# 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:

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?

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:

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

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