If with reference to a right handed system of mutually perpendicular unit vectors ` hat i , hat j and hat k`, we have ` vecalpha=3 hat i- hat j` ,and `vecbeta=2 hat i+ hat j-3 hat kdot` then express ` vecbeta` in the form `vecbeta= vecbeta_1+ vecbeta_2`, where `vecbeta_1` is parallel to ` vecalpha and vecbeta_2` is perpendicular to ` vecalpha`

Options

A. `vecbeta_1=3/2hati-1/2hat j` ,and `vecbeta_2=1/2 hat i+ 3/2hat j-3 hat kdot`
B. `vecbeta_1=3/2hati+1/2hat j` ,and `vecbeta_2=1/3 hat i+ 3/2hat j-3 hat kdot`
C. `vecbeta_1=1/2hati-1/2hat j` ,and `vecbeta_2=1/2 hat i+ 3/2hat j-3 hat kdot`
D. `vecbeta_1=1/2hatj-1/2hat k` ,and `vecbeta_2=1/2 hat i+ 3/2hat j+3 hat kdot`

Answer

To express the vector \(\vec{\beta}\) in the form \(\vec{\beta} = \vec{\beta_1} + \vec{\beta_2}\), where \(\vec{\beta_1}\) is parallel to \(\vec{\alpha}\) and \(\vec{\beta_2}\) is perpendicular to \(\vec{\alpha}\), we will follow these steps:

### 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 write:
\[ \vec{\beta_1} = \lambda \vec{\alpha} \]
for some scalar \(\lambda\).

### Step 3: Express \(\vec{\beta_2}\)
Using the relation \(\vec{\beta} = \vec{\beta_1} + \vec{\beta_2}\), we can express \(\vec{\beta_2}\) as:
\[ \vec{\beta_2} = \vec{\beta} - \vec{\beta_1} = \vec{\beta} - \lambda \vec{\alpha} \]

### Step 4: Ensure \(\vec{\beta_2}\) is perpendicular to \(\vec{\alpha}\)
For \(\vec{\beta_2}\) to be perpendicular to \(\vec{\alpha}\), we need:
\[ \vec{\alpha} \cdot \vec{\beta_2} = 0 \]
Substituting \(\vec{\beta_2}\):
\[ \vec{\alpha} \cdot (\vec{\beta} - \lambda \vec{\alpha}) = 0 \]
This simplifies to:
\[ \vec{\alpha} \cdot \vec{\beta} - \lambda \vec{\alpha} \cdot \vec{\alpha} = 0 \]

### Step 5: Calculate \(\vec{\alpha} \cdot \vec{\beta}\) and \(\vec{\alpha} \cdot \vec{\alpha}\)
Calculating \(\vec{\alpha} \cdot \vec{\beta}\):
\[ \vec{\alpha} \cdot \vec{\beta} = (3\hat{i} - \hat{j}) \cdot (2\hat{i} + \hat{j} - 3\hat{k}) = 3 \cdot 2 + (-1) \cdot 1 + 0 = 6 - 1 = 5 \]

Calculating \(\vec{\alpha} \cdot \vec{\alpha}\):
\[ \vec{\alpha} \cdot \vec{\alpha} = (3\hat{i} - \hat{j}) \cdot (3\hat{i} - \hat{j}) = 3^2 + (-1)^2 = 9 + 1 = 10 \]

### Step 6: Solve for \(\lambda\)
Substituting back into the equation:
\[ 5 - \lambda \cdot 10 = 0 \]
This gives:
\[ \lambda = \frac{5}{10} = \frac{1}{2} \]

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

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

### Final Result
Thus, we can express \(\vec{\beta}\) as:
\[ \vec{\beta} = \left(\frac{3}{2}\hat{i} - \frac{1}{2}\hat{j}\right) + \left(\frac{1}{2}\hat{i} + \frac{3}{2}\hat{j} - 3\hat{k}\right) \]