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:

enter image description here

From this illustration, I can analyse the problem as follows:

  1. Draw plane EBP

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

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

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

  5. 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?


2 Answers 2


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:


imposing that B $\in$ EPB:

$$-12 \cdot 6 = -72 = d$$

and finally:


that is equivalent to (dividing both side by -3):


once you have the plane equation the distance of the plane EPB from the origin A is given by:

$$\frac {|24|}{\sqrt{4^2+3^2+8^2}}=\frac {24}{\sqrt{89}}=\frac {24}{89}\sqrt{89}$$

  • $\begingroup$ How can we find the equation for plane EBP? $\endgroup$
    – akusaja
    Nov 27, 2017 at 18:08
  • 1
    $\begingroup$ I've added the full solution for the problemwith two different ways to find the EPB plane $\endgroup$
    – user
    Nov 27, 2017 at 21:34
  • $\begingroup$ Thanks, now i understand it. Thank you $\endgroup$
    – akusaja
    Nov 28, 2017 at 1:41

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 $\Delta{AOE}$ ".



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