>>7998485
the length of (1, -1, 0) is sqrt(2)
the length of (0, sqrt(2), 2) is sqrt(5)

so it's sqrt5
2,23

unless you're starting at (0,0,0), then all that shit gives you is a direction.

If it's the former, use your trig.

>>7998485
sqrt(6)

http://mathworld.wolfram.com/VectorMagnitude.html

>>7998485
Take it's norm (square all components, add them up and then find the square root of that).
So that's the square root of 6.

>>7998491
you don't even need trig, just pythagorean theorem in 3 dimensions

|d| = sqrt(1^2 + (-1)^2 + 2^2) = sqrt(1 + 1 + 4) = sqrt(6)

>>7998485
a length of a vector is the square root of the sum of the squares of all of the vector's thingies.

[math]\sqrt{6}[/math] assuming an orthonormal base.

>>7998485
4 by the taxicab theorem

you will need at least three set of Pytagoras to make such a calculation

2,45 is only the beginning, my friend

keep making more drawings

>>7998485
take the square of its lengt:
[math]d\times d=(1+ -1+ 2)^2
\\
=(1-1+2-1+1-2+2-2+4)\\
=(4)[/math]
Then you need to divide by the operator norm:
[math]Tr(d)=(1+-1+2)=2\\
length(d)=\frac{4}{2}=2[/math]
Bob's your unle

>>7998490
What are you even doing? Can't see how you came to those leaps.

>>7998485
Its the magnitude, given by 3d Pythagorean: the length is the square root of the sum of the squares of the vector components

Hence:
|d| = √(1 + 1 + 4) ≈ 2.45