plane perpendicular to the straight line I have this problem to find the equation of the plane perpendicular to the straight line joining the points $(1, 3, 5)$ and $(4, 3 ,2)$ at its middle points.  How would you solve this?
 A: Any plane can be expressed uniquely by a normal direction, and a point through which it passes. You have been given both these things, only you need to extract them.
The normal direction $\hat{n} = \left(\frac{1}{\sqrt{2}}, 0, -\frac{1}{\sqrt{2}}\right)$
The point through which it passes $\vec{O} = \left(\frac{5}{2}, 3, \frac{7}{2}\right)$
Now assume the plane has the equation
$$ax + by + cz = d$$
Here, the tuple $(a,b,c)$ correspond to the plane normal, and will be parallel to it, and $d$ is used to ensure it passes through $\vec{O}$
$$\pi: x - z = -1$$
A: The formula for it is: $(X-X_0)\cdot \vec{n}=0$ whereas $X = (x,y,z), X_0 = \left(\frac{1+4}{2}, \frac{3+3}{2}, \frac{2+5}{2}\right)=\left(\frac{5}{2}, 3, \frac{7}{2}\right)$ , and $\vec{n}= \dfrac{1}{\sqrt{3^2+0^2+(-3)^2}}\left(4-1,3-3,2-5\right)=\dfrac{1}{2\sqrt{3}}\left(3,0,-3\right)=\left(\frac{\sqrt{3}}{2}, 0,\dfrac{-\sqrt{3}}{2}\right)$. Can you complete the equation?
A: The line has direction vector $(4-1,3-3,2-5)=(3,0,-3)$.
The midpoint is $\dfrac{(1,3,5)+(4,3,2)}2=(2.5,3,3.5)$.
Therefore, the equation of the plane is $3x+0y+-3z=3(2.5)-3(3.5)$;
i.e., $3x-3z=-3 $, or $x-z=-1$.
