What is the distance from the point (1,2,4) to the plane 2x+y+z=12 ?
I got 5√6 / 6
Anybody agree?
The distance from the point $\displaystyle (x_0,y_0,z_0)$ to the plane $\displaystyle ax+by+cz=d $ is given by $\displaystyle \dfrac{|ax_0+by_0+cz_0-d|}{\sqrt{a^2+b^2+c^2}}$. Plugging in the numbers for this problem, I get $\displaystyle 4\sqrt6/6$.