Given, `vec(a)= 3hat(i) - hat(j) and vec(b)= 2hat(i) + hat(j) - 3hat(k)`. Express `vec(b) = vec(b)_(1) + vec(b)_(2)` where `vec(b)_(1)` is parallel to `vec(a) and vec(b)_(2)` is perpendicular to `vec(a)`

Answer

To solve the problem, we need to express the vector \(\vec{b}\) as the sum of two vectors: \(\vec{b}_1\) (which is parallel to \(\vec{a}\)) and \(\vec{b}_2\) (which is perpendicular to \(\vec{a}\)).

Given:
\[ \vec{a} = 3\hat{i} - \hat{j} \]
\[ \vec{b} = 2\hat{i} + \hat{j} - 3\hat{k} \]

### Step 1: Find the unit vector of \(\vec{a}\)
First, we need to find the unit vector of \(\vec{a}\):
\[ |\vec{a}| = \sqrt{(3)^2 + (-1)^2} = \sqrt{9 + 1} = \sqrt{10} \]
The unit vector \(\hat{a}\) is:
\[ \hat{a} = \frac{\vec{a}}{|\vec{a}|} = \frac{3\hat{i} - \hat{j}}{\sqrt{10}} \]

### Step 2: Express \(\vec{b}_1\)
Let \(\vec{b}_1 = k\hat{a}\), where \(k\) is a scalar. Thus:
\[ \vec{b}_1 = k\left(\frac{3\hat{i} - \hat{j}}{\sqrt{10}}\right) = \frac{3k}{\sqrt{10}}\hat{i} - \frac{k}{\sqrt{10}}\hat{j} \]

### Step 3: Express \(\vec{b}_2\)
Now, we can express \(\vec{b}\) as:
\[ \vec{b} = \vec{b}_1 + \vec{b}_2 \]
Let \(\vec{b}_2 = p\hat{i} + q\hat{j} + r\hat{k}\).

### Step 4: Set up the equation
Substituting \(\vec{b}_1\) and \(\vec{b}_2\) into the equation gives:
\[ 2\hat{i} + \hat{j} - 3\hat{k} = \left(\frac{3k}{\sqrt{10}} + p\right)\hat{i} + \left(-\frac{k}{\sqrt{10}} + q\right)\hat{j} + r\hat{k} \]

### Step 5: Compare coefficients
From the equation, we can compare coefficients:
1. For \(\hat{i}\):
\[ \frac{3k}{\sqrt{10}} + p = 2 \quad \text{(1)} \]
2. For \(\hat{j}\):
\[ -\frac{k}{\sqrt{10}} + q = 1 \quad \text{(2)} \]
3. For \(\hat{k}\):
\[ r = -3 \quad \text{(3)} \]

### Step 6: Use the perpendicular condition
Since \(\vec{b}_2\) is perpendicular to \(\vec{a}\), we have:
\[ \vec{a} \cdot \vec{b}_2 = 0 \]
This gives:
\[ 3p - q = 0 \quad \text{(4)} \]

### Step 7: Solve the equations
From equation (4), we can express \(q\) in terms of \(p\):
\[ q = 3p \]

Substituting \(q = 3p\) into equation (2):
\[ -\frac{k}{\sqrt{10}} + 3p = 1 \quad \text{(5)} \]

Now substituting \(p\) from equation (1):
\[ p = 2 - \frac{3k}{\sqrt{10}} \]
Substituting this into equation (5):
\[ -\frac{k}{\sqrt{10}} + 3\left(2 - \frac{3k}{\sqrt{10}}\right) = 1 \]
This simplifies to:
\[ -\frac{k}{\sqrt{10}} + 6 - \frac{9k}{\sqrt{10}} = 1 \]
\[ 6 - \frac{10k}{\sqrt{10}} = 1 \]
\[ \frac{10k}{\sqrt{10}} = 5 \implies k = \frac{5\sqrt{10}}{10} = \frac{\sqrt{10}}{2} \]

### Step 8: Find \(p\) and \(q\)
Substituting \(k\) back into the equations:
\[ p = 2 - \frac{3(\sqrt{10}/2)}{\sqrt{10}} = 2 - \frac{3}{2} = \frac{1}{2} \]
\[ q = 3p = 3 \cdot \frac{1}{2} = \frac{3}{2} \]

### Step 9: Write \(\vec{b}_2\)
Now we have:
\[ \vec{b}_2 = \frac{1}{2}\hat{i} + \frac{3}{2}\hat{j} - 3\hat{k} \]

### Step 10: Write \(\vec{b}_1\)
Substituting \(k\) into \(\vec{b}_1\):
\[ \vec{b}_1 = \frac{\sqrt{10}}{2}\left(\frac{3\hat{i} - \hat{j}}{\sqrt{10}}\right) = \frac{3}{2}\hat{i} - \frac{1}{2}\hat{j} \]

### Final Result
Thus, we can express \(\vec{b}\) as:
\[ \vec{b} = \vec{b}_1 + \vec{b}_2 \]
Where:
\[ \vec{b}_1 = \frac{3}{2}\hat{i} - \frac{1}{2}\hat{j} \]
\[ \vec{b}_2 = \frac{1}{2}\hat{i} + \frac{3}{2}\hat{j} - 3\hat{k} \]