### PowerPoint 5

```Even More Vectors
Scalar Equation of Lines in the
Plane
The standard form of the equation of a line in the plane, Ax+By+C=0,
can be developed using one point on the line and a slope. This form
is also known as a scalar equation of the line or as a Cartesian
equation of the line.
An alternative development of this equation makes use of vector
concepts.
Perpendicular vectors
A normal (vector) to a line l is a vector n which is perpendicular to the
line.
u
l
u
u  u   | u || u  |  cos  90 
u  u  0
 xu   xu

   y  y   0
u
u
u  ( xu , yu )  u    ( yu , xu )
Understanding
Develop a scalar equation of the line through the point P0(4, -1)
with normal n=(3,5)
We first pick any point P(x,y) on the line
Because P0 and P are on both on the line, we can determine the
vector Po P that points in the direction of the line.
So, Po P is perpendicular to n
Thus, Po P  n  0
But, Po P   x  4, y  1
So,
 x  4, y  1   3,5  0
3  x  4   5  y  1  0
3x  5 y  7  0
This tells us that 3x+5y-7=0 is
a scalar equation of the line
through the point PO(4,-1) with
normal n=(3,5)
Scalar Equations
A scalar equation of the line through Po  xo , yo  with normal n   n1, n2  is given by:
n1x  n2 y  C  0
For a line given by Ax  By  C  0, the vector ( A, B) is a normal for the line.
Understanding
Find a scalar equation of the line through the point P0(4, -1) with
normal n=(3,5)
Because we are given the normal, the scalar equation is in
the form: 3x+5y+C=0
To determine C, we need only substitute the coordinates of a point
on the line: P0(4, -1)
3x  5 y  C  0
3  4   5  1  C  0
7C  0
C  7
Therefore, 3x+5y-7=0 is a scalar equation of the line.
Understanding
Find a normal vector for each line:
a) 3x+2y=7
b) (x-2, y+4)=r(3,-1)
a) Placing 3x+2y=7 in the form Ax+By+C=0:
3x  2 y  7  0
Then : n  3,2
b) (3, -1) is a direction vector:
Then : n  1,3
nd  0
or
 n1, n2    3, 1  0
3n1  n2  0
Therefore we can choose n1=1 and n2=3
Understanding
Rewrite the equation (x,y)=(2,-5)+t(2,-3) as a scalar equation for the line
The normal:
The equation:
3x  2 y  C  0
3  2   2  5  C  0
C4
3x  2 y  4  0
or
 x, y    2, 5   t  2, 3
 x  2, y  5  t  2, 3
n  3,2
or
x2 y5

2
3
3  x  2   2  y  5 
3 x  6  2 y  10
3x  2 y  4  0
 x, y    2, 5  t  2, 3
 x, y    3, 2    2, 5   3, 2   t  2, 3   3, 2 
3x  2 y  6  10  t  0 
3x  2 y  4  0
Understanding
Determine the vector equation of a line that is perpendicular to 2x-7y+5=0 and
has the same y-intercept as the line (x,y)=(2,-5)+t(2,-3).
Vector Equation:
Direction vector:
 x, y   P0  td
d   2, 7 
Point: (find t that makes x=0) t  1,  P0   0, 2
(x,y)=(0,-2)+t(2,-7)
Intersection of Line
One method for determining the intersection, if any,
of two lines in the plane involves solving a system to
two scalar equations. However, the other forms of
equations of lines in the plane and in space suggest
alternative techniques for investigating intersections.
Understanding
Investigate the intersection of the lines l1 and l2 for:
l1 :
x5 y2 z 7


3
2
6
l2 :
x
y6 z 3

5
1
Write in parametric form
Solve using any 2
equations and verify
with the 3rd
l1 :
x  5  3s
l2 :
xt
y  2  2s
y  6  5t
z  7  6s
z  3  t
Since the direction vectors: (3, 2, 6) for l1 and (1, -5, -1) for l2, the lines are
not parallel, thus any point of intersection has the property:
5  3s  t
2  2 s  6  5t
7  6 s  3  t
3s  t  5
2 s  5t  8
6s  t  4
s 1
t  2
Substitute into
either l1 on l2
x  2
y4
z  1
Understanding
Investigate the point of intersection of the lines l1 and l2 for:
l1 : OP   2,1,0  s 1,3,7 
x3 y 3 4 z
l2 :


5
4
2
Direction vectors: l1=(1,3,7)
Therefore these lines are not parallel
l2=(5,-4,-2)
Parametric:
l1 :
x  2  s
Therefore
y  1  3s
z  7s
l2 :
x  3  5t
y  3  4t
z  4  2t
2  s  3  5t
1  3s  3  4t
7 s  4  2t
s0
Solving (1)
and (2)
t  1
Sub into (3)
7  0  4  2  1
Therefore
these lines do
not intersect.
