\) in terms of \(y\):
\[ x = \frac{8y + 6}{12} = \frac{2y + \frac{3}{2}}{3} \quad \text{(8)} \]

Substituting (8) into (7):
\[ 6\left(\frac{2y + \frac{3}{2}}{3}\right) + 25y = 6 \]
This simplifies to:
\[ 4y + 3 + 25y = 6 \Rightarrow 29y = 3 \Rightarrow y = \frac{3}{29} \]

### Step 6: Find \(x\) and \(z\)
Substituting \(y\) back into (8):
\[ x = \frac{2\left(\frac{3}{29}\right) + \frac{3}{2}}{3} = \frac{\frac{6}{29} + \frac{3}{2}}{3} \]
Finding a common denominator:
\[ x = \frac{\frac{6}{29} + \frac{43.5}{29}}{3} = \frac{\frac{49.5}{29}}{3} = \frac{49.5}{87} = \frac{7}{29} \]

Now using (5) to find \(z\):
\[ z = \frac{4\left(\frac{3}{29}\right)}{3} = \frac{4}{29} \]

### Final vector \(\vec{b}\)
Thus, we have:
\[ \vec{b} = \frac{7}{29}\hat{i} + \frac{3}{29}\hat{j} + \frac{4}{29}\hat{k} \]

### Conclusion
The final answer for vector \(\vec{b}\) is:
\[ \vec{b} = \frac{1}{29}(7\hat{i} + 3\hat{j} + 4\hat{k}) \]