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1.6: Lines and Planes

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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.

linpln20.gif

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

x,y,z=a,b,c+tv.

Example 1.6.1

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+t4,2,1=1+4t,22t,3+t

or

x(t)=1+4t,

y(t)=22t,

z(t)=3+t.

Exercise 1.6.1

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).

Planes

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

xa,yb,zc

is parallel to the plane, hence

nxa,yb,zc=0.

This defines the equation of the plane.

linpln21.gif

Example 1.6.2

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

Solution

We have

1,2,3x2,y1,z0=0

so that

1(x2)+2(y1)+3z=0

or

x+2y+3z=4.

Example 1.6.3

Find the equation of the plane through the points

  • P=(0,0,1)
  • Q=(2,1,0)
  • R=(1,1,1)
Solution

Let

v=QP=2,1,1

and

w=RP=1,1,0

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

v×w=|ˆiˆiˆi211110|=ˆiˆj+ˆk

or

1,1,1.

We can now use the formula:

1,1,1x,y,z1=0

or

xy+z1=0

or

xy+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

ProjnPQ=||PQn||||n||.

Example 1.6.4

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

2xy2z=5.

Solution

The normal vector can be read off from the equation as

n=2,1,2.

linpln25.gif

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

Q=1,7,3

and

nPQ=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 113.

The Angle Between Two Planes

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

Example 1.6.5

Find the angle between the two planes

3x2y+5z=1

and

4x+2yz=4.

We have the two normal vectors are

n=3,2,5

and

m=4,2,1.

We have

nm=3

||n||=38

||m||=21

hence the angle is

cos1(33821)=1.46rad.


This page titled 1.6: Lines and Planes is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by Larry Green.

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