In a triangle with vertices A (4; -14; 8), B (2; -18; 12), C (12; -8; 12). find the length of the height lowered

In a triangle with vertices A (4; -14; 8), B (2; -18; 12), C (12; -8; 12). find the length of the height lowered from the vertex C to the side AB.

Find the lengths of the sides of the triangle:

a = | AB | = √ (2 ^ 2 + 4 ^ 2 + (-4) ^ 2) = √36 = 6

b = | BC | = √ (10 ^ 2 + (-10) ^ 2 +0) = 10 * √2

c = | AC | = √ ((- 8) ^ 2 + (-6) ^ 2 + (-4) ^ 2) = √116 = 2 √29

Let’s designate the point of intersection of the height with the side | AB | via M

Then, by the Pythagorean theorem:

CM ^ 2 + AM ^ 2 = AC ^ 2

CM ^ 2 + MB ^ 2 = CB ^ 2

We denote the height of CM by h, | AM | through x. We get the system of equations:

x ^ 2 + h ^ 2 = (2√29) ^ 2

h ^ 2 + (x – 6) ^ 2 = (10√2) ^ 2

x = √116 – h ^ 2

Let’s substitute the second:

h ^ 2 + (√ (116 – h ^ 2)) ^ 2 = 116

h = 8√2

Answer: height 8√2.