Find the direction cosines of the unit vector perpendicular to the plane ` -> rdot(6 hat i-3 hat j-2 hat k)+1=0`passing through the origin.

Explanation

To find the direction cosines of the unit vector perpendicular to the plane given by the equation \( \mathbf{r} \cdot (6 \hat{i} - 3 \hat{j} - 2 \hat{k}) + 1 = 0 \) and passing through the origin, we will follow these steps:

### Step 1: Identify the normal vector of the plane
The equation of the plane can be rewritten in the form \( \mathbf{r} \cdot \mathbf{n} + d = 0 \), where \( \mathbf{n} \) is the normal vector to the plane. From the equation \( \mathbf{r} \cdot (6 \hat{i} - 3 \hat{j} - 2 \hat{k}) + 1 = 0 \), we can identify the normal vector \( \mathbf{n} = 6 \hat{i} - 3 \hat{j} - 2 \hat{k} \).

### Step 2: Calculate the magnitude of the normal vector
To find the unit vector, we first need to calculate the magnitude of the normal vector \( \mathbf{n} \):
\[
...