Let `alpha=3i+j" and "beta=2i-j+3k." Suppose "beta=beta_(1)-beta_(2)," where "beta_(1)` is parallel to `alpha" and "beta_(2)` is perpendicular to `alpha." If "beta_(1) times beta_(2)=-3/2i+aj+bk`, then a + b = ______

Answer

To solve the problem step by step, we will break down the vectors and their relationships as described in the question.

### Step 1: Define the vectors
Let:
- \(\alpha = 3\mathbf{i} + \mathbf{j}\)
- \(\beta = 2\mathbf{i} - \mathbf{j} + 3\mathbf{k}\)

### Step 2: Express \(\beta\) in terms of \(\beta_1\) and \(\beta_2\)
We have:
\[ \beta = \beta_1 - \beta_2 \]
where \(\beta_1\) is parallel to \(\alpha\) and \(\beta_2\) is perpendicular to \(\alpha\).

### Step 3: Express \(\beta_1\)
Since \(\beta_1\) is parallel to \(\alpha\), we can express it as:
\[ \beta_1 = \lambda \alpha \]
for some scalar \(\lambda\).

### Step 4: Find \(\beta_2\)
From the expression \(\beta = \beta_1 - \beta_2\), we can rearrange to find \(\beta_2\):
\[ \beta_2 = \beta_1 - \beta \]

### Step 5: Use the property of perpendicular vectors
Since \(\beta_2\) is perpendicular to \(\alpha\), we have:
\[ \beta_2 \cdot \alpha = 0 \]

### Step 6: Substitute \(\beta_1\) into the equation
Substituting \(\beta_1\) into the equation for \(\beta_2\):
\[ \beta_2 = \lambda \alpha - \beta \]
Now, substituting \(\alpha\):
\[ \beta_2 = \lambda (3\mathbf{i} + \mathbf{j}) - (2\mathbf{i} - \mathbf{j} + 3\mathbf{k}) \]

### Step 7: Calculate the dot product
Now, we can calculate the dot product:
\[ (\lambda (3\mathbf{i} + \mathbf{j}) - (2\mathbf{i} - \mathbf{j} + 3\mathbf{k})) \cdot (3\mathbf{i} + \mathbf{j}) = 0 \]
Expanding this, we get:
\[ (\lambda(3) - 2) \cdot 3 + (\lambda(1) + 1) \cdot 1 + (-3\lambda) \cdot 0 = 0 \]
This simplifies to:
\[ 3\lambda - 6 + \lambda + 1 = 0 \]
\[ 4\lambda - 5 = 0 \]
\[ \lambda = \frac{5}{4} \]

### Step 8: Find \(\beta_1\) and \(\beta_2\)
Now, substituting \(\lambda\) back to find \(\beta_1\):
\[ \beta_1 = \frac{5}{4}(3\mathbf{i} + \mathbf{j}) = \frac{15}{4}\mathbf{i} + \frac{5}{4}\mathbf{j} \]
Now substituting \(\beta_1\) into the equation for \(\beta_2\):
\[ \beta_2 = \left(\frac{15}{4}\mathbf{i} + \frac{5}{4}\mathbf{j}\right) - (2\mathbf{i} - \mathbf{j} + 3\mathbf{k}) \]
Calculating this gives:
\[ \beta_2 = \left(\frac{15}{4} - 2\right)\mathbf{i} + \left(\frac{5}{4} + 1\right)\mathbf{j} - 3\mathbf{k} \]
\[ = \left(\frac{15}{4} - \frac{8}{4}\right)\mathbf{i} + \left(\frac{5}{4} + \frac{4}{4}\right)\mathbf{j} - 3\mathbf{k} \]
\[ = \frac{7}{4}\mathbf{i} + \frac{9}{4}\mathbf{j} - 3\mathbf{k} \]

### Step 9: Calculate the cross product
Now we need to find \(\beta_1 \times \beta_2\):
\[ \beta_1 \times \beta_2 = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ \frac{15}{4} & \frac{5}{4} & 0 \\ \frac{7}{4} & \frac{9}{4} & -3 \end{vmatrix} \]
Calculating this determinant gives:
\[ = \mathbf{i}\left(\frac{5}{4} \cdot (-3) - 0\right) - \mathbf{j}\left(\frac{15}{4} \cdot (-3) - 0\right) + \mathbf{k}\left(\frac{15}{4} \cdot \frac{9}{4} - \frac{5}{4} \cdot \frac{7}{4}\right) \]
\[ = -\frac{15}{4}\mathbf{i} + \frac{45}{4}\mathbf{j} + \left(\frac{135}{16} - \frac{35}{16}\right)\mathbf{k} \]
\[ = -\frac{15}{4}\mathbf{i} + \frac{45}{4}\mathbf{j} + \frac{100}{16}\mathbf{k} \]
\[ = -\frac{15}{4}\mathbf{i} + \frac{45}{4}\mathbf{j} + \frac{25}{4}\mathbf{k} \]

### Step 10: Compare coefficients
We know that:
\[ \beta_1 \times \beta_2 = -\frac{3}{2}\mathbf{i} + a\mathbf{j} + b\mathbf{k} \]
Comparing coefficients gives:
\[ -\frac{15}{4} = -\frac{3}{2} \implies a = \frac{45}{4}, \quad b = \frac{25}{4} \]

### Step 11: Find \(a + b\)
Now, we calculate:
\[ a + b = \frac{45}{4} + \frac{25}{4} = \frac{70}{4} = \frac{35}{2} \]

Thus, the final answer is:
\[ \boxed{\frac{35}{2}} \]