# Shortest distance between parallel line and plane

I've been doing questions regarding the shortest distance between lines/planes and points , and I've come across a question asking to find the shortest distance between a line and a plane which are both parallel to each other.

Now I haven't seen this type of question before so I drew a diagram to work out the shortest distance which I believed to be the perpendicular distance between the line and the plane. In the picture $$R$$ is a point on L such that $$\vec{BR}.\vec{AR} = 0$$ and $$A$$ and $$B$$ are known points on the line and plane.

The equations for the plane and line are given as:

$$\prod r = \begin{pmatrix} 1\\ 3\\ 4\\ \end{pmatrix} + u\begin{pmatrix} 4\\ 1\\ 2\\ \end{pmatrix} + v\begin{pmatrix} 3\\ 2\\ -1\\ \end{pmatrix}$$

$$L\: r = \begin{pmatrix} 2\\ 1\\ -3\\ \end{pmatrix} +t \begin{pmatrix} 2\\ 3\\ -4\\ \end{pmatrix}$$

Where $$\vec{OR} = \begin{pmatrix} 2 + 2t\\ 1 + 3t\\ -(3 + 4t)\\ \end{pmatrix}$$

So I found $$\vec{BR}$$ and $$\vec{AR}$$ and dotted them together to solve for t where I ended up with this: $$29t^{2} + 24t = 0 \\ t(29t + 24) = 0 \\ t = 0\:\:\:t = -\frac{24}{29}$$

So when I subbed the value of t $$-\frac{24}{29}$$ back into $$\vec{AR}$$ and found the modulus I was greeted with the anwser of $$\sqrt{\frac{990}{29}}$$ which is wrong because the correct anwser is $$2\sqrt{6}$$.

I reworked my answer 3 times but arrived at the same answer everything hence I can only conclude that my logic in my diagram is flawed. If my answer is not the shortest distance then what exactly did I just work out? Can someone please tell me why my method doesn't work?

Thank you

The flaw is in your diagram. For your method to work, the point $A$ must lie on the projection of the line onto the plane, otherwise the vector $\vec{AR}$ will be oblique to the plane.

While drawing the figure you have assumed that the point $A$ is lying on the projection of $L$ onto $\Pi$ while it's actually not.

The only thing I can think of that is wrong with your method is how you obtained point R and thus BR and AR. For finding R, I would have solved for the intersection between line L and OR. Then you can find the lines pretty easily. Besides that I am really not sure what could have gone wrong

The method I have always used to solve this problem is as following:

1) First convert the plane equation to Cartesian form (vector form is pretty useless to be honest). To do this you first find the normal vector of the plane (cross product of the two direct vectors) $$<4,1,2> X <3,2,-1> = <-5,10,5>$$ Thus, the equation of the plane becomes $$-5x + 10y + 5z + d = 0$$ We are given a point on the plane so we can solve for d.
Using the point (1,3,4),
$$-5(1) + 10(3) + 5(4) + d = 0$$ Thus d is 45 and the equation of the plane is $$-5x + 10y + 5z + 45 = 0$$ 2) From here, we can take a point on the given line, and make a vector in the normal direction of the plane. The intersection of this line and the plane will be at a right angle and we thus use the intersection to find the distance. Using (2,1,-3), we get vector: $$r = (2,1,-3) + t<-5,10,5>$$ or $$r = <2-5t,1+10t,-3 +5t>$$ The point of intersection of this vector and plane must satisfy the equation of the plane (because the intersection point is on the plane). Thus, I can plug the generalizations of the x,y, and z coordinates into the plane. We get: $$-5(2-5t) + 10(1+10t) +5(-3+5t) + 45 = 0$$ Solve this for t and you will get the point on the plane where the vector "r" and plane intersect. Thus, if you find the distance between the found point and (2,1,-3) you will have the distance between the two.

If you need a vector perpendicular to the plane you could use the cross-product of $<4, 1, 2>$ and $<3, 2, -1>$.

• Yes this is what part of the question asked to me to find , showing that the plane and line were parallel to each other. I did cross the direction vectors of the plane together to get the perpendicular vector. – Nubcake May 13 '14 at 18:32