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

Answer

To find a unit vector perpendicular to the vectors \(2\vec{A} + \vec{B}\) and \(\vec{A} - 2\vec{B}\), where \(\vec{A} = \hat{i} + 2\hat{j} - \hat{k}\) and \(\vec{B} = \hat{i} + \hat{j} + \hat{k}\), we can follow these steps:

### Step 1: Calculate \(2\vec{A} + \vec{B}\)

First, we need to calculate \(2\vec{A}\) and then add \(\vec{B}\):

\[ 2\vec{A} = 2(\hat{i} + 2\hat{j} - \hat{k}) = 2\hat{i} + 4\hat{j} - 2\hat{k} \]

Now, add \(\vec{B}\):

\[ \vec{B} = \hat{i} + \hat{j} + \hat{k} \]

So,

\[ 2\vec{A} + \vec{B} = (2\hat{i} + 4\hat{j} - 2\hat{k}) + (\hat{i} + \hat{j} + \hat{k}) = (2 + 1)\hat{i} + (4 + 1)\hat{j} + (-2 + 1)\hat{k} = 3\hat{i} + 5\hat{j} - \hat{k} \]

### Step 2: Calculate \(\vec{A} - 2\vec{B}\)

Next, we calculate \(-2\vec{B}\):

\[ -2\vec{B} = -2(\hat{i} + \hat{j} + \hat{k}) = -2\hat{i} - 2\hat{j} - 2\hat{k} \]

Now, add \(\vec{A}\):

\[ \vec{A} - 2\vec{B} = (\hat{i} + 2\hat{j} - \hat{k}) + (-2\hat{i} - 2\hat{j} - 2\hat{k}) = (1 - 2)\hat{i} + (2 - 2)\hat{j} + (-1 - 2)\hat{k} = -\hat{i} + 0\hat{j} - 3\hat{k} \]

### Step 3: Find the cross product \((2\vec{A} + \vec{B}) \times (\vec{A} - 2\vec{B})\)

Now we compute the cross product:

\[ \vec{U} = 2\vec{A} + \vec{B} = 3\hat{i} + 5\hat{j} - \hat{k} \]
\[ \vec{V} = \vec{A} - 2\vec{B} = -\hat{i} - 3\hat{k} \]

We can set up the determinant for the cross product:

\[ \vec{U} \times \vec{V} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 5 & -1 \\ -1 & 0 & -3 \end{vmatrix} \]

Calculating the determinant:

\[ = \hat{i} \begin{vmatrix} 5 & -1 \\ 0 & -3 \end{vmatrix} - \hat{j} \begin{vmatrix} 3 & -1 \\ -1 & -3 \end{vmatrix} + \hat{k} \begin{vmatrix} 3 & 5 \\ -1 & 0 \end{vmatrix} \]

Calculating each of these 2x2 determinants:

1. \(\begin{vmatrix} 5 & -1 \\ 0 & -3 \end{vmatrix} = (5)(-3) - (0)(-1) = -15\)
2. \(\begin{vmatrix} 3 & -1 \\ -1 & -3 \end{vmatrix} = (3)(-3) - (-1)(-1) = -9 - 1 = -10\)
3. \(\begin{vmatrix} 3 & 5 \\ -1 & 0 \end{vmatrix} = (3)(0) - (5)(-1) = 0 + 5 = 5\)

Putting it all together:

\[ \vec{U} \times \vec{V} = -15\hat{i} + 10\hat{j} + 5\hat{k} \]

### Step 4: Find the unit vector

Now, we need to find the magnitude of the vector \(-15\hat{i} + 10\hat{j} + 5\hat{k}\):

\[ \|\vec{U} \times \vec{V}\| = \sqrt{(-15)^2 + (10)^2 + (5)^2} = \sqrt{225 + 100 + 25} = \sqrt{350} = 5\sqrt{14} \]

Now, the unit vector is given by:

\[ \hat{n} = \frac{\vec{U} \times \vec{V}}{\|\vec{U} \times \vec{V}\|} = \frac{-15\hat{i} + 10\hat{j} + 5\hat{k}}{5\sqrt{14}} = \frac{-3\hat{i} + 2\hat{j} + \hat{k}}{\sqrt{14}} \]

### Final Answer

Thus, the required unit vector perpendicular to both \(2\vec{A} + \vec{B}\) and \(\vec{A} - 2\vec{B}\) is:

\[ \hat{n} = \frac{-3\hat{i} + 2\hat{j} + \hat{k}}{\sqrt{14}} \]