I have a problem while trying to calculating the distance form point $B(0,3,4)$ to the y-axis in xyz coordinate system. Here is how I calculate it:
There is point $Y(0,t,0)$ belonging to the y-axis so the vector $YB$ will be $(0,3-t,4)$.
Vector $YB$ must be perpendicular to y-axis which mean the dot product of $YB(0,3-t,4)$ with $OY(0,t,0)$ must be zero: $$ t(3-t) = 0 $$ So $t=0$ or $t=3$. The problem is here, by using graph I can calculate the distance between $B$ and y-axis is 4 which means $$ \sqrt{(3-t)^2+4^2} = 4 $$ So $t$ must be 3 only, but the result shows that $t=0$ or $t=3$. There is no way $t=0$, but I don't find any problem with the method I used.
If $t=0$ then $OY$ is $(0,0,0)$. So $OY.v = 0$ for all vectors $v$, but that doesn't make $OY$ a solution. Instead of using $OY=(0,t,0)$, you should have used a guaranteed non-zero vector parallel to the $y$-axis, such as (for instance) $(0,1,0)$.