These are
skew lines
Distance between two points
y
A
dist( A, B)  || B  A ||
yA

B
yB
O
xA
xB
 B  A   B  A
 ( xB  xA )2  ( yB  y A ) 2
x
Distance between Point and Line
The distance from point Q(x1,y1) to line Ax+By+C=0
Q
let v represent the direction vector for line l

b  a  b a cos  
Recall:
n
l
proja b  b cos  
proja b 
ba
P0
a
v
Q’ = P0 +tv
Q  Q  v
we need to find the length d of the projection of QP0 on n
QQ '  d 
P0Q  n
Now:

n

 x1  x0 , y1  y0    A, B 
A B
Ax1  By1  C
2
A2  B 2
2

Ax1  By1    Ax0  By0 
A2  B 2
Since Po(xo,yo) is on the line it satisfies
the equation so Axo+By0+C=0 therefore
C=-Axo-Byo
Understanding
Find the distance from the point Q(2, -3) to the line 4x+5y-6=0
d
d

Ax1  By1  C
A2  B2
4x  5 y  6
4 2  52
4  2   5  3  6
13

41
41
Distance in Space
A
Find the distance from the point
A(1,2,3) to the line l where l is given
by l: (x,y,z)= (0,1,5) + t(3,4,1)
l

Since Po(0, 1, 5) lies on the line.
Vector PoA is (1-0, 2-1, 3-5)=(1, 1, -2)
Since
Then
P0
d  P0 A sin  
P0 A  v  P0 A v sin  
Therefore
d
P0 A  v
d

v
Q’ = P0 +tv
v
1,1, 2    3, 4,1
32  42  12
 9, 7,1
32  42  12
9   7   12
2
2
Can
Why
youdid
use
itallnot
this
work?as
Try
itvectors
Try
writing
technique
for d(-5,
2D lines
3D vectors
4, 0)as
in the previous example?

 2.2
26
Understanding
A
Find the distance from the point
A(1,2,3) to the line l where l is given
by l: (x,y,z)= (0,1,5) + t(3,4,1)
D
We need the length of the projection of
line segment APO on the direction
vector d=(3,4,1).
AP0  d
p
d

5

26
Now apply the Pythagorean theorem
AP0   D   p 2
2
2
D
 0  1,1  2,5  3   3, 4,1
32  42  12
Q’
p
P0
l

2
AP0  p

 1   1   2 
 6
 2.2
(0-1,1-2,5-3)
2
2
25
26
2
2

2
 5 


 26 
2
Distance between two lines in 3D
l1  Q1  su
Q1
l1
P1
l2  Q2  tv
u
d
l2
Q2
v
P2
The distance is attained between two points Q1 and Q2
so that (Q1 – Q2)  u and (Q1 – Q2)  v
Distance between two lines in 3D
l1  Q1  su
Q1
l1
P1
l2  Q2  tv
u
d
l2
Q2
v
P2
So we will project the position vector Q1Q2 onto a vector  to both u and v
That is, onto the normal vector u  v
This will give us the shortest distance between both lines
D  projn Q1Q2 
Q1Q2  n
n
Understanding
For the lines l1 and l2 given by the following equations
l1: x-1 = y = z
l2: x/2 = 1-y = z
a) Show that l1 and l2 are skew lines
b) Find the distance between l1 and l2
These are not parallel, so
the lines are not parallel
Check the
direction
vectors:
Determine a
point of
intersection
d1  1,1,1 and d2   2, 1,1
x  1 t
yt
z t
x  2s
y  1 s
zs
1  t  2s
t  1 s
ts
No solution
Since the lines do not intersect, these are skew lines
Understanding
For the lines l1 and l2 given by the following equations
l1: x-1 = y = z
l2: x/2 = 1-y = z
a) Show that l1 and l2 are skew lines
b) Find the distance between l1 and l2
Check the
direction
vectors:
Determine
the normal
vector
Find A1A2
Point has
to satisfy
equation
d1  1,1,1 and d2   2, 1,1
n  1,1,1   2, 1,1
  2,1, 3
Q1Q2   0  1,1  0,0  0 
  1,1,0 
