Find the unit vector perpendicular the plane `rcdot(2hat(i)+hat(j)+2hat(k))=5`.

Answer

To find the unit vector perpendicular to the plane given by the equation \( \mathbf{r} \cdot (2\hat{i} + \hat{j} + 2\hat{k}) = 5 \), we can follow these steps:

### Step 1: Identify the normal vector of the plane
The equation of the plane can be expressed in the form \( \mathbf{r} \cdot \mathbf{n} = d \), where \( \mathbf{n} \) is the normal vector to the plane. From the given equation, we can identify:
\[ \mathbf{n} = 2\hat{i} + \hat{j} + 2\hat{k} \]

### Step 2: Calculate the magnitude of the normal vector
The magnitude of the normal vector \( \mathbf{n} \) is calculated using the formula:
\[ |\mathbf{n}| = \sqrt{n_x^2 + n_y^2 + n_z^2} \]
where \( n_x, n_y, n_z \) are the components of the normal vector. Thus, we have:
\[ |\mathbf{n}| = \sqrt{2^2 + 1^2 + 2^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3 \]

### Step 3: Find the unit vector
The unit vector \( \mathbf{u} \) in the direction of the normal vector is given by:
\[ \mathbf{u} = \frac{\mathbf{n}}{|\mathbf{n}|} \]
Substituting the values we found:
\[ \mathbf{u} = \frac{2\hat{i} + \hat{j} + 2\hat{k}}{3} \]
This simplifies to:
\[ \mathbf{u} = \frac{2}{3}\hat{i} + \frac{1}{3}\hat{j} + \frac{2}{3}\hat{k} \]

### Final Answer
The unit vector perpendicular to the plane is:
\[ \mathbf{u} = \frac{2}{3}\hat{i} + \frac{1}{3}\hat{j} + \frac{2}{3}\hat{k} \]

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