Find a unit vector perpendicular to each of the vectors `veca+vecb,veca-vecb" where vectors a "veca=3hati+2hatj+2hatkandvecb=hati+2hatj-2hatk`

Answer

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

### Step 1: Define the vectors
Given:
\[ \vec{a} = 3\hat{i} + 2\hat{j} + 2\hat{k} \]
\[ \vec{b} = \hat{i} + 2\hat{j} - 2\hat{k} \]

### Step 2: Calculate \(\vec{a} + \vec{b}\)
\[ \vec{a} + \vec{b} = (3\hat{i} + 2\hat{j} + 2\hat{k}) + (\hat{i} + 2\hat{j} - 2\hat{k}) \]
Combine the components:
\[ = (3 + 1)\hat{i} + (2 + 2)\hat{j} + (2 - 2)\hat{k} = 4\hat{i} + 4\hat{j} + 0\hat{k} \]
Thus,
\[ \vec{a} + \vec{b} = 4\hat{i} + 4\hat{j} \]

### Step 3: Calculate \(\vec{a} - \vec{b}\)
\[ \vec{a} - \vec{b} = (3\hat{i} + 2\hat{j} + 2\hat{k}) - (\hat{i} + 2\hat{j} - 2\hat{k}) \]
Combine the components:
\[ = (3 - 1)\hat{i} + (2 - 2)\hat{j} + (2 + 2)\hat{k} = 2\hat{i} + 0\hat{j} + 4\hat{k} \]
Thus,
\[ \vec{a} - \vec{b} = 2\hat{i} + 4\hat{k} \]

### Step 4: Find the cross product of \(\vec{a} + \vec{b}\) and \(\vec{a} - \vec{b}\)
Let \(\vec{u} = \vec{a} + \vec{b} = 4\hat{i} + 4\hat{j} + 0\hat{k}\) and \(\vec{v} = \vec{a} - \vec{b} = 2\hat{i} + 0\hat{j} + 4\hat{k}\).

The cross product \(\vec{u} \times \vec{v}\) is given by the determinant:
\[ \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & 4 & 0 \\ 2 & 0 & 4 \end{vmatrix} \]

Calculating this determinant:
\[ = \hat{i} \begin{vmatrix} 4 & 0 \\ 0 & 4 \end{vmatrix} - \hat{j} \begin{vmatrix} 4 & 0 \\ 2 & 4 \end{vmatrix} + \hat{k} \begin{vmatrix} 4 & 4 \\ 2 & 0 \end{vmatrix} \]
Calculating each of these 2x2 determinants:
\[ = \hat{i} (4 \cdot 4 - 0 \cdot 0) - \hat{j} (4 \cdot 4 - 0 \cdot 2) + \hat{k} (4 \cdot 0 - 4 \cdot 2) \]
\[ = 16\hat{i} - 16\hat{j} - 8\hat{k} \]

Thus, the cross product is:
\[ \vec{c} = 16\hat{i} - 16\hat{j} - 8\hat{k} \]

### Step 5: Find the magnitude of \(\vec{c}\)
\[ |\vec{c}| = \sqrt{(16)^2 + (-16)^2 + (-8)^2} = \sqrt{256 + 256 + 64} = \sqrt{576} = 24 \]

### Step 6: Find the unit vector
The unit vector \(\hat{n}\) in the direction of \(\vec{c}\) is given by:
\[ \hat{n} = \frac{\vec{c}}{|\vec{c}|} = \frac{16\hat{i} - 16\hat{j} - 8\hat{k}}{24} \]
Simplifying this:
\[ = \frac{16}{24}\hat{i} - \frac{16}{24}\hat{j} - \frac{8}{24}\hat{k} = \frac{2}{3}\hat{i} - \frac{2}{3}\hat{j} - \frac{1}{3}\hat{k} \]

### Final Answer:
The unit vector perpendicular to both \(\vec{a} + \vec{b}\) and \(\vec{a} - \vec{b}\) is:
\[ \hat{n} = \frac{2}{3}\hat{i} - \frac{2}{3}\hat{j} - \frac{1}{3}\hat{k} \]