Select a
point on
each line
A1 1,0,0 
A2  0,1,0 
Determine distance
D
Q1Q2  n
 0.3
n

 1,1,0    2,1, 3
2
2
2
 2  1   3
Equations of Planes
In order to determine a plane it is enough to specify either of the following
sets of information:
1) Three non-collinear points on the plane
2) One point (fix location)on the plane and two-non-collinear direction
vectors (defines plane’s slant)
A vector equation of the plane through point P0 with direction vectors d and e
is P0 P  sd  te for real numbers s and t
P
e
P0
te
sd
d

Understanding
Develop a vector equation for the plane through the point P0(-3,5,0) with
direction vectors d1=(1,2, -1) and d2=(3, -1,4)
Pick a point P on the plane:
Now apply:
Position
vector form
Parametric
form
P  x, y, z 
P0 P  sd1  td2
 x  3, y  5, z  0  s 1,2, 1  t 3, 1,4
s, t 
 x, y, z    3,5,0  s 1,2, 1  t 3, 1,4
s, t 
x  3  s  3t
y  5  2s  t
z  0  s  4t
Understanding
Find a vector and parametric equations of the plane through points A(1,7,2),
B(4, 0, -1), and C(1, 2, 3)
One direction vector of the plane is:
d  AB   4  1,0  7, 1  2    3, 7, 3
Another direction vector is:
e  AC  1  1,2  7,3  2    0, 5,1
Using A as the point:
OP  1,7,2  s  3, 7, 3  t  0, 5,1
Parametric:
x  1  3s
y  7  7 s  5t
z  2  3s  t
Understanding
Find three points on each plane.
OP   0,4,2   s  5, 2,3  t 1,0,1
a) The plane with vector equation :
b) The plane with parametric equation: x  1  4r  3s
y  3r  s
z  1  2r
Each point on the plane corresponds to a pair of values for each of the parameters
a)
b)
(0,4,2)+1(5,-2,3)+1(1,0,1)=(6,2,6)
(0,4,2)+1(5,-2,3)+0(1,0,1)=(5,2,5)
(0,4,2) -1(5,-2,3)+3(1,0,1)=(-2,6,2)
4=1+4(0)+3(1)
1=-3(0)+1(1)
1=1-2(0)
8=1+4(1)+3(1)
-2=-3(1)+1(1)
-1=1-2(1)
-3=1+4(-1)+3(0)
3=-3(-1)+1(0)
3=1-2(-1)
Normal Vector to a Plane
It becomes useful to be able to distinguish between different equations of the
same plane and equations of different planes. By recognizing that each point
on the plane must correspond to a value of the parameter, it is possible to
compare such equations to decide whether they describe to same plane.
First we determine whether or not the planes are parallel.
A normal (vector) to a plane is a vector n that is perpendicular to every
vector in the plane
This is not as difficult as might first appear to check that a given vector is
normal to a specific plane. In fact, if a vector n is perpendicular to any two
independent vectors in a plane then it is normal to the plane
Understanding
Determine whether or not both equations describe the same plane
 1 :  x  1, y  6, z   r  3, 2,1  s 1,0,1
 2 : x  5  3 p  4q
y  4  2 p  2q
z  2  p  2q
To decide where planes π1 and π2 are parallel, we consider n1, a normal to π1, and
n2, a normal to π2. We obtain these normals by finding the cross product of the
direction vectors of each plane
n1   3, 2,1  1,0,1
  2, 2, 2 
n2   3, 2,1   4, 2, 2 
  2, 2, 2 
Because these normals are collinear, the planes may be identical or parallel.
