Vtyjkyuk

How do I find the vector length for high dimensions?.We can find vector length for 3d with the formula $sqrtv_1^2+v_2^2+v_3^2$
Likewise how to find the vector magnitude for high dimensions?

The same way:$$bigl|(x_1,ldots,x_n)bigr|=sqrtx_1^2+x_2^2+cdots+x_n^2.$$This is the usual norm in $mathbbR^n$.

Given a vector $vecv=(v_1,v_2,v_3,v_4,...,v_n)$ in $n$ dimensions, the formula for the magnitude (or length) remains the same: it the the square root of the sum of the squares of the components of the vector:$$|vecv|=sqrtv_1^2+v_2^2+v_3^2+v_4^2+...+v_n^2$$

As a side answer, since the OP do not understand how "orthogonal in high dimensions" can be represented...

Well, it is normal. We are living in a seemingly 3D spatial dimensions world, so we can easily represent ourselves in such a space, and understand concepts with parallels in this 3D spatial world.

However, it is, for our brain, much more difficult to have a spatial representation of something in 4D, 5D etc. This is why we are using more formalism in order to be able to go beyond the limits of our perception.

Imagine you are a kind of microscopic worm living at the surface of a giant leaf. For you, everything is in 2D (remember that ourselves we believed that the earth - not our entire world - was 2D at some point in time, because when you are at the surface, you don't seeit otherwise). You can go forward, left, right, backward, but that's it... You don't know if there are other dimensions, and, should you have a brain caable of abstract thought, it would nevertheless be puzzling to think spatially about a space in 3D.

When you want to do higher level mathematics (and in particular geometry), you need to distance yourself in a certain extent to the notions you gained with analogies with the real world. I say to a certain extent, for doing parallels to the 3D case sometimes can bring sometimes insight about a problem (and sometimes it is misleading).