Question

The distance between the point (2,1,1) and the plane x-2y=5 is​

Answer 1

The given plane has normal vector

$x-2y=5\implies\mathbf n=\langle1,-2,0\rangle$

Scaling n by a real number t gives a set of vectors that span an entire line through the origin. Translating this line by adding the vector <2, 1, 1> makes it so that this line passes through the point (2, 1, 1). So this line has equation

$\mathbf r(t)=\langle1,-2,0\rangle t+\langle2,1,1\rangle=\langle 2+t, 1-2t, 1\rangle$

This line passes through (2, 1, 1) when t = 0, and the line intersects with the plane when

$x-2y=5\implies(2+t)-2(1-2t)=5\implies5t=5\implies t=1$

which corresponds the point (3, -1, 1) (simply plug t = 1 into the coordinates of $\mathbf r(t)$).

So the distance between the plane and the point is the distance between the points (2, 1, 1) and (3, -1, 1):

$\sqrt{(2-3)^2+(1-(-1))^2+(1-1)^2}=\boxed{\sqrt5}$