Question

orthocentre of 1 0 0 , 0 2 0 and 0 0 3

Answer 1

(36/49, 18/49, 12/49)

Answer 2

Let the vertices of the triangle be A = (1, 0, 0), B = (0, 2, 0), and C = (0, 0, 3).
Let H = (x, y, z) be the orthocentre of the triangle ABC. The orthocentre is the intersection point of the altitudes of the triangle. An altitude from a vertex is perpendicular to the opposite side.

1. The altitude from A to BC is perpendicular to the vector $\vec{BC} = C - B = (0, 0, 3) - (0, 2, 0) = (0, -2, 3)$.
   The vector $\vec{AH} = H - A = (x - 1, y, z)$.
   Since $\vec{AH}$ is perpendicular to $\vec{BC}$, their dot product is zero:
   $\vec{AH} \cdot \vec{BC} = (x - 1)(0) + y(-2) + z(3) = 0$
   $-2y + 3z = 0$ (Equation 1)

2. The altitude from B to AC is perpendicular to the vector $\vec{AC} = C - A = (0, 0, 3) - (1, 0, 0) = (-1, 0, 3)$.
   The vector $\vec{BH} = H - B = (x, y - 2, z)$.
   Since $\vec{BH}$ is perpendicular to $\vec{AC}$, their dot product is zero:
   $\vec{BH} \cdot \vec{AC} = x(-1) + (y - 2)(0) + z(3) = 0$
   $-x + 3z = 0$ (Equation 2)

3. The altitude from C to AB is perpendicular to the vector $\vec{AB} = B - A = (0, 2, 0) - (1, 0, 0) = (-1, 2, 0)$.
   The vector $\vec{CH} = H - C = (x, y, z - 3)$.
   Since $\vec{CH}$ is perpendicular to $\vec{AB}$, their dot product is zero:
   $\vec{CH} \cdot \vec{AB} = x(-1) + y(2) + (z - 3)(0) = 0$
   $-x + 2y = 0$ (Equation 3)

From Equation 2, we have $x = 3z$.
From Equation 3, we have $x = 2y$.
Thus, $2y = 3z$. This is consistent with Equation 1 ($3z - 2y = 0$).

We have two independent equations relating x, y, and z: $x = 3z$
$y = \frac{3}{2}z$

The orthocentre H must lie on the plane containing the triangle ABC. The equation of the plane passing through A(1, 0, 0), B(0, 2, 0), and C(0, 0, 3) can be found using the intercept form: $\frac{x}{1} + \frac{y}{2} + \frac{z}{3} = 1$
Multiplying by the LCM of 1, 2, and 3 (which is 6), we get:
$6x + 3y + 2z = 6$ (Equation 4)

Now, substitute the expressions for x and y in terms of z from the altitude conditions into the plane equation: $6(3z) + 3(\frac{3}{2}z) + 2z = 6$
$18z + \frac{9}{2}z + 2z = 6$
Multiply the entire equation by 2 to eliminate the fraction:
$36z + 9z + 4z = 12$
$49z = 12$
$z = \frac{12}{49}$

Now find x and y using the value of z:
$x = 3z = 3 \times \frac{12}{49} = \frac{36}{49}$
$y = \frac{3}{2}z = \frac{3}{2} \times \frac{12}{49} = \frac{3 \times 6}{49} = \frac{18}{49}$

So, the orthocentre H is $(\frac{36}{49}, \frac{18}{49}, \frac{12}{49})$.

Alternatively, for a triangle with vertices on the coordinate axes at (a, 0, 0), (0, b, 0), (0, 0, c), the orthocentre (x, y, z) is given by the formula: $x = \frac{1/a}{1/a^2 + 1/b^2 + 1/c^2}$
$y = \frac{1/b}{1/a^2 + 1/b^2 + 1/c^2}$
$z = \frac{1/c}{1/a^2 + 1/b^2 + 1/c^2}$
In this problem, a = 1, b = 2, c = 3.
$1/a = 1$, $1/b = 1/2$, $1/c = 1/3$.
$1/a^2 = 1$, $1/b^2 = 1/4$, $1/c^2 = 1/9$.
The denominator is $D = \frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2} = 1 + \frac{1}{4} + \frac{1}{9} = \frac{36 + 9 + 4}{36} = \frac{49}{36}$.
$x = \frac{1/a}{D} = \frac{1}{49/36} = \frac{36}{49}$
$y = \frac{1/b}{D} = \frac{1/2}{49/36} = \frac{1}{2} \times \frac{36}{49} = \frac{18}{49}$
$z = \frac{1/c}{D} = \frac{1/3}{49/36} = \frac{1}{3} \times \frac{36}{49} = \frac{12}{49}$
The orthocentre is $(\frac{36}{49}, \frac{18}{49}, \frac{12}{49})$.