Find the shortest distance between line and plane in this cuboid? Given a cuboid ABCD.EFGH with length AB=6 cm, BC=4 cm, and AE= 3 cm. P is located in the middle of line segment DC. Find the shortest distance from A to EBP?
My attempt:
At the first, I think I need to draw a illustration for this problem, such:
 
From this illustration, I can analyse the problem as follows:


*

*Draw plane EBP 

*Draw a line from A that intersect plane EBP, line AC. This line intersect EBP at O

*Draw a line from intersection of O and EBP, line EO.

*Then, for the last, draw a perpendicular line from A to EO, line AX

*line AX= shortest distance between A to EBP
I try to find the answer, and I got $\frac{12\sqrt{13}}{17}$ as the answer. But, the key says that the answer should be $\frac{24\sqrt{89}}{89}$
Can someone try to provide me another way to solve this kind of problem and state where is my mistake to solve this problem? 
 A: set the origin in A and then calculate the equation of the plane EBP finally calculate the distance of the plane from the origin A
if the plane equation is ax+by+cz=d the distance from the origin is given by
$$\frac {|d|}{\sqrt{a^2+b^2+c^2}}$$
Assuming the origin of the axis in A, AB=x-axis, AD=y-axis and AE= z-axis the equation of the plane EBP can be done at least in two ways:
1) by direct calculation imposing that E,B and P $\in%$ plane (you can assume wlog that d=1):
$$E=(0,0,3): a\cdot 0 + b \cdot 0 + c \cdot 3 = 1$$
$$B=(6,0,0): a\cdot 6 + b \cdot 0 + c \cdot 0 = 1$$
$$P=(3,4,0): a\cdot 3 + b \cdot 4 + c \cdot 0 = 1$$
you obtain the following linear system of three equation in three unknown:
$$3c= 1$$
$$6a= 1$$
$$3a+ 4b = 1$$
and then:$$c=1/3,a=1/6,b=1/8$$
finally the EBP plane equation:
$$\frac{1}{6}x+\frac{1}{8}y+\frac{1}{3}z=1$$
that is equivalent to:
$$4x+3y+8z=24$$
2) by cross product of two vectors BE and BP (but you can use another pair)
in this case you obtain: BE=(-6,0,3) and BP=(-3,4,0)
$$\begin{vmatrix}
i & j & k \\ 
-6 & 0 & 3 \\ 
-3 & 4 & 0
\end{vmatrix}$$
$$= -12 \cdot i- 9 \cdot j-24 \cdot k$$
that is a normal vector to the plane EBP which components coincide with the coefficients a,b,c of the plane EBP, thus the equation of the plane EPB is:
$$-12x-9y-24z=d$$
imposing that B $\in$ EPB:
$$-12 \cdot 6 = -72 = d$$
and finally:
$$-12x-9y-24z=-72$$
that is equivalent to (dividing both side by -3):
$$4x+3y+8z=24$$
once you have the plane equation the distance of the plane EPB from the origin A is given by:
$$\frac {|24|}{\sqrt{4^2+3^2+8^2}}=\frac {24}{\sqrt{89}}=\frac {24}{89}\sqrt{89}$$
A: I know im late for this, but i think it's interesting to know how to do this without vector and plane equation.
So first make a plane through $A$ perpendicular to $BE$ and intersect $EBP$ at some other line through point $P$ (let's say it intersect line $PQ$ at point $R$ and $BE$ at point $S$, so plane 1 intersect $EBP$ at line $RS$) (cross section plane EBP isnt a triangle because line $EP$ is a space diagonal, it doesnt intersect any straight face of the cube), make a second plane through $A$ and $E$ perpendicular to $BP$ (the intersection of plane 2 and $EBP$ is line $EO$), the distance between $A$ to $EBP$ is a line segment length of A onto an intersection point of $EO$ and $RS$ (let's say point $T$, so that the distance of $A$ onto $EBP$ is length of $AT$), check perpendicularity of the line with slope (a line is perpendicular when $m_{1}.m_{2} = -1$, with the slope is the usual $\Delta{\frac{y}{x}}$ but in $3D$ it can also be $\Delta{\frac{z}{x}}$ and $\Delta{\frac{z}{y}}$, but this method doesnt work if your line is a flying line in 3D space and it doesnt lie on a straight plane without inclination).So in order to find the length of $AT$ find the height of $\Delta{AOE}$ through point $A$ using "the two version way to find area of $\Delta{AOE}$ ".

