If `veca =hati + hatj-hatk, vecb = - hati + 2hatj + 2hatk and vecc = - hati +2hatj -hatk` , then a unit vector normal to the vectors `veca + vecb and vecb -vecc`, is

Options

A. `hati`
B. `hatj`
C. `hatk`
D. none of these

Explanation

To find a unit vector normal to the vectors \(\vec{a} + \vec{b}\) and \(\vec{b} - \vec{c}\), we will follow these steps:

### Step 1: Calculate \(\vec{a} + \vec{b}\)

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

Now, we add the two vectors:
\[ \vec{a} + \vec{b} = (\hat{i} - \hat{i}) + (\hat{j} + 2\hat{j}) + (-\hat{k} + 2\hat{k}) \]
\[ = 0\hat{i} + 3\hat{j} + 1\hat{k} \]
\[ = 3\hat{j} + \hat{k} \]

### Step 2: Calculate \(\vec{b} - \vec{c}\)

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

Now, we subtract \(\vec{c}\) from \(\vec{b}\):
\[ \vec{b} - \vec{c} = (-\hat{i} + 2\hat{j} + 2\hat{k}) - (-\hat{i} + 2\hat{j} - \hat{k}) \]
\[ = (-\hat{i} + 2\hat{j} + 2\hat{k}) + \hat{i} - 2\hat{j} + \hat{k} \]
\[ = 0\hat{i} + 0\hat{j} + 3\hat{k} \]
\[ = 3\hat{k} \]

### Step 3: Find the cross product \((\vec{a} + \vec{b}) \times (\vec{b} - \vec{c})\)

Now we find the cross product:
\[ (3\hat{j} + \hat{k}) \times (3\hat{k}) \]

Using the determinant method for the cross product:
\[ \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 3 & 1 \\ 0 & 0 & 3 \end{vmatrix} \]

Calculating the determinant:
\[ = \hat{i} \begin{vmatrix} 3 & 1 \\ 0 & 3 \end{vmatrix} - \hat{j} \begin{vmatrix} 0 & 1 \\ 0 & 3 \end{vmatrix} + \hat{k} \begin{vmatrix} 0 & 3 \\ 0 & 0 \end{vmatrix} \]
\[ = \hat{i} (3 \cdot 3 - 0 \cdot 1) - \hat{j} (0 - 0) + \hat{k} (0 - 0) \]
\[ = 9\hat{i} + 0\hat{j} + 0\hat{k} \]
\[ = 9\hat{i} \]

### Step 4: Normalize the resulting vector

To find the unit vector, we divide by its magnitude:
\[ \text{Magnitude} = \sqrt{9^2} = 9 \]
\[ \text{Unit vector} = \frac{9\hat{i}}{9} = \hat{i} \]

### Final Answer

The unit vector normal to the vectors \(\vec{a} + \vec{b}\) and \(\vec{b} - \vec{c}\) is:
\[ \hat{i} \]