Lines and Planes

Lines

Our goal is to come up with the equation of a line given a vector v parallel to the line and a point (a,b,c) on the line.  The figure (shown in 2D for simplicity) shows that if P is a point on the line then

<x,y> = P + tv

for some number t. The picture is the same for 3D.  The formula is given below.

 Parametric Equations of a Line The parametric equations for the line through the point (a,b,c) and parallel to the vector v are       = + tv

Example:

Find the parametric equations of the line that passes through the point (1, 2, 3) and is parallel to the vector <4, -2, 1>

Solution:

We write:

<x, y, z> = <1, 2, 3> + t <4, -2, 1> = <1 + 4t, 2 - 2t, 3 + t>

or

x(t) = 1 + 4t,   y(t) = 2 - 2t,   z(t) = 3 + t

Exercise

Find the parametric equations of the line through the two points (2,1,7) and (1,3,5).

Hint: a vector parallel to the line has tail at (2,1,7) and head at (1,3,5).

If S is a plane then a vector n is normal (perpendicular) to the plane if it is orthogonal to every vector that lies on the plane.  Suppose that n is a normal vector to a plane and (a,b,c) is a point on the plane.  Let (x,y,z) be a general point on the plane, then

<x - a, y - b, z - c> is parallel to the plane, hence

 n . = 0

this defines the equation of the plane.

Example:

Find the equation of the plane that contains the point (2,1,0) and has normal vector <1,2,3>

Solution:

We have

<1,2,3> . <x - 2,y - 1,z - 0>  =  0

so that

1(x - 2) + 2(y - 1) + 3z  =  0

or

x + 2y + 3z  =  4

Example

Find the equation of the plane through the points

P  =  (0,0,1)      Q  =  (2,1,0)     and      R  =  (1,1,1)

Solution

Let

v
=  Q - P  =  <2, 1, -1>

and

w
= R - P = <1, 1, 0>

then to find a vector normal to the plane, we find the cross product of v and w: or

<1, -1, 1>

We can now use the formula:

<1, -1, 1> . <x, y, z - 1>  =  0

or

x - y + z - 1  =  0

or

x - y + z  =  1

Distance Between a Point and a Plane

Let P be a point and Q be a point on a plane with normal vector n, then the distance between P and the plane is given by

 Distance Between a Point P and a Plane With Normal Vector n Let Q be a point on the plane with normal vector n.  The the distance from the point P to this plane is given by                                    ||PQ . n||               ProjnPQ  =                                                          ||n||

Example Find the distance between the point (1,2,3) and the plane

2x - y - 2z = 5

Solution

The normal vector can be read off from the equation as

n  =  <2, -1, -2>

Now find a convenient point on the plane such as Q  =  (0, -5, 0).  We have

PQ  =  <-1, -7, -3>

and

n . PQ  =  -2 + 7 + 6  =  11

We find the magnitude of n by taking the square root of the sum of the squares.  The sum is

4 + 1 + 4  =  9

so

|| n ||  =  3

Hence the distance from the point to the plane is 11/3.

The Angle Between 2 Planes

The angle between two planes is given by the angle between the normal vectors.

Example

Find the angle between the two planes

3x - 2y + 5z  =  1      and      4x + 2y - z  =  4

We have the two normal vectors are

n = <3,-2,5>      and      m = <4,2,-1>

We have

n . m  =  3, hence the angle is 