Parametric Equations for a line in space
Linear equation for plane in space
Sketching planes given equations
Finding distance between points, planes, and lines in space

Use intercepts to find intersections with the coordinate axes
(traces)

Equation of a line
VECTOR VALUE FUNCTION, PARAMETRIC
EQUATION, SYMMETRIC EQUATION, STANDARD
FORM, AND GENERAL FORM

A line corresponds to the endpoints of a set of 2dimensional position vectors.

Find a vector equation for the line that is parallel to the vector
<0, 1, -3> and passes through the point <3, -2, 0>

30 LECT URE 4. LINES AND CURVES IN 3D
Figure 4.2: T he vect or r ( t ) = ( f ( t ) , g ( t ) , h ( t )) describes t he posit ion as a funct ion of t ime.
const ant speed v , t he whole t ime, t hen t he dist ance between P lengt h of t he vect or P
0
P . Since speed is di stance/ ti me ,
0
−−→
P
0
P v = t or
−−→
P
0
P = vt and P is t he t he
(4.3)
(4.4)
If we define the veclocit y vect or as a vector of length v point ing in our direct ion of mot ion, v = v
ˆv t hen it must be t rue t hat v is parallel t o
0
=
−−→
P
0
P , so t hat
−−→
P
0
P = ( v ˆv) t
(4.5)
(4.6)
31
Since t he coordinat es of our posit ion vect or r , are, by definition, the same as the
LECT URE 4. LINES AND CURVES IN 3D coordinat es of t he point P (see equat ion (4.2)),
Figure 4.3: The equat ion of a line is paramet rized by.
31
LECTURE 4. LINES AND CURVES IN 3D r = r
0
+ v t (4.7) where we have defined r
0 t o be our posit ion vect or at t ime t = 0. Equat ion (4.7) is t he equat ion of a line t hrough r
0 Figure 4.3: The equat ion of a line is paramet rized by.
Figure 4.3: T he equat ion of a line is paramet rized by.
T he equation of a line through the point r
0 and parallel to the vector v = 0 is given by r = r
0
+ v t. I f u = 0 is any vect or par all el ( or ant i-par all el) t o v then r = r
0
+ u t gives an equation for the same line.
If we denot e our coordinat es at any t ime t by ( x, y, z ), and our velocity vect or by ( v x
, v y
, v z
), t hen
( x, y, z ) = ( x
0
, y
0
, z
0
) + ( v x
, v y
, v z
) t
= ( x
0
, y
0
, z
0
) + ( v x t, v y t, v z t )
0
+ v x t, y
0
+ v y t, z
0
+ v z t ) (4.8)
31 x = x
0
+ v x t, y = y
0
+ v y t, z = z
0
+ v z t Mat h 250, Fall 2006 (4.9)
If two vect ors are equal t hen each of t heir component s must be equal,
Equat ion (4.9) gives t he par am et r ic equat ion of a l ine. The numbers v x
, v y
, and v z are called t he dir ect i on numb er s of t he line.
+ v x t, y = y
0
+ v y t, z = z
0
+ v z t (4.9) x = x
0
+ v x t, y = y
0
+ v y t, z = z
0
+ v z t (4.9) parametric equation of a line.
Ex am ple 4.1
Find the parametri c equation of a line through the points (2 , − 1 , 5) v x
, v y
, and v z are called t he dir ect i on number s of t he line.
1 and v z are called t he dir ect i on numb er s of t he line.
1 and (7 , − 2 , 3) .
v x
, v y
,
Solution.
We can de fine
Ex am ple 4.1
Find the parametri c equation of a line through the points (2 , − 1 , 5)
Ex ample 4.1
Find the parametri c equation of a line through the points (2 , − 1 , 5) and (7 , − 2 , 3) .
and (7 , − 2 , 3) .
Solution.
We can de fine r r
0
= (2 , − 1 , 5)
Solution.
We can de fine
1
= (7 , − 2 , 3) r
0
= (2 , − 1 , 5) r
1
= (7 , − 2 , 3) so t hat v = r
1
− r
0
= (7 , − 2 , 3) − (2 , − 1 , 5) = (5 , r r
0
= (2 , − 1 , 5)
− 1 ,
= − 2)
(7 , − 2 , 3) so t hat v = r
1
− r
0
= (7 , − 2 , 3) − (2 , − 1 , 5) = (5 , − 1 , − 2) v = r
1
− r
0
= (7 , − 2 , 3) − (2 , − 1 , 5) = (5 , − 1 , − 2) r = (2 , − 1 , 5) + (5 , − 1 , − 2) t = (2 + 5 t, − 1 − t, 5 − 2 t ) is a vect or parallel t o t he line. Hence t he equat ion of t he line is r = (2 , − 1 , 5) + (5 , − 1 , − 2) t = (2 + 5 t, − 1 − t, 5 − 2 t ) r = (2 , − 1 , 5) + (5 , − 1 , − 2) t = (2 + 5 t, − 1 − t, 5 − 2 t ) x = 2 + 5 t, y = − 1 − t, z = 5 − 2 t.
The paramet ric equat ions of t he line are
T he paramet ric equat ions of t he line are
If v x
= 0 , v y
= 0, and v z
= 0, t hen we can solve for t he variable t in each of x = 2 + 5 t, y = − 1 − t, z = 5 − 2 equat ions (4.9),
If v x
= 0 , v y
= 0, and v z
= 0, t hen we can solve for t he variable t in each of equat ions (4.9), t = ( x − x
0
) / v x t = ( y − y
0
) / v y t = ( z − z
0
) / v z
If v x
= 0 , v y
= 0, and v z
= 0, t hen we can solve for t he variable t in each of t = ( x − x
0
) / v x
(4.10)
T he t erm
(4.10)
(4.11) t t =
= (
( y z
− y
− z
0
0
)
) / v
/ v y z t t = ( x − x
= ( y − y
0
0
)
) / v x
/ v y
(4.11)
(4.12)
(4.10)
(4.11)
1 t = ( z − z
0
) / v z
(4.12) is ment ioned prominent ly in t he t ext .
(4.12)
1
T he t erm “ direct ion numbers” is rarely used and should probably be avoided, even though it
1
T he t erm Mat h 250, Fall 2006 is ment ioned prominent ly in t he t ext .
is ment ioned prominent ly in t he t ext .
Revised December 6, 2006.
Mat h 250, Fall 2006 Revised December 6, 2006.
Mat h 250, Fall 2006 Revised December 6, 2006.

