A plane is placed through the center of the connecting line of points $A(2,0,1)$ and $B(3,1,3)$ such that it's perpendicular to the connecting line. At which point does this plane cut the $y-$axis?
This got me very confused. Could someone elaborate on what's going on here?
On
Note that the midpoint of $A(2,0,1)$ and $B(3,1,3)$ is $M(\frac52,\frac12,2)$ and the vector $\vec{AB}(1,1,2)$. Let $P(x,y,z)$ be any point in the plane. The given that the plain is perpendicular to the line $AB$ leads to
$$\vec{PM}\cdot\vec{AB}=0\implies (x-\frac52, y-\frac12,z-2)\cdot(1,1,2)=0$$
which yields the equation of the plane
$$x+y+2z=7$$
Set $x=z=0$ to obtain the point of $y$-intersection at $(0,7,0)$.
This question is essentially asking you to find two things: a plane which is perpendicular to and bisects the given line segment, and then the point at which this plane intersects the y-axis.
The first part requires you to find a vector normal to the plane. Given the conditions of the question, can you think of one?
The second part requires you to find a point on the y-axis which is also on the plane. What must be true of x and z if it's on the y-axis?
Hope this helps!