In the following, determine the direction-cosines of the normal to the plane and the distance from the origin :
(i) z = 2 (ii) 5y + 8 = 0.

Answer

To solve the problem, we need to determine the direction cosines of the normal to the given planes and the distance from the origin to these planes.

### (i) For the plane \( z = 2 \):

1. **Identify the normal vector**:
The equation \( z = 2 \) can be rewritten in the standard form of a plane equation as \( 0x + 0y + 1z - 2 = 0 \). Here, the coefficients of \( x \), \( y \), and \( z \) give us the normal vector \( \mathbf{n} = (0, 0, 1) \).

2. **Calculate the direction cosines**:
The direction cosines \( l, m, n \) of the normal vector are given by the ratios of the components of the normal vector to its magnitude. The magnitude of the normal vector \( \mathbf{n} \) is:
\[ |\mathbf{n}| = \sqrt{0^2 + 0^2 + 1^2} = 1 \]
Therefore, the direction cosines are:
\[ l = \frac{0}{1} = 0, \quad m = \frac{0}{1} = 0, \quad n = \frac{1}{1} = 1 \]
Thus, the direction cosines of the normal to the plane \( z = 2 \) are \( (0, 0, 1) \).

3. **Calculate the distance from the origin**:
The distance \( d \) from the origin to the plane can be calculated using the formula:
\[ d = \frac{|ax_1 + by_1 + cz_1 + d|}{\sqrt{a^2 + b^2 + c^2}} \]
For our plane \( 0x + 0y + 1z - 2 = 0 \), we have \( a = 0, b = 0, c = 1, d = -2 \) and the coordinates of the origin \( (x_1, y_1, z_1) = (0, 0, 0) \):
\[ d = \frac{|0 \cdot 0 + 0 \cdot 0 + 1 \cdot 0 - 2|}{\sqrt{0^2 + 0^2 + 1^2}} = \frac{|0 - 2|}{1} = \frac{2}{1} = 2 \]
Therefore, the distance from the origin to the plane \( z = 2 \) is \( 2 \) units.

### (ii) For the plane \( 5y + 8 = 0 \):

1. **Identify the normal vector**:
The equation \( 5y + 8 = 0 \) can be rewritten as \( 0x + 5y + 0z + 8 = 0 \). Here, the coefficients give us the normal vector \( \mathbf{n} = (0, 5, 0) \).

2. **Calculate the direction cosines**:
The magnitude of the normal vector \( \mathbf{n} \) is:
\[ |\mathbf{n}| = \sqrt{0^2 + 5^2 + 0^2} = 5 \]
Therefore, the direction cosines are:
\[ l = \frac{0}{5} = 0, \quad m = \frac{5}{5} = 1, \quad n = \frac{0}{5} = 0 \]
Thus, the direction cosines of the normal to the plane \( 5y + 8 = 0 \) are \( (0, 1, 0) \).

3. **Calculate the distance from the origin**:
Using the same distance formula, for the plane \( 0x + 5y + 0z + 8 = 0 \), we have \( a = 0, b = 5, c = 0, d = -8 \):
\[ d = \frac{|0 \cdot 0 + 5 \cdot 0 + 0 \cdot 0 + 8|}{\sqrt{0^2 + 5^2 + 0^2}} = \frac{|8|}{5} = \frac{8}{5} = 1.6 \]
Therefore, the distance from the origin to the plane \( 5y + 8 = 0 \) is \( 1.6 \) units.

### Summary of Results:
- For the plane \( z = 2 \):
- Direction cosines: \( (0, 0, 1) \)
- Distance from the origin: \( 2 \) units.

- For the plane \( 5y + 8 = 0 \):
- Direction cosines: \( (0, 1, 0) \)
- Distance from the origin: \( 1.6 \) units.