Let `vec(alpha)=3hat(i)+hat(j), vec(beta)=2hat(i)-hat(j)+3hat(k)` and `vec(beta)=vec(beta)_(1)-vec(beta)_(2)`, such that `vec(beta)_(1)` is parallel to `vec(alpha)` and `vec(beta)_(2)` is perpendicular to `alpha`. Find `vec(beta)_(1)xx vec(beta)_(2)`.

Options

A. `(1)/(2)(hat(i)-9hat(j)+8hat(k))`
B. `(1)/(2)(hat(i)-3hat(j)+4hat(k))`
C. `(1)/(2)(-3hat(i)+9hat(j)+10 hat(k))`
D. `(3)/(2)(3hat(i)+9hat(j)+10hat(k))`

Answer

To solve the problem, we need to find the vectors \(\vec{\beta}_1\) and \(\vec{\beta}_2\) such that \(\vec{\beta} = \vec{\beta}_1 - \vec{\beta}_2\), where \(\vec{\beta}_1\) is parallel to \(\vec{\alpha}\) and \(\vec{\beta}_2\) is perpendicular to \(\vec{\alpha}\). Finally, we will compute the cross product \(\vec{\beta}_1 \times \vec{\beta}_2\).

### Step-by-Step Solution:

1. **Identify the vectors**:
\[ \vec{\alpha} = 3\hat{i} + \hat{j} \]
\[ \vec{\beta} = 2\hat{i} - \hat{j} + 3\hat{k} \]

2. **Express \(\vec{\beta}_1\)**:
Since \(\vec{\beta}_1\) is parallel to \(\vec{\alpha}\), we can express it as:
\[ \vec{\beta}_1 = \lambda \vec{\alpha} = \lambda (3\hat{i} + \hat{j}) = 3\lambda \hat{i} + \lambda \hat{j} \]

3. **Express \(\vec{\beta}_2\)**:
From the equation \(\vec{\beta} = \vec{\beta}_1 - \vec{\beta}_2\), we can express \(\vec{\beta}_2\) as:
\[ \vec{\beta}_2 = \vec{\beta}_1 - \vec{\beta} = (3\lambda \hat{i} + \lambda \hat{j}) - (2\hat{i} - \hat{j} + 3\hat{k}) \]
Simplifying this gives:
\[ \vec{\beta}_2 = (3\lambda - 2)\hat{i} + (\lambda + 1)\hat{j} - 3\hat{k} \]

4. **Set up the perpendicularity condition**:
Since \(\vec{\beta}_2\) is perpendicular to \(\vec{\alpha}\), we have:
\[ \vec{\beta}_2 \cdot \vec{\alpha} = 0 \]
This gives:
\[ (3\lambda - 2) \cdot 3 + (\lambda + 1) \cdot 1 = 0 \]
Expanding this results in:
\[ 9\lambda - 6 + \lambda + 1 = 0 \implies 10\lambda - 5 = 0 \implies \lambda = \frac{1}{2} \]

5. **Substitute \(\lambda\) back into \(\vec{\beta}_1\) and \(\vec{\beta}_2\)**:
\[ \vec{\beta}_1 = 3\left(\frac{1}{2}\right)\hat{i} + \left(\frac{1}{2}\right)\hat{j} = \frac{3}{2}\hat{i} + \frac{1}{2}\hat{j} \]
\[ \vec{\beta}_2 = (3\left(\frac{1}{2}\right) - 2)\hat{i} + \left(\frac{1}{2} + 1\right)\hat{j} - 3\hat{k} = \left(\frac{3}{2} - 2\right)\hat{i} + \left(\frac{1}{2} + 1\right)\hat{j} - 3\hat{k} \]
Simplifying \(\vec{\beta}_2\):
\[ \vec{\beta}_2 = -\frac{1}{2}\hat{i} + \frac{3}{2}\hat{j} - 3\hat{k} \]

6. **Calculate the cross product \(\vec{\beta}_1 \times \vec{\beta}_2\)**:
Using the determinant method:
\[ \vec{\beta}_1 \times \vec{\beta}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \frac{3}{2} & \frac{1}{2} & 0 \\ -\frac{1}{2} & \frac{3}{2} & -3 \end{vmatrix} \]
Expanding this determinant:
\[ = \hat{i} \left(\frac{1}{2} \cdot (-3) - 0 \cdot \frac{3}{2}\right) - \hat{j} \left(\frac{3}{2} \cdot (-3) - 0 \cdot -\frac{1}{2}\right) + \hat{k} \left(\frac{3}{2} \cdot \frac{3}{2} - \frac{1}{2} \cdot -\frac{1}{2}\right) \]
\[ = -\frac{3}{2}\hat{i} + \frac{9}{2}\hat{j} + \left(\frac{9}{4} + \frac{1}{4}\right)\hat{k} \]
\[ = -\frac{3}{2}\hat{i} + \frac{9}{2}\hat{j} + \frac{10}{4}\hat{k} = -\frac{3}{2}\hat{i} + \frac{9}{2}\hat{j} + \frac{5}{2}\hat{k} \]

7. **Final result**:
The cross product is:
\[ \vec{\beta}_1 \times \vec{\beta}_2 = \frac{1}{2} \left(-3\hat{i} + 9\hat{j} + 5\hat{k}\right) \]