Let `veca -2hati + hatj +hatk , vecb =hati + 2hatj +hatk and vecc = 2hati -3hatj +4hatk . A " vector " vecr " satisfying " vecr xx vecb = vecc xx vecb and vecr . Veca =0` is

Options

A. ` -2hati =2hatj +2hatk`
B. ` -2hati +hatj +3hatk`
C. ` 3hati+2hatj +4hatk`
D. ` hati - 5hatj +3hatk`

Explanation

To solve the problem step by step, we need to find a vector \(\vec{r}\) that satisfies the conditions \(\vec{r} \times \vec{b} = \vec{c} \times \vec{b}\) and \(\vec{r} \cdot \vec{a} = 0\).

### Step 1: Write down the given vectors
We have:
\[ \vec{a} = -2\hat{i} + \hat{j} + \hat{k} \]
\[ \vec{b} = \hat{i} + 2\hat{j} + \hat{k} \]
\[ \vec{c} = 2\hat{i} - 3\hat{j} + 4\hat{k} \]

### Step 2: Calculate \(\vec{c} \times \vec{b}\)
To find \(\vec{c} \times \vec{b}\), we can use the determinant of a matrix formed by the unit vectors and the components of \(\vec{c}\) and \(\vec{b}\):
\[ \vec{c} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -3 & 4 \\ 1 & 2 & 1 \end{vmatrix} \]

Calculating the determinant:
\[ = \hat{i} \begin{vmatrix} -3 & 4 \\ 2 & 1 \end{vmatrix} - \hat{j} \begin{vmatrix} 2 & 4 \\ 1 & 1 \end{vmatrix} + \hat{k} \begin{vmatrix} 2 & -3 \\ 1 & 2 \end{vmatrix} \]
Calculating each of these 2x2 determinants:
\[ = \hat{i}((-3)(1) - (4)(2)) - \hat{j}((2)(1) - (4)(1)) + \hat{k}((2)(2) - (-3)(1)) \]
\[ = \hat{i}(-3 - 8) - \hat{j}(2 - 4) + \hat{k}(4 + 3) \]
\[ = -11\hat{i} + 2\hat{j} + 7\hat{k} \]

### Step 3: Set up the equation \(\vec{r} \times \vec{b} = \vec{c} \times \vec{b}\)
Now we have:
\[ \vec{r} \times \vec{b} = -11\hat{i} + 2\hat{j} + 7\hat{k} \]

### Step 4: Use the condition \(\vec{r} \cdot \vec{a} = 0\)
This means that \(\vec{r}\) is perpendicular to \(\vec{a}\). If we let:
\[ \vec{r} = x\hat{i} + y\hat{j} + z\hat{k} \]
Then:
\[ \vec{r} \cdot \vec{a} = (x)(-2) + (y)(1) + (z)(1) = 0 \]
This simplifies to:
\[ -2x + y + z = 0 \quad \text{(Equation 1)} \]

### Step 5: Solve for \(\vec{r}\)
We can express \(z\) in terms of \(x\) and \(y\) using Equation 1:
\[ z = 2x - y \]

### Step 6: Substitute \(\vec{r}\) into the cross product equation
We need to find \(\vec{r} \times \vec{b}\):
\[ \vec{r} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ x & y & z \\ 1 & 2 & 1 \end{vmatrix} \]
Calculating this determinant:
\[ = \hat{i} \begin{vmatrix} y & z \\ 2 & 1 \end{vmatrix} - \hat{j} \begin{vmatrix} x & z \\ 1 & 1 \end{vmatrix} + \hat{k} \begin{vmatrix} x & y \\ 1 & 2 \end{vmatrix} \]
Calculating each of these 2x2 determinants gives us:
\[ = \hat{i}(y \cdot 1 - z \cdot 2) - \hat{j}(x \cdot 1 - z \cdot 1) + \hat{k}(x \cdot 2 - y \cdot 1) \]
Substituting \(z = 2x - y\):
\[ = \hat{i}(y - 2(2x - y)) - \hat{j}(x - (2x - y)) + \hat{k}(2x - y) \]
Simplifying gives:
\[ = \hat{i}(y - 4x + 2y) - \hat{j}(-x + 2x - y) + \hat{k}(2x - y) \]
\[ = \hat{i}(3y - 4x) + \hat{j}(x - y) + \hat{k}(2x - y) \]

### Step 7: Set the components equal to those from \(\vec{c} \times \vec{b}\)
We have:
\[ 3y - 4x = -11 \quad \text{(Equation 2)} \]
\[ x - y = 2 \quad \text{(Equation 3)} \]
\[ 2x - y = 7 \quad \text{(Equation 4)} \]

### Step 8: Solve the system of equations
From Equation 3:
\[ y = x - 2 \]
Substituting \(y\) into Equation 4:
\[ 2x - (x - 2) = 7 \]
\[ 2x - x + 2 = 7 \]
\[ x + 2 = 7 \implies x = 5 \]
Then substituting back to find \(y\):
\[ y = 5 - 2 = 3 \]
And substituting \(x\) and \(y\) back into Equation 1 to find \(z\):
\[ z = 2(5) - 3 = 10 - 3 = 7 \]

### Final Answer
Thus, the vector \(\vec{r}\) is:
\[ \vec{r} = 5\hat{i} + 3\hat{j} + 7\hat{k} \]