All we need to do is check that a point on one plane is on the other plane.
Let r=s=0, then the point (1,6,0) is on π1,
We need to check if this point is on π2
Understanding
The point (1,6,0) is on π1,Is this point on π2 ? If so
1  5  3 p  4q
6  4  2 p  2q
0  2  p  2q
4  3 p  4q
2  2 p  2q
2  p  2q
1
 2
 3
0  p
 2   3
p0
Substituting p=0 into (3), we obtain q=-1
Now substitute p=0, q=-1 into (1) to check:
4  3 0   4  1
4  4
Therefore the point (1, 6, 0) is on both planes, thus the two plane are identical.
Scalar Equations of Planes
We have already discussed the scalar equation of the line Ax+By+C=0, We will
continue this process and define Ax+By+Cz+D=0 as the scalar or Cartesian
equation of the plane.
Let’s develop a scalar equation of the plane through point P0(4,-1,3) with
normal n=(3,5,2)
We pick any point P(x,y,z) on the plane. Since both P0 and P are on the plane,
the vector P0P is a vector in the direction of the plane. So P0P is perpendicular
to n. Thus
P0 P  n  0
 x  4, y  1, z  3   3,5, 2   0
3  x  4   5  y  1  2  z  3  0
3x  12  5 y  5  2 z  6  0
3x  5 y  2 z  13  0
Note: the coefficients of
x, y, z are the terms of
the normal
A scalar equation of the plane through the point P0  x0 , y0 , z0  with
normal n   n1 , n2 , n3  is given by: n1 x  n2 y  n3 z  D  0
Understanding
Find a scalar equation of the plane through the point P0  4, 1,3 with normal n  3,5,2.
Because n  3, 5, 2 is normal to this plane, then a scalar equation is in the form:
3x  5 y  2 z  D  0
The point P0(4, -1,3) satisfies the equation
3  4   5  1  2  3  D  0
12  5  6  D  0
D  13
Therefore: 3x+5y+2z-13=0 is a scalar equation of the plane
Understanding
Rewrite the equation (x,y,z)=(2,-5,1)+s(2,-3,0)+n(1,1,-1) as a scalar equation for
the plane.
First we find a normal by the cross product of two non-collinear direction vectors
n   2, 3,0   1,1, 1
  3, 2,5 
Therefore a scalar equation of the plane is of the form 3x+2y+5z+D=0
The point (2, -5, 1) satisfies this equation
3  2   2  5  5 1  D  0
6  10  5  D  0
D  1
Therefore: 3x+2y+5z-1=0 is a scalar equation of the plane
or
Understanding
Rewrite the equation (x,y,z)=(2,-5,1)+s(2,-3,0)+n(1,1,-1) as a scalar equation for
the plane.
Another solution is found by finding the dot product of both sides with the
normal to the plane
 x, y, z    2, 5,1  s  2, 3,0   t 1,1,, 1
 x, y, z    3, 2,5   2, 5.1   3, 2,5  s  2, 3,0    3, 2, 5  t 1,1,, 1   3, 2,5
3x  2 y  5 z   2, 5.1   3, 2,5   0  0
3x  2 y  5 z  6  10  5  0  0
3x  2 y  5 z  1
3x  2 y  5 z  1  0
Will always be
zero because
(3,2,5) is normal
to the plane
Understanding
Determine the scalar equation of the plane passing through the point (2,-5,1) and
parallel to the lines (x,y,z)=(1,6,-1)+s(2,-3,0) and (x,y,z)=(4,5,3)+n(3,2,-1) .
