# Math Help - Smallest distance between a line and a plane

1. ## Smallest distance between a line and a plane

I have a plane: 1 x + -5 y + 5 z = -7

and a line: (x,y,z) = (0,0,0) + t(-1,-3,2)

and i need to find thesmallest distance between the two.

My work:

Plane : normal = (1,-5,5) & point P = (-7,0,0)
Line : direction vector = (-1,-3,2) & point Q = (0,0,0)

Need to find vector PQ = (7,0,0)
The distance = ||Projection of PQ onto normal n ||

( ( (7,0,0)(1,-5,5) ) / ||(1,-5,5)|| ) * (1,-5,5)

= (7/51) * sqrt51 or 7/sqrt51

This is wrong, and after looking online and through my textbook, i think it’s because I’m not getting the shortest distance, but rather the distance to P
I can’t get a hold of my prof and the help room at my school thought it looked fine but it isn’t right

I’ve searched online for the past three days and i just KNOW this type of question will be on a test

2. Hello, williamb!

Is this a trick question?

Given the plane: . $x -5y + 5z \:=\: -7$

and the line: . $(x,y,z) \:= \:(0,0,0) + t(\text{-}1,\text{-}3,2)$

find the smallest distance between the two.

The normal vector of the plane is: . $\vec{n} \:=\:\langle 1,\text{-}5,5\rangle$

The direction vector of the line is: . $\vec{v} \:=\:\langle \text{-}1, \text{-}3, 2\rangle$

Since $\vec{n}\cdot\vec{v}\,\neq\,0$, the plane and line are not parallel.

Therefore, they intersect. .Their minimum distance is 0.

3. ## little confused?

I'm a little confused i was taught that if a dot b equals zero they are perpendicular.

4. If the line is perpendicular to the normal then it is parallel to the plane. If not then not.