I am trying to find the shortest distance from the point $(3,0,-2)$ to the plane $x+y+z = 2$ and I keep getting the same incorrect solution.

Question

Can someone tell me where I am going wrong?

Answer 1

Hint: $$d=\frac{|Ax_0+By_0+Cz_0+D|}{\sqrt{A^2+B^2+C^2}},$$ where $(x_0,y_0,z_0)$ are the coordinates of the point, and $Ax+By+Cz+D=0$ is the equation of the plane

Answer 2

Here is the best way to find the distance from the poinnt $P_1(x_1,y_1,z_1)$ to the plane $\Pi$ with equation $$(1) \text { } \text { } \text { }\text { }\text { }\text { }\text { }\text { }\text { }\text { } ax+by+cz+d=0.$$ First drop a perpendicular $L$ from the point $P_1$ to the plane $\Pi.$ The normal vector to the plane $\Pi$ is $[a,b,c],$ so parametric equations of $L$ are $$(2) \text { } \text { } \text { }\text { }\text { }\text { }\text { }\text { }\text { }\text { }x=x_1+at,y=y_1+bt,z=z_1+ct.$$ Substitute these expressions for $x,y \text { and }z$ into (1) and solve for $t$ in terms of $a,b,c \text { and }d. $ Substitute this value of $t$ into the expressions in (2) for $x,y \text { and }z$ to give $$x_*=x_1+at,y_*=y_1+bt,z_*=z_1+ct.$$ The line $L$ intersects the plane $\Pi$ in the point $P_*(x_*,y_*,z_*)$ , which is the foot of the perpendicular dropped from the point $P_1$ to the plane $\Pi.$ The shortest distance from the point $P_1$ to the plane $\Pi$ is the distance between the points $P_1$ and $P_*.$