A unit vector coplanar with `i+j+2k` and `i+2j+k` and perpendicular to `i+j+k` is.......

Answer

To find a unit vector that is coplanar with the vectors \( \mathbf{a} = \mathbf{i} + \mathbf{j} + 2\mathbf{k} \) and \( \mathbf{b} = \mathbf{i} + 2\mathbf{j} + \mathbf{k} \), and perpendicular to \( \mathbf{c} = \mathbf{i} + \mathbf{j} + \mathbf{k} \), we can follow these steps:

### Step 1: Define the vectors
Let:
- \( \mathbf{a} = \mathbf{i} + \mathbf{j} + 2\mathbf{k} \)
- \( \mathbf{b} = \mathbf{i} + 2\mathbf{j} + \mathbf{k} \)
- \( \mathbf{c} = \mathbf{i} + \mathbf{j} + \mathbf{k} \)

### Step 2: Find a vector coplanar with \( \mathbf{a} \) and \( \mathbf{b} \)
A vector \( \mathbf{r} \) that is coplanar with \( \mathbf{a} \) and \( \mathbf{b} \) can be expressed as:
\[ \mathbf{r} = \mathbf{a} + \lambda \mathbf{b} \]
where \( \lambda \) is a scalar.

### Step 3: Write the expression for \( \mathbf{r} \)
Substituting \( \mathbf{a} \) and \( \mathbf{b} \) into the equation:
\[ \mathbf{r} = (\mathbf{i} + \mathbf{j} + 2\mathbf{k}) + \lambda (\mathbf{i} + 2\mathbf{j} + \mathbf{k}) \]
This simplifies to:
\[ \mathbf{r} = (1 + \lambda)\mathbf{i} + (1 + 2\lambda)\mathbf{j} + (2 + \lambda)\mathbf{k} \]

### Step 4: Set up the perpendicular condition
Since \( \mathbf{r} \) is perpendicular to \( \mathbf{c} \), we have:
\[ \mathbf{r} \cdot \mathbf{c} = 0 \]
Calculating the dot product:
\[ (1 + \lambda)(1) + (1 + 2\lambda)(1) + (2 + \lambda)(1) = 0 \]
This expands to:
\[ (1 + \lambda) + (1 + 2\lambda) + (2 + \lambda) = 0 \]
Combining like terms gives:
\[ 4 + 4\lambda = 0 \]

### Step 5: Solve for \( \lambda \)
From the equation \( 4 + 4\lambda = 0 \):
\[ 4\lambda = -4 \implies \lambda = -1 \]

### Step 6: Substitute \( \lambda \) back into \( \mathbf{r} \)
Substituting \( \lambda = -1 \) into the expression for \( \mathbf{r} \):
\[ \mathbf{r} = (1 - 1)\mathbf{i} + (1 - 2)\mathbf{j} + (2 - 1)\mathbf{k} = 0\mathbf{i} - 1\mathbf{j} + 1\mathbf{k} \]
Thus, we have:
\[ \mathbf{r} = -\mathbf{j} + \mathbf{k} \]

### Step 7: Find the unit vector
To find the unit vector \( \mathbf{u} \) in the direction of \( \mathbf{r} \):
1. Calculate the magnitude of \( \mathbf{r} \):
\[ |\mathbf{r}| = \sqrt{0^2 + (-1)^2 + 1^2} = \sqrt{0 + 1 + 1} = \sqrt{2} \]
2. The unit vector \( \mathbf{u} \) is given by:
\[ \mathbf{u} = \frac{\mathbf{r}}{|\mathbf{r}|} = \frac{-\mathbf{j} + \mathbf{k}}{\sqrt{2}} = -\frac{1}{\sqrt{2}}\mathbf{j} + \frac{1}{\sqrt{2}}\mathbf{k} \]

### Final Answer
The required unit vector is:
\[ \mathbf{u} = -\frac{1}{\sqrt{2}}\mathbf{j} + \frac{1}{\sqrt{2}}\mathbf{k} \]