If we place in Scalar form, we need only find a vector perpendicular to (2,-3,0)
and (3, 2,-1).
d1   2, 3, 0 
d 2   3, 2, 1
n  d1  d 2
  3, 2,13
Now for D:
3 A  2 B  13C  D  0
3  2   2  5   13 1  D  0
D  9
3x  2 y  13z  9  0
Understanding
Find the distance from the point Q(3, -1, 1) to the plane 4x-8y-z=-41
Find normal:
n   4, 8, 1
Find Point:
P0   0,0,41
QP0   0  3,0  (1),41  1   3,1,40 
Then:
projn QP0 

QP0  n
n
 3,1, 40    4, 8, 1
2
2
2
 4    8   1
12  8  40
9
60

9

Q
n
Q’
P0

Intersection of a Line and a Plane
Find the point of intersection of the line
x  3 y  2 z 1
l:


4
3
2
and the plane
 : x  y  2z  0
Write the symmetric equations for l as parametric equations
x  3  4t
y  2  3t
z  1  2t
If a point (x,y,z) lies on the plane, it must satisfy the equation of the plane
x  3  4 1  7
 3  4t    2  3t   2  1  2t   0
3  3t  0
t 1
Hence
y  2  3 1  5
z  1  2 1  1
Distance from a Point to a Plane
The distance from the point Q(x1, y1, z1) to the plane Ax+By+Cz+D=0 is:
d
Ax1  By1  Cz1  D
A2  B 2  C 2
Understanding
Find the distance from the point Q(1, 3, -2) to the plane 4x-y-z+6=0
d
d
d
d
Ax1  By1  Cz1  D
A2  B 2  C 2
4x  y  z  6
42   1   1
2
2
4 1   3   2  6
18
9
18
Understanding
Find the point of intersection between the line and the plane
l :  x, y, z   1,0, 1  t 3,1, 7
 : 1,3,1   x, y, z   2
If (x,y,z) is the point of intersection
1,3,1   x, y, z   2
1,3,1   1,0, 1  t  3,1, 7    2
1,3,1  1,0, 1  1,3,1  t  3,1, 7   2
1  0  1  3t  3t  7t  2
t  2
 x, y, z   1,0, 1   2  3,1, 7 
 x, y, z    5, 2,13
Understanding
Find the point of intersection of the line
l:
x2 y5 z 6


3
1
8
and the plane
 : 5x  y  2 z  2  0
Write the symmetric equations for l as parametric equations
x  2  3t
y  5  t
z  6  8t
If a point (x,y,z) lies on the plane, it must satisfy the equation of the plane
5  2  3t    5  t   2  6  8t   2  0
5  0
There is no solution, no
point of intersection
Understanding
Find the point of intersection of the line
l:
x  2 y 1 z  3


1
2
5
and the plane
 : 3x  19 y  7 z  8  0
Write the symmetric equations for l as parametric equations
x  2t
y  1  2t
z  3  5t
If a point (x,y,z) lies on the plane, it must satisfy the equation of the plane
3 2  t   19  1  2t   7  3  5t   8  0
00
There is infinite solutions,
the line is on the plane
Understanding
u  v   w  0
Three vectors are coplanar if:
Given:
u  1,1,1
Determine z, that makes them coplanar
v   0,1, 1
w  1, 0, z 
Therefore:
Now:
u  v    2,1,1
 2,1,1  1, 0, z   0
2  0  z  0
z2
Intersection of Two Planes
If two distinct planes meet, their intersection is a line.
Because the line of intersection of the planes lies in
each plane, the direction vector d of the line is
perpendicular to both the normals of each plane
d  n1  n2
What we need to find now is a point that is in both
planes (not easy) and combine this with the
direction vector.
Or simpler, use elimination to solve for their
intersection.
Only if the planes are parallel can you determine
the distance between them.
