Multivariable Calculus
Summary 3 : LINES AND PLANES IN SPACE
Equation of lines in space
1) Vector form:
Let (x0, y0, z0) and (x1, y1, z1) be points of a line L in space and let (x, y, z ) be any

other point on the line. Consider the vectors u  x0 , y 0 , z 0


and v  x1 - x0 , y1 - y 0 , z1 - z 0 and w  x, y, z . The equation of the line L is given by
  
w  u  tv
where t  real numbers

v is called the direction vector of the line.
2) Parametric form of the equation of a line.
Let {a, b, c} be a set of direction numbers for the line L, where
a = x1-x0 , b = y1 - y0 and c = z1 -z0. For any point (x, y, z) on the line
 x  x 0  t(x 1 - x 0 )

 y  y 0  t(y 1 - y 0 )
 z  z  t(z - z )
0
1
0

or
 x  x 0  at

 y  y 0  at
 z  z  at
0

where t is a scalar
3) Symmetric form: solving for t we obtain
x  xo y  y o z  z 0


, provided a  o, b  o and c  0
a
b
c
Two lines in space:
Two lines in space are
1) Coincident if they are the same line
2) Parallel if they have no points in common but belong to the same plane.
3) Intersecting if they have only one point in common.
4) Skew lines if they have no points in common and belong to different planes.
 
Area of a parallelogram: Area= u  v

u
h

A  bh , but sin     h  u sin 
u
 
 
Therefore , A  v u sin   u  v
hh

v
Area of a triangle =
Pr oof :
1
2
 
u v
The proof follows from the area of a parallelogram because the area of a triangle is half the
area of the parallelogram.
Problem. Find the area of the triangle ABC in space determined by the point A=(2, -3, 5),
B=(-2, 1, 4) and C=(4, -1, -2)
Solution:
i
j k




 
u  AB   4, 4,  1 and v  AC  2, 2, 7 , u v   4 4  1  26i  30 j  16k
2 2 7
Area= (1 / 2) ( 26)2   302  ( 16)2  (1 / 2) 1832  458  21.4
1
  
Volume of a parallelepiped w  u  v 
Volume=(area of the base)(height)
  
w
 u  v 

h  Pr oju  v w 
 
u v
  
  w  u  v    
Volume  u  v
 w  u  v 
 
u v
 
u v

w

v

u
Planes in space: any equation of the form
Vector representation of a plane: If  x0 , y0 , z 0  is a point in a plane with normal vector
A, B, C , then any point (x, y, z) of the plane satisfies the vector equation
A, B, C  x  x0 , y  y 0 , z  z 0  0
Parametric representation of a plane:
If  x0 , y0 , z 0  is a point in a plane, and a1 , b1 , c1 and a2 , b2 , c2 are vectors in the plane,
then any point (x, y, z) can be represented parametrically by the system of equation
 x  a1r  a 2s  x 0

y  b1r  b 2s  y 0 where r and s are scalar numbers
z  c r  c s  z
1
2
0

Rectangular representation of a plane:
Ax + By + Cz + D =0 or a(x - x0) + B(y - y0) + C(z - z0) = 0 where (x0, y0, z0) is a given point of
the plane and the vector Ai +Bj + Ck is normal (perpendicular) vector to the plane.
How to determine the equationof a plane

1) Find two non-parallel vectors u and v on the plane.
 
2) u  v determines a vector perpendicular or normal to the plane.
The coordinates of such vector determine A, B, C.
3) Use one of the points to determine D.
How to determine the equation of a plane
 given three noncollinear points P, Q, R
1) Determine vectors u from P to Q and v from P to R ( or R to Q)
 
2) u x v determines a normal vector (A, B, C) to the plane.
3) Use the normal vector and any of the points to find the plane.
Parallel planes:
two planes A1x +B1y +C1z + D1 = 0 and A2x +B2y +C2z + D2 = 0 are parallel
if there exist a real number k such that A1=kA2, B1=kB2, C1=kC2, D1=kD2.
The normal vectors of two parallel planes are normal
2
Perpendicular planes:
If wo planes A1x +B1y +C1z + D1 = 0 and A2x +B2y +C2z + D2 = 0 are perpendicular then
A1A2 + B1B2 + C1C2 =0 (the dot product of their normal lines is zero)
How to find the distance between a point and a plane:
Let Ax  By  Cz  D  0 and (x1, y1, z1) be a po int .

Let u  x  x1 , y  y1 z  z1 be a vector from the given point to a point on the line.

The vector v  A, B, C is normal to the plane.
The desired distance is given by
 
A( x  x1 )  By  y1   Cz  z1 
 uv

Pr oj v u   
v
A 2  B 2  C2

Ax  By  Cz  Ax1  By1  Cz1
Dis tan ce 
A 2  B 2  C2
Ax1  By1  Cz1  D

 D  Ax1  By1  Cz1
A 2  B 2  C2
A 2  B 2  C2
Note: to find the distance between two parallel planes in space, take any point of one of the planes and
find the distance between the point and the other plane.
Formula for the distance between a line and a point in
R2
2
By a similar argument for the distance between a plane and a point in R ,
If follows that
Ax1  By1  C
Dis tan ce 
A2  B2
Note: to find the distance between two parallel lines in the plane, take any point of one of the lines and
find the distance between the point and the other line.
x  x1 y  y1 z  z1
Line normal to a plane: The line
is perpendicular


A
B
C
to the plane Ax + By +Cz +D =0, provided A≠0, B≠0and C≠0.
How to find the distance between two lines in space?
1. If the lines are coincident then d=0
2. If the lines intersect at one point, d=0.
3
3. Distance between two parallel lines. If the lines are parallel,
 
u v
d 
u
where

u
is a
vector in the direction of the line and v is a vector from one point on the line to a point on the
other line. This formula can also be used to find the distance from a point to a line.
R
d

v

u
P


Let  be the angle between u and v
 
 
u v sin  u  v

d
sin     d  v sin . Therefore d 
 

v
u
u
  
w  (u  v )
4. Distance between two skewed lines. If the lines are skew, d 
 
u v
a direction of L1,

v
is a direction of L2,

w
is a vector containing a point of each line.
 


u  v is perpendicular to both u and v
  
w  u  v 

d  Pr oju v w 
 
u v
Proof: it follows from the fact that
The distance
where
4
and therefore

u
is