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: one vector \(\vec{b_1}\) that is parallel to \(\vec{a} = \hat{i} + \hat{j}\) and another vector \(\vec{b_2}\) that is perpendicular to \(\vec{a}\). We will then find the cross product \(\vec{b_1} \times \vec{b_2}\).

### Step-by-Step Solution:

1. **Identify the vectors:**
- Given \(\vec{b} = 3\hat{i} + 4\hat{k}\)
- Given \(\vec{a} = \hat{i} + \hat{j}\)

2. **Express \(\vec{b_1}\) in terms of \(\vec{a}\):**
Since \(\vec{b_1}\) is parallel to \(\vec{a}\), we can express it as:
\[ \vec{b_1} = \lambda \vec{a} = \lambda (\hat{i} + \hat{j}) = \lambda \hat{i} + \lambda \hat{j} \]
where \(\lambda\) is a scalar.

3. **Express \(\vec{b_2}\):**
Since \(\vec{b_2}\) is perpendicular to \(\vec{a}\), we have:
\[ \vec{b_2} = \vec{b} - \vec{b_1} \]

4. **Set up the equation:**
From the previous steps, we have:
\[ \vec{b} = \vec{b_1} + \vec{b_2} \]
Substituting \(\vec{b_1}\):
\[ 3\hat{i} + 4\hat{k} = (\lambda \hat{i} + \lambda \hat{j}) + \vec{b_2} \]

5. **Dot product to find \(\lambda\):**
Taking the dot product of \(\vec{b}\) with \(\vec{a}\):
\[ \vec{b} \cdot \vec{a} = (3\hat{i} + 4\hat{k}) \cdot (\hat{i} + \hat{j}) = 3(1) + 4(0) = 3 \]
On the other side:
\[ \vec{b_1} \cdot \vec{a} = \lambda \vec{a} \cdot \vec{a} = \lambda (1 + 1) = 2\lambda \]
Setting them equal gives:
\[ 3 = 2\lambda \implies \lambda = \frac{3}{2} \]

6. **Find \(\vec{b_1}\):**
Substitute \(\lambda\) back into the expression for \(\vec{b_1}\):
\[ \vec{b_1} = \frac{3}{2}(\hat{i} + \hat{j}) = \frac{3}{2}\hat{i} + \frac{3}{2}\hat{j} \]

7. **Find \(\vec{b_2}\):**
Now substitute \(\vec{b_1}\) into the equation for \(\vec{b_2}\):
\[ \vec{b_2} = (3\hat{i} + 4\hat{k}) - \left(\frac{3}{2}\hat{i} + \frac{3}{2}\hat{j}\right) \]
Simplifying gives:
\[ \vec{b_2} = \left(3 - \frac{3}{2}\right)\hat{i} - \frac{3}{2}\hat{j} + 4\hat{k} = \frac{3}{2}\hat{i} - \frac{3}{2}\hat{j} + 4\hat{k} \]

8. **Calculate \(\vec{b_1} \times \vec{b_2}\):**
Now we can compute the cross product:
\[ \vec{b_1} \times \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \frac{3}{2} & \frac{3}{2} & 0 \\ \frac{3}{2} & -\frac{3}{2} & 4 \end{vmatrix} \]

Expanding the determinant:
\[ = \hat{i} \left(\frac{3}{2} \cdot 4 - 0 \cdot -\frac{3}{2}\right) - \hat{j} \left(\frac{3}{2} \cdot 4 - 0 \cdot \frac{3}{2}\right) + \hat{k} \left(\frac{3}{2} \cdot -\frac{3}{2} - \frac{3}{2} \cdot \frac{3}{2}\right) \]
\[ = \hat{i} \cdot 6 - \hat{j} \cdot 6 + \hat{k} \cdot \left(-\frac{9}{4} - \frac{9}{4}\right) \]
\[ = 6\hat{i} - 6\hat{j} - \frac{9}{2}\hat{k} \]

### Final Answer:
\[ \vec{b_1} \times \vec{b_2} = 6\hat{i} - 6\hat{j} - \frac{9}{2}\hat{k} \]