Find the direction-cosines of the perpendicular from the origin to the plane `vec(r). (hati + 2 hatj - 2 hatk)` = 18 .

Answer

To find the direction cosines of the perpendicular from the origin to the plane given by the equation \(\vec{r} \cdot (\hat{i} + 2\hat{j} - 2\hat{k}) = 18\), we can follow these steps:

### Step 1: Identify the normal vector of the plane
The equation of the plane can be rewritten in the form \(\vec{r} \cdot \vec{n} = p\), where \(\vec{n}\) is the normal vector of the plane. From the given equation, we can see that:
\[ \vec{n} = \hat{i} + 2\hat{j} - 2\hat{k} \]

### Step 2: Find the direction ratios of the normal vector
The direction ratios of the normal vector \(\vec{n}\) are simply the coefficients of \(\hat{i}\), \(\hat{j}\), and \(\hat{k}\) in the normal vector. Therefore, the direction ratios are:
\[ 1, 2, -2 \]

### Step 3: Calculate the magnitude of the normal vector
To find the direction cosines, we need the magnitude of the normal vector \(\vec{n}\):
\[ |\vec{n}| = \sqrt{1^2 + 2^2 + (-2)^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3 \]

### Step 4: Find the direction cosines
The direction cosines are given by the ratios of the components of the normal vector to its magnitude. Thus, the direction cosines \(l\), \(m\), and \(n\) are:
\[ l = \frac{1}{3}, \quad m = \frac{2}{3}, \quad n = \frac{-2}{3} \]

### Final Result
The direction cosines of the perpendicular from the origin to the plane are:
\[ \left( \frac{1}{3}, \frac{2}{3}, -\frac{2}{3} \right) \]