Let `alpha=3i+j" and "beta=2i-j+3k." If "beta=beta_(1)-beta_(2), " where "beta_(1)` is parallel to `alpha" and "beta_(2)` is perpendicular to `alpha`, then `beta_(1) times beta_(2)` is equal to

Options

A. `1/2(-3i+9j+5k)`
B. `1/2(3i-9j+5k)`
C. `3i-9j-5k`
D. `-3i+9j+5k`

Answer

To solve the problem, we need to find the cross product of the vectors \(\beta_1\) and \(\beta_2\), where \(\beta_1\) is parallel to \(\alpha\) and \(\beta_2\) is perpendicular to \(\alpha\).

Given:
\[ \alpha = 3\mathbf{i} + \mathbf{j} \]
\[ \beta = 2\mathbf{i} - \mathbf{j} + 3\mathbf{k} \]

### Step 1: Express \(\beta_1\) and \(\beta_2\)
Since \(\beta_1\) is parallel to \(\alpha\), we can express \(\beta_1\) as:
\[ \beta_1 = \lambda \alpha = \lambda (3\mathbf{i} + \mathbf{j}) = 3\lambda \mathbf{i} + \lambda \mathbf{j} \]

### Step 2: Express \(\beta_2\)
We know that:
\[ \beta = \beta_1 - \beta_2 \implies \beta_2 = \beta_1 - \beta \]
Substituting for \(\beta_1\):
\[ \beta_2 = (3\lambda \mathbf{i} + \lambda \mathbf{j}) - (2\mathbf{i} - \mathbf{j} + 3\mathbf{k}) \]
\[ = (3\lambda - 2)\mathbf{i} + (\lambda + 1)\mathbf{j} - 3\mathbf{k} \]

### Step 3: Use the perpendicular condition
Since \(\beta_2\) is perpendicular to \(\alpha\), we have:
\[ \beta_2 \cdot \alpha = 0 \]
Calculating the dot product:
\[ (3\lambda - 2)(3) + (\lambda + 1)(1) + (-3)(0) = 0 \]
\[ 9\lambda - 6 + \lambda + 1 = 0 \]
\[ 10\lambda - 5 = 0 \implies \lambda = \frac{1}{2} \]

### Step 4: Find \(\beta_1\) and \(\beta_2\)
Now substituting \(\lambda\) back into \(\beta_1\):
\[ \beta_1 = 3\left(\frac{1}{2}\right)\mathbf{i} + \left(\frac{1}{2}\right)\mathbf{j} = \frac{3}{2}\mathbf{i} + \frac{1}{2}\mathbf{j} \]

Now substituting \(\lambda\) into \(\beta_2\):
\[ \beta_2 = \left(3\left(\frac{1}{2}\right) - 2\right)\mathbf{i} + \left(\frac{1}{2} + 1\right)\mathbf{j} - 3\mathbf{k} \]
\[ = \left(\frac{3}{2} - 2\right)\mathbf{i} + \left(\frac{3}{2}\right)\mathbf{j} - 3\mathbf{k} \]
\[ = -\frac{1}{2}\mathbf{i} + \frac{3}{2}\mathbf{j} - 3\mathbf{k} \]

### Step 5: Calculate the cross product \(\beta_1 \times \beta_2\)
Now we compute the cross product:
\[ \beta_1 \times \beta_2 = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ \frac{3}{2} & \frac{1}{2} & 0 \\ -\frac{1}{2} & \frac{3}{2} & -3 \end{vmatrix} \]

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

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