The vertices of `DeltaABC` are (2, 0, 0), B(0, 1, 0), C(0, 0, 2). Its orthocentre is H and circumcentre is S. P is a point equidistant from A, B, C and the origin O.
Q. The z-coordinate of H is :

Options

A. 1
B. `1//2`
C. `1//6`
D. `1//3`

Answer

To find the z-coordinate of the orthocenter \( H \) of triangle \( ABC \) with vertices \( A(2, 0, 0) \), \( B(0, 1, 0) \), and \( C(0, 0, 2) \), we will follow these steps:

### Step 1: Find the equation of line AC
The line \( AC \) can be expressed using the parametric form. The coordinates of points \( A \) and \( C \) are:
- \( A(2, 0, 0) \)
- \( C(0, 0, 2) \)

Using the formula for the equation of a line:
\[ \frac{x - x_1}{x_2 - x_1} = \frac{y - y_1}{y_2 - y_1} = \frac{z - z_1}{z_2 - z_1} \]
Substituting the coordinates of \( A \) and \( C \):
\[ \frac{x - 2}{0 - 2} = \frac{y - 0}{0 - 0} = \frac{z - 0}{2 - 0} \]
This simplifies to:
\[ \frac{x - 2}{-2} = \frac{z}{2} \]
Letting this equal a parameter \( k \):
\[ x = -2k + 2, \quad y = 0, \quad z = 2k \]

### Step 2: Find the coordinates of point D on line AC
Point \( D \) lies on line \( AC \) and is such that the distance from point \( B \) to point \( D \) is minimized. The coordinates of point \( D \) in terms of \( k \) are:
\[ D(-2k + 2, 0, 2k) \]

### Step 3: Calculate the distance BD
The coordinates of point \( B \) are \( (0, 1, 0) \). The distance \( BD \) is given by:
\[ BD = \sqrt{((-2k + 2) - 0)^2 + (0 - 1)^2 + (2k - 0)^2} \]
This simplifies to:
\[ BD = \sqrt{(-2k + 2)^2 + (-1)^2 + (2k)^2} \]
Expanding this:
\[ = \sqrt{(4k^2 - 8k + 4) + 1 + 4k^2} = \sqrt{8k^2 - 8k + 5} \]

### Step 4: Minimize the distance BD
To minimize \( BD \), we will differentiate the expression inside the square root:
Let \( f(k) = 8k^2 - 8k + 5 \).
The derivative is:
\[ f'(k) = 16k - 8 \]
Setting the derivative to zero for minimization:
\[ 16k - 8 = 0 \implies k = \frac{1}{2} \]

### Step 5: Find the coordinates of D using k
Substituting \( k = \frac{1}{2} \) into the coordinates of \( D \):
\[ D = (-2 \cdot \frac{1}{2} + 2, 0, 2 \cdot \frac{1}{2}) = (1, 0, 1) \]

### Step 6: Find the z-coordinate of the orthocenter H
Since \( BD \) is perpendicular to \( AC \) at point \( D \), and the z-coordinate of point \( D \) is \( 1 \), the z-coordinate of the orthocenter \( H \) is also \( 1 \).

Thus, the z-coordinate of \( H \) is:
\[ \boxed{1} \]