On the line $l$ find the point equidistant from $A$ and $B$.
$A(3,2,0), \quad B(1,1,1), \quad l : \begin{cases} x = 2t + 1 \\ y = t + 2 \\ z = 2t - 1 \end{cases} t \in \mathbb{R}$
So directional vector is $\vec{l} = [2,1,2]$ and $\vec{AB} = [-2,-1,1]$
This is first time I encounter such task, can someone help me solve it?
On
You started good.
The simplest way to proceed further is considering that the equidistant point shall be on the plane bisecting $AB$ and on the line given. So
Find the bisecting plane of the segment $AB$:
- mid point is ...
- two normal vectors to segment $AB$ are ..
- the parametric equation of the bisecting plane (in the parameters $u,v$ different than $t$) is ...
Find the the intersection of the bisecting plane with the given line : three linear equations in three unknowns ...
You have to solve the equation $(2t+1-3)^{2}+(t+2-2)^{2}+(2t-1)^{2} =(2t+1-1)^{2}+(t+2-1)^{2}+(2t-2)^{2}$. When you expand the sqquares you will get $9t^{2}$ on both sides. Once you cancel this you can easily find $t$. Finally you get the desired point $(x,y,z)$ by plugging in the value of $t$.