geometry
I am stuck with the following problem that says:
Let $A,B$ be the ends of the longest diagonal of the unit cube . The length of
the shortest path from $A$ to $B$ along the surface is :
My Try: 
So, the length of the longest diagonal $AB=\sqrt{3}$. If I reach from $A$ to $B$ along the surface line $AC+CD+BD$, then it gives $3$ units. But the answer is given to be option 3.
Can someone explain? Thanks in advance for your time.
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:
$$\frac {|24|}{\sqrt{4^2+3^2+8^2}}=\frac {24}{\sqrt{89}}=\frac {24}{89}\sqrt{89}$$
Here's the box:
Clearly the only unfolding that matters is with the two adjacent point-bearing sides adjacent.
Then it is clear the distance is $$d = \sqrt{(5/4)^2 + (1/2)^2}$$
There are only three cases:
Best Answer
The path goes through the middle point of common opposite side considering two squares only.