Answer
$\dfrac{3}{5\sqrt 2} i+\dfrac{4}{5\sqrt 2}j -\dfrac{1}{\sqrt 2 }k$ and $(\dfrac{1}{2},3,\dfrac{5}{2})$
Work Step by Step
Let us consider $P_1$ and $P_2$ as vectors and, a vector
$u=P_2-P_1=\lt 2,5,0 \gt -\lt -1,1,5 \gt =\lt 3,4,-5 \gt$
or, $u=3 i+4j-5k$
and $|u|=\sqrt{(3)^2+(4)^2+(-5)^2}=\sqrt {50}=5 \sqrt 2$
The unit vector $\hat{\textbf{u}}$ can be calculated as: $\hat{\textbf{u}}=\dfrac{u}{|u|}$
Now, $\hat{\textbf{u}}=\dfrac{3 i+4j-5k}{5 \sqrt 2}=\dfrac{3}{5\sqrt 2} i+\dfrac{4}{5\sqrt 2}j -\dfrac{1}{\sqrt 2 }k$
The mid-point can be found as: $(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2},\dfrac{z_1+z_2}{2})$
Thus, $(\dfrac{2+(-1)}{2},\dfrac{5+1}{2},\dfrac{0+5}{2})=(\dfrac{1}{2},3,\dfrac{5}{2})$