Understanding
Find the parametric form of the line of intersection of the planes:
1 : 4 x  5 y  2 z  1  0
 2 : x  y  2z  0
Eliminate x
Eliminate y
 4x  5 y  2z 1  4  x  y  2z   0
y  6z 1
 4 x  5 y  2 z  1  5  x  y  2 z   0
x  8z  1
Let z=t
A parametric form of the line of
intersection
x  1  8t
y  1  6t
z  0t
Understanding
Find the parametric form of the line of intersection of the
planes (using direction vectors)
1 : 4 x  5 y  2 z  1  0
 2 : x  y  2z  0
Find
direction
vector
d  n1  n2
  4, 5, 2   1, 1, 2 
  12, 10,1
Solve these two
equations for x, and y
Therefore:
Pick an axis
to zero, say
z-axis
4x  5 y 1  0
4x  4 y
0
x  1  12t
y  1  10t
z  0t
4x  5 y 1  0
 y 1  0
x y 0
y  1
x  1
Understanding
Find the parametric form of the line of intersection of the planes:
 1 x  4  s
 2 x  8  6u  2v
y  s  3t
y  3u  5v
z   s  2t
z  u v
n1  1,1, 1   0,3, 2 
 1, 2,3
  8,8, 24 
Scalar equation:
 x  4, y, z   1, 2,3  0
x  2 y  3z  4  0
Eliminate x
8  x  2 y  3z  4    8x  8 y  24 z  64 
y  4  2 z
Eliminate y
4  x  2 y  3z  4    8 x  8 y  24 z  64 
x  4 z
 x  8, y, z    8,8,24  0
8x  8 y  24 z  64  0
n2   6,3,1   2,5, 1
Let z=t
x  4t
y  4  2t
z  0t
Understanding
Find the parametric form of the line of intersection of the planes:
 1 x  2 y  3z  6  0
 2 4 x  8 y  12 z  25  0
Let’s find the direction vector of the line
d  1, 2,3   4,8,12 
  0,0,0 
Therefore these planes are
either parallel or the same.
Since plane 1 is not a multiple
of plane 2, they are not the
same and thus are parallel and
do not intersect.
Understanding
Find the vector equation of the lines that passes through the point (2, -1,7)
and is parallel to the line of intersection of the planes:
1 : x  2 y  3z  6
 2 : 3x  y  2 z  4
Let’s find the direction vector of the line
d  1, 2, 3   3, 1, 2 
 1, 11, 7 
The vector equation is:
r   2, 1,7  t 1, 11, 7 ,
t
Intersection of Three Planes
Case 1:
The system has a unique solution, in which case the
three planes intersect at only one point
Case 2:
The system has an infinite number of solutions
described by one parameter, in which the three
planes intersect in a line
Case 3:
Case 4:
The system has an infinite number of solutions
described by two parameter, in which the three
planes are coincident (same plane) and the
solution consists of the coordinates of all points in
the plane.
The system has no solution, that is it is
inconsistent. This will happen if at least two of the
planes are parallel and distinct. It will also happen
if the three lines of intersection of pairs of planes
are parallel.
Case 1
Determine the intersection of the three planes:
 2 1 1 4 
 0 3 2 2 


 3 1 2 7 
424 44 
12 10 1 011 1 12
 0 13 3 
  22 
0
2

2

2
6
2

  
 0 100.51111/0.5
34244 / 13
2 x  y  z  4
3y  2z  2
3 x  y  2 z  7
Therefore the planes intersect at (x,y,z)=(1, -2, -4)
Case 2
Determine the intersection of the three planes:
1 1 2  2 
3 1 14 6 


1 2 0 5 
1 0 4 1 
 0 1  2 3 


 0 0 0 0 
Gauss-Jordan
elimination
x  y  2 z  2
3x  y  14 z  6
x  2 y  5
This reduces to:
x+4z=1
y-2z=-3
Or
(x,y,z)=(1-4t, -3+2t,t)
The line passes
through (1, -3, 0) with
direction (-4, 2, 1)
Case 3
Determine the intersection of the three planes:
 1 1 2 2 
 3 3 6 6 


5 5 10 10 
Gauss-Jordan
elimination
x  y  2 z  2
3x  3 y  6 z  6
5 x  5 y  10 z  10
This reduces to:
x+y+2z=-2
 1 1 2 2 
0 0 0 0 


 0 0 0 0 
The planes are the same
Case 4
Determine the intersection of the three planes:
 1 2 3 9 
 1 1 1 4 


 2 4 6 5 
9 
 1 2 3
 0 3 4 5 


 0 0
0 13
Gauss-Jordan
elimination
x  2 y  3z  9
x yz  4
2x  4 y  6z  5
The last line states :
0=-13
Which has no solution.
Case 4
Determine the intersection of the three planes:
 1 1 4 5 
3 1 1 2 


5 1 9 1 
5 
 1 1 4
 0 4 11 17 


 0 0
0
7 
Gauss-Jordan
elimination
x  y  4z  5
3 x  y  z  2
5x  y  9 z  1
The last line states :
0=7
Which has no solution.
```