If the vector `vecb = 3hati + 4hatk` is written as the sum of a vector `vecb_(1)` parallel to `veca = hati + hatj` and a vector `vecb_(2)`, perpendicular to `veca` then `vecb_(1) xx vecb_(2)` is equal to:

Options

A. `3hati - 3hatj + 9hatk`
B. `-3hati + 3hatj + 9hatk`
C. `-6hati + 6hatj -9/2 hatk`
D. `6hati - 6hatj + 9/2 hatk`

Answer

To solve the problem, we need to express the vector \(\vec{b} = 3\hat{i} + 4\hat{k}\) as the sum of two vectors: \(\vec{b_1}\), which is parallel to \(\vec{a} = \hat{i} + \hat{j}\), and \(\vec{b_2}\), which is perpendicular to \(\vec{a}\). We then need to find the cross product \(\vec{b_1} \times \vec{b_2}\).

### Step 1: Express \(\vec{b_1}\) and \(\vec{b_2}\)

Since \(\vec{b_1}\) is parallel to \(\vec{a}\), we can write:
\[ \vec{b_1} = k(\hat{i} + \hat{j}) = k\hat{i} + k\hat{j} \]
for some scalar \(k\).

### Step 2: Express \(\vec{b_2}\)

Since \(\vec{b_2}\) is perpendicular to \(\vec{a}\), we can express \(\vec{b_2}\) as:
\[ \vec{b_2} = p\hat{i} + q\hat{j} + r\hat{k} \]
where \(p\), \(q\), and \(r\) are scalars.

### Step 3: Set up the equation

We know that:
\[ \vec{b} = \vec{b_1} + \vec{b_2} \]
Substituting the expressions for \(\vec{b_1}\) and \(\vec{b_2}\):
\[ 3\hat{i} + 4\hat{k} = (k\hat{i} + k\hat{j}) + (p\hat{i} + q\hat{j} + r\hat{k}) \]

### Step 4: Combine like terms

Combining the terms gives:
\[ 3\hat{i} + 4\hat{k} = (k + p)\hat{i} + (k + q)\hat{j} + r\hat{k} \]

### Step 5: Set up the system of equations

From the above equation, we can set up the following system:
1. \(k + p = 3\) (coefficient of \(\hat{i}\))
2. \(k + q = 0\) (coefficient of \(\hat{j}\))
3. \(r = 4\) (coefficient of \(\hat{k}\))

### Step 6: Solve for \(k\), \(p\), \(q\), and \(r\)

From equation 3, we have:
\[ r = 4 \]

From equation 2, we can express \(q\) in terms of \(k\):
\[ q = -k \]

Substituting \(q\) into equation 1:
\[ k + p = 3 \implies p = 3 - k \]

### Step 7: Substitute values

Now we have:
- \(p = 3 - k\)
- \(q = -k\)
- \(r = 4\)

### Step 8: Find \(\vec{b_1}\) and \(\vec{b_2}\)

Substituting these into the expressions for \(\vec{b_1}\) and \(\vec{b_2}\):
\[ \vec{b_1} = k\hat{i} + k\hat{j} \]
\[ \vec{b_2} = (3 - k)\hat{i} - k\hat{j} + 4\hat{k} \]

### Step 9: Calculate the cross product \(\vec{b_1} \times \vec{b_2}\)

Now we can calculate the cross product:
\[ \vec{b_1} \times \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ k & k & 0 \\ 3 - k & -k & 4 \end{vmatrix} \]

### Step 10: Evaluate the determinant

Calculating the determinant:
\[ \vec{b_1} \times \vec{b_2} = \hat{i} \begin{vmatrix} k & 0 \\ -k & 4 \end{vmatrix} - \hat{j} \begin{vmatrix} k & 0 \\ 3 - k & 4 \end{vmatrix} + \hat{k} \begin{vmatrix} k & k \\ 3 - k & -k \end{vmatrix} \]

Calculating each of these:
1. For \(\hat{i}\):
\[ k \cdot 4 - 0 \cdot (-k) = 4k \]
2. For \(-\hat{j}\):
\[ k \cdot 4 - 0 \cdot (3 - k) = 4k \]
3. For \(\hat{k}\):
\[ k \cdot (-k) - k \cdot (3 - k) = -k^2 - 3k + k^2 = -3k \]

Putting it all together:
\[ \vec{b_1} \times \vec{b_2} = 4k\hat{i} - 4k\hat{j} - 3k\hat{k} \]

### Final Result

Thus, the final expression for \(\vec{b_1} \times \vec{b_2}\) is:
\[ \vec{b_1} \times \vec{b_2} = 4k\hat{i} - 4k\hat{j} - 3k\hat{k} \]