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 solve the problem, we need to find the vector \( \vec{b} \) given the vector \( \vec{a} = 2\hat{i} + 3\hat{j} + 4\hat{k} \), the dot product \( \vec{a} \cdot \vec{b} = 2 \), and the cross product \( \vec{a} \times \vec{b} = 2\hat{i} - \hat{k} \).

### Step 1: Define the vector \( \vec{b} \)
Let \( \vec{b} = x\hat{i} + y\hat{j} + z\hat{k} \).

### Step 2: Use the dot product
The dot product \( \vec{a} \cdot \vec{b} \) can be calculated as follows:
\[ \vec{a} \cdot \vec{b} = (2\hat{i} + 3\hat{j} + 4\hat{k}) \cdot (x\hat{i} + y\hat{j} + z\hat{k}) = 2x + 3y + 4z \]
Given that \( \vec{a} \cdot \vec{b} = 2 \), we have:
\[ 2x + 3y + 4z = 2 \quad \text{(Equation 1)} \]

### Step 3: Use the cross product
The cross product \( \vec{a} \times \vec{b} \) can be calculated using the determinant:
\[ \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 4 \\ x & y & z \end{vmatrix} \]
Calculating this determinant, we get:
\[ \vec{a} \times \vec{b} = \hat{i}(3z - 4y) - \hat{j}(2z - 4x) + \hat{k}(2y - 3x) \]
Given that \( \vec{a} \times \vec{b} = 2\hat{i} - \hat{k} \), we can equate the coefficients:
1. For \( \hat{i} \): \( 3z - 4y = 2 \) (Equation 2)
2. For \( \hat{j} \): \( 2z - 4x = 0 \) (Equation 3)
3. For \( \hat{k} \): \( 2y - 3x = -1 \) (Equation 4)

### Step 4: Solve the equations
From Equation 3, we can express \( z \) in terms of \( x \):
\[ 2z = 4x \implies z = 2x \]

Substituting \( z = 2x \) into Equation 2:
\[ 3(2x) - 4y = 2 \implies 6x - 4y = 2 \implies 3x - 2y = 1 \quad \text{(Equation 5)} \]

Now we have two equations:
1. \( 2x + 3y + 4(2x) = 2 \) (from Equation 1)
2. \( 3x - 2y = 1 \) (from Equation 5)

Substituting \( z = 2x \) into Equation 1:
\[ 2x + 3y + 8x = 2 \implies 10x + 3y = 2 \quad \text{(Equation 6)} \]

### Step 5: Solve Equations 5 and 6
We now have:
1. \( 10x + 3y = 2 \) (Equation 6)
2. \( 3x - 2y = 1 \) (Equation 5)

Let's multiply Equation 5 by 3:
\[ 9x - 6y = 3 \quad \text{(Equation 7)} \]
Now, we can multiply Equation 6 by 2:
\[ 20x + 6y = 4 \quad \text{(Equation 8)} \]

Adding Equations 7 and 8:
\[ (9x - 6y) + (20x + 6y) = 3 + 4 \implies 29x = 7 \implies x = \frac{7}{29} \]

### Step 6: Find \( y \) and \( z \)
Substituting \( x \) back into Equation 5:
\[ 3\left(\frac{7}{29}\right) - 2y = 1 \implies \frac{21}{29} - 2y = 1 \implies -2y = 1 - \frac{21}{29} = \frac{29 - 21}{29} = \frac{8}{29} \implies y = -\frac{4}{29} \]

Now, substituting \( x \) back to find \( z \):
\[ z = 2x = 2\left(\frac{7}{29}\right) = \frac{14}{29} \]

### Step 7: Write the vector \( \vec{b} \)
Thus, we have:
\[ \vec{b} = x\hat{i} + y\hat{j} + z\hat{k} = \frac{7}{29}\hat{i} - \frac{4}{29}\hat{j} + \frac{14}{29}\hat{k} \]
Factoring out \( \frac{1}{29} \):
\[ \vec{b} = \frac{1}{29}(7\hat{i} - 4\hat{j} + 14\hat{k}) \]

### Final Answer
The vector \( \vec{b} \) is:
\[ \vec{b} = \frac{1}{29}(7\hat{i} - 4\hat{j} + 14\hat{k}) \]