Find the parametric equation of a line through the points
(2, -1, 5) and (7, -2, 3)

Solving for t
This gives the symmetric equation of a line.
Write the line L through the point P = (2, 3, 5) and parallel to the vector v=<4, -1, 6>, in the following forms: a) Vector function b) Parametric c) Symmetric d) Find two points on L distinct from P.

We can obtain an especially useful form of a line if we notice that
Substitute v into the equation for a line and reduce…

Equation of a Plane
STANDARD EQUATION, GENERAL FORM,
FUNCTIONAL FORM (*NOT IN BOOK)

Given any plane, there must be at least one nonzero vector n = <a, b, c> that is perpendicular to every vector v parallel to the plane.

Standard Form or Point Normal Form
By regrouping terms, you obtain the general form of the equation of a plane: ax+by+cz+d=0
(Standard form and general form are NOT unique!!!)
Solving for “z” will get you the functional form
. (unique)

Find the equation of the plane with normal n = <1, 2, 7> which contains the point (5, 3, 4). Write in standard, general, and functional form.

Find the equation of the plane passing through
(1, 2, 2), (4, 6, 1), and (0, 5 4) in standard and functional form.
Note: using points in different order may result in a different normal and standard equation but the functional form will be the same.

Does it matter which point we use to plug into our standard equation?

Find the equation of the plane containing the point (1, 2, 2) and the line L(t) = (4t+8, t+7, -3t-2)

Angle:
Line:
Find the angle between the planes x+2y-z=0 and x-y+3z+4=0

Examples of intersections of planes
(note: these are not scalar multiples of each other…therefore NOT parallel!
1a. Write an equation for the line of intersection of the planes x + y - z = 2 and 3x - 4y + 5z = 6
1b. find the angle between the planes.
2a. Write an equation for the line of intersection of the planes 5x-3y+z-10=0 and
2x+4y-x+3=0
2b. Find the angle between the planes
Example of parallel planes will be in a future slide---for those problems, we only find the distance between the planes.

*This formula is from your cross product and sine formula.

Given line L that goes through the points (-3, 1, -4) and (4, 4, -6), find the distance d from the point P = (1, 1, 1) to the line L.

Find the distance between the two parallel planes given by
3x-y+2z -6=0 and 6x-2y+4z+4=0

Find the distance between the two parallel planes given by
10x+2y-2z -6=0 and 5x+y-z-1=0

Homework:
PG. 759/#1-7ODD, 8, 9-13ODD, 14-19,
21, 25-33ODD, 37-51ODD, 63, 67-81
ODD