Let `vecalpha = 3hati + hatj` and `beta= 2hati - hatj + 3hatk`. If `vecbeta = vecbeta_(1) - vecbeta_(2)`, where `vecbeta_(1)` is parallel to `vecalpha` and `vecbeta_(2)` is perpendicular to `vecalpha`, then `vecbeta_(1) xx vecbeta_(2)` is equal to:

Options

A. `1/2(3hati - 9hatj + 5hatk)`
B. `1/2(-3hati+ 9hatj + 5hatk)`
C. `-3hati + 9hatj + 5hatk`
D. `3hati - 9hatj - 5hatk`

Answer

To solve the problem, we need to find the cross product of two vectors, \(\vec{\beta_1}\) and \(\vec{\beta_2}\), where \(\vec{\beta_1}\) is parallel to \(\vec{\alpha}\) and \(\vec{\beta_2}\) is perpendicular to \(\vec{\alpha}\). Let's go through the solution step by step.

### Step 1: Define the vectors
Given:
\[ \vec{\alpha} = 3\hat{i} + \hat{j} \]
\[ \vec{\beta} = 2\hat{i} - \hat{j} + 3\hat{k} \]

### Step 2: Express \(\vec{\beta_1}\) in terms of \(\vec{\alpha}\)
Since \(\vec{\beta_1}\) is parallel to \(\vec{\alpha}\), we can express it as:
\[ \vec{\beta_1} = \lambda \vec{\alpha} = \lambda (3\hat{i} + \hat{j}) \]
where \(\lambda\) is a scalar.

### Step 3: Express \(\vec{\beta_2}\)
From the problem statement, we have:
\[ \vec{\beta} = \vec{\beta_1} - \vec{\beta_2} \]
This implies:
\[ \vec{\beta_2} = \vec{\beta_1} - \vec{\beta} \]

### Step 4: Use the condition for perpendicularity
Since \(\vec{\beta_2}\) is perpendicular to \(\vec{\alpha}\), we have:
\[ \vec{\beta_2} \cdot \vec{\alpha} = 0 \]
Substituting \(\vec{\beta_2}\):
\[ (\vec{\beta_1} - \vec{\beta}) \cdot \vec{\alpha} = 0 \]

### Step 5: Substitute \(\vec{\beta_1}\) and \(\vec{\beta}\)
Substituting \(\vec{\beta_1}\):
\[ (\lambda (3\hat{i} + \hat{j}) - (2\hat{i} - \hat{j} + 3\hat{k})) \cdot (3\hat{i} + \hat{j}) = 0 \]
This simplifies to:
\[ (3\lambda - 2)\hat{i} + (\lambda + 1)\hat{j} - 3\hat{k} \cdot (3\hat{i} + \hat{j}) = 0 \]

### Step 6: Calculate the dot product
Calculating the dot product:
\[ (3\lambda - 2) \cdot 3 + (\lambda + 1) \cdot 1 + (-3) \cdot 0 = 0 \]
This gives:
\[ 9\lambda - 6 + \lambda + 1 = 0 \]
\[ 10\lambda - 5 = 0 \]
Thus,
\[ \lambda = \frac{1}{2} \]

### Step 7: Find \(\vec{\beta_1}\)
Substituting \(\lambda\) back into the equation for \(\vec{\beta_1}\):
\[ \vec{\beta_1} = \frac{1}{2}(3\hat{i} + \hat{j}) = \frac{3}{2}\hat{i} + \frac{1}{2}\hat{j} \]

### Step 8: Find \(\vec{\beta_2}\)
Now substituting \(\vec{\beta_1}\) into the equation for \(\vec{\beta_2}\):
\[ \vec{\beta_2} = \left(\frac{3}{2}\hat{i} + \frac{1}{2}\hat{j}\right) - (2\hat{i} - \hat{j} + 3\hat{k}) \]
Calculating this gives:
\[ \vec{\beta_2} = \left(\frac{3}{2} - 2\right)\hat{i} + \left(\frac{1}{2} + 1\right)\hat{j} - 3\hat{k} \]
\[ = -\frac{1}{2}\hat{i} + \frac{3}{2}\hat{j} - 3\hat{k} \]

### Step 9: Calculate \(\vec{\beta_1} \times \vec{\beta_2}\)
Now we need to find the cross product \(\vec{\beta_1} \times \vec{\beta_2}\):
\[ \vec{\beta_1} \times \vec{\beta_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \frac{3}{2} & \frac{1}{2} & 0 \\ -\frac{1}{2} & \frac{3}{2} & -3 \end{vmatrix} \]

### Step 10: Expand the determinant
Calculating the determinant:
\[ = \hat{i} \left( \frac{1}{2} \cdot (-3) - 0 \cdot \frac{3}{2} \right) - \hat{j} \left( \frac{3}{2} \cdot (-3) - 0 \cdot -\frac{1}{2} \right) + \hat{k} \left( \frac{3}{2} \cdot \frac{3}{2} - \frac{1}{2} \cdot -\frac{1}{2} \right) \]
\[ = -\frac{3}{2}\hat{i} + \frac{9}{2}\hat{j} + \left( \frac{9}{4} + \frac{1}{4} \right)\hat{k} \]
\[ = -\frac{3}{2}\hat{i} + \frac{9}{2}\hat{j} + \frac{10}{4}\hat{k} \]
\[ = -\frac{3}{2}\hat{i} + \frac{9}{2}\hat{j} + \frac{5}{2}\hat{k} \]

### Final Result
Thus, the result of \(\vec{\beta_1} \times \vec{\beta_2}\) is:
\[ \vec{\beta_1} \times \vec{\beta_2} = -\frac{3}{2}\hat{i} + \frac{9}{2}\hat{j} + \frac{5}{2}\hat{k} \]