If `vec(a)=2hat(i)-hat(j)-2hat(k)` and `vec(b)=7hat(i)+2hat(j)-3hat(k)`, then express `vec(b)` in the form `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 express the vector \(\vec{b}\) in the form \(\vec{b} = \vec{b}_{1} + \vec{b}_{2}\), where \(\vec{b}_{1}\) is parallel to \(\vec{a}\) and \(\vec{b}_{2}\) is perpendicular to \(\vec{a}\), we will follow these steps:

### Step 1: Identify the given vectors
We have:
\[ \vec{a} = 2\hat{i} - \hat{j} - 2\hat{k} \]
\[ \vec{b} = 7\hat{i} + 2\hat{j} - 3\hat{k} \]

### Step 2: Express \(\vec{b}_{1}\) in terms of \(\vec{a}\)
Since \(\vec{b}_{1}\) is parallel to \(\vec{a}\), we can write:
\[ \vec{b}_{1} = \lambda \vec{a} \]
for some scalar \(\lambda\). Thus,
\[ \vec{b}_{1} = \lambda(2\hat{i} - \hat{j} - 2\hat{k}) = 2\lambda\hat{i} - \lambda\hat{j} - 2\lambda\hat{k} \]

### Step 3: Express \(\vec{b}_{2}\)
Since \(\vec{b}_{2} = \vec{b} - \vec{b}_{1}\), we can write:
\[ \vec{b}_{2} = \vec{b} - \vec{b}_{1} = (7\hat{i} + 2\hat{j} - 3\hat{k}) - (2\lambda\hat{i} - \lambda\hat{j} - 2\lambda\hat{k}) \]
This simplifies to:
\[ \vec{b}_{2} = (7 - 2\lambda)\hat{i} + (2 + \lambda)\hat{j} + (-3 + 2\lambda)\hat{k} \]

### Step 4: Apply the condition for perpendicularity
Since \(\vec{b}_{2}\) is perpendicular to \(\vec{a}\), we have:
\[ \vec{b}_{2} \cdot \vec{a} = 0 \]
Calculating the dot product:
\[ (7 - 2\lambda)(2) + (2 + \lambda)(-1) + (-3 + 2\lambda)(-2) = 0 \]
Expanding this gives:
\[ 14 - 4\lambda - 2 - \lambda + 6 - 4\lambda = 0 \]
Combining like terms:
\[ 18 - 9\lambda = 0 \]
Thus,
\[ 9\lambda = 18 \implies \lambda = 2 \]

### Step 5: Find \(\vec{b}_{1}\) and \(\vec{b}_{2}\)
Substituting \(\lambda = 2\) into \(\vec{b}_{1}\):
\[ \vec{b}_{1} = 2(2\hat{i} - \hat{j} - 2\hat{k}) = 4\hat{i} - 2\hat{j} - 4\hat{k} \]
Now substituting \(\lambda = 2\) into \(\vec{b}_{2}\):
\[ \vec{b}_{2} = (7 - 2(2))\hat{i} + (2 + 2)\hat{j} + (-3 + 2(2))\hat{k} \]
This simplifies to:
\[ \vec{b}_{2} = (7 - 4)\hat{i} + (2 + 2)\hat{j} + (-3 + 4)\hat{k} = 3\hat{i} + 4\hat{j} + 1\hat{k} \]

### Step 6: Write the final expression for \(\vec{b}\)
Thus, we can express \(\vec{b}\) as:
\[ \vec{b} = \vec{b}_{1} + \vec{b}_{2} = (4\hat{i} - 2\hat{j} - 4\hat{k}) + (3\hat{i} + 4\hat{j} + 1\hat{k}) \]

### Final Result
\[ \vec{b} = (4 + 3)\hat{i} + (-2 + 4)\hat{j} + (-4 + 1)\hat{k} = 7\hat{i} + 2\hat{j} - 3\hat{k} \]
This confirms that \(\vec{b} = \vec{b}_{1} + \vec{b}_{2}\).