In the following cases, find the co-ordinates of the foot of the perpendicular drawn from the origin to the plane :
(i) 2x + 3y + 4z - 12 = 0
(ii) 3y + 4z - 6 = 0
(iii) x + y + z = 1
(iv) 5y + 8 = 0.

Answer

To find the coordinates of the foot of the perpendicular drawn from the origin (0, 0, 0) to the given planes, we will use the formula derived from the general equation of a plane \(Ax + By + Cz + D = 0\). The coordinates of the foot of the perpendicular can be calculated as follows:

\[ x_1 = -\frac{AD}{A^2 + B^2 + C^2}, \quad y_1 = -\frac{BD}{A^2 + B^2 + C^2}, \quad z_1 = -\frac{CD}{A^2 + B^2 + C^2} \]

Where \(A\), \(B\), \(C\) are the coefficients of \(x\), \(y\), and \(z\) in the plane equation, and \(D\) is the constant term.

### (i) For the plane \(2x + 3y + 4z - 12 = 0\):

1. Identify coefficients:
- \(A = 2\), \(B = 3\), \(C = 4\), \(D = -12\)

2. Calculate \(x_1\):
\[ x_1 = -\frac{2 \times (-12)}{2^2 + 3^2 + 4^2} = -\frac{-24}{4 + 9 + 16} = -\frac{-24}{29} = \frac{24}{29} \]

3. Calculate \(y_1\):
\[ y_1 = -\frac{3 \times (-12)}{2^2 + 3^2 + 4^2} = -\frac{-36}{29} = \frac{36}{29} \]

4. Calculate \(z_1\):
\[ z_1 = -\frac{4 \times (-12)}{2^2 + 3^2 + 4^2} = -\frac{-48}{29} = \frac{48}{29} \]

**Coordinates of the foot of the perpendicular**: \(\left(\frac{24}{29}, \frac{36}{29}, \frac{48}{29}\right)\)

### (ii) For the plane \(3y + 4z - 6 = 0\):

1. Identify coefficients:
- \(A = 0\), \(B = 3\), \(C = 4\), \(D = -6\)

2. Calculate \(x_1\):
\[ x_1 = -\frac{0 \times (-6)}{0^2 + 3^2 + 4^2} = 0 \]

3. Calculate \(y_1\):
\[ y_1 = -\frac{3 \times (-6)}{0^2 + 3^2 + 4^2} = \frac{18}{25} \]

4. Calculate \(z_1\):
\[ z_1 = -\frac{4 \times (-6)}{0^2 + 3^2 + 4^2} = \frac{24}{25} \]

**Coordinates of the foot of the perpendicular**: \(\left(0, \frac{18}{25}, \frac{24}{25}\right)\)

### (iii) For the plane \(x + y + z = 1\):

1. Identify coefficients:
- \(A = 1\), \(B = 1\), \(C = 1\), \(D = -1\)

2. Calculate \(x_1\):
\[ x_1 = -\frac{1 \times (-1)}{1^2 + 1^2 + 1^2} = \frac{1}{3} \]

3. Calculate \(y_1\):
\[ y_1 = -\frac{1 \times (-1)}{1^2 + 1^2 + 1^2} = \frac{1}{3} \]

4. Calculate \(z_1\):
\[ z_1 = -\frac{1 \times (-1)}{1^2 + 1^2 + 1^2} = \frac{1}{3} \]

**Coordinates of the foot of the perpendicular**: \(\left(\frac{1}{3}, \frac{1}{3}, \frac{1}{3}\right)\)

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

1. Identify coefficients:
- \(A = 0\), \(B = 5\), \(C = 0\), \(D = -8\)

2. Calculate \(x_1\):
\[ x_1 = -\frac{0 \times (-8)}{0^2 + 5^2 + 0^2} = 0 \]

3. Calculate \(y_1\):
\[ y_1 = -\frac{5 \times (-8)}{0^2 + 5^2 + 0^2} = \frac{40}{25} = \frac{8}{5} \]

4. Calculate \(z_1\):
\[ z_1 = -\frac{0 \times (-8)}{0^2 + 5^2 + 0^2} = 0 \]

**Coordinates of the foot of the perpendicular**: \(\left(0, \frac{8}{5}, 0\right)\)

### Summary of Results:
1. \(\left(\frac{24}{29}, \frac{36}{29}, \frac{48}{29}\right)\)
2. \(\left(0, \frac{18}{25}, \frac{24}{25}\right)\)
3. \(\left(\frac{1}{3}, \frac{1}{3}, \frac{1}{3}\right)\)
4. \(\left(0, \frac{8}{5}, 0\right)\)