After thinking longing i can't figure it out no matter what..
So i have 3d line starts (0,0,0) and ends (3.5,3.5,2.5) so therefore has length of about 5.
Now how do i find out vector that is completely perpendicular to any vector like this one (1,2,3) assume all vectors start at 0,0,0.
If it was 2d i'd have used slope to find perpendicular line.
Please use more English and if you put letters like t or stuff explain it along too.
On
There exists a subspace of perpendicular vectors for any given vector. To find a perpendicular vector to any two vectors you can take their cross-product. To obtain a desired length, normalize and multiply by the desired length.
Consider the inner product: $\vec{u}\cdot\vec{v}=|\vec{u}||\vec{v}|\cos\theta$. What conditions can you impose to derive unknown components?
The vectors orthogonal to $\mathbf{u} \in \mathbb{R}^n$ are the vectors $\mathbf{v} \in \mathbb{R}^n$ such that $\mathbf{u} \cdot \mathbf{v}=0$, where $\cdot$ is the dot product. In the case $\mathbf{u}=(3.5,3.5,2.5)$, we can find all such $\mathbf{v}=(a,b,c) \in \mathbb{R}^3$ by solving the equation defined by $$(3.5,3.5,2.5) \cdot (a,b,c)=0.$$ That is, all vectors $(a,b,c)$ such that $$3.5a+3.5b+2.5c=0$$ will be orthogonal to $(3.5,3.5,2.5)$.