12.4
Cross Product
Geometric description of the cross product of the vectors u and v
The cross product of two vectors is a vector!
• u x v is perpendicular to u and v
• The length of u x v is u  v  u v sin 
• The direction is given by the right hand side rule
Right  hand rule
Place your 4 fingers in the
direction of the first vector,
curl them in the direction
of the second vector,
Your thumb will point in the
direction of the cross product
Algebraic description of the cross product of the vectors u and v
The cross product of u  u1 , u2 , u3 and v  v1 , v2 , v3 is
u  v  u2v3  u3v2 , u3v1  u1v3 , u1v2  u2v1
check: (u  v)  u  u2 v3  u3v2 , u3v1  u1v3 , u1v2  u2v1  u1 , u2 , u3
 u2 v3u1  u3v2u1  u3v1u2  u1v3u2  u1v2u3  u2v1u3  0
similary: (u  v)  v  0
length......
An easier way to remember the formula for the cross products is in terms of determinants:
a b
2x2 determinant:
 ad  bc
c d
1 2
 4  6  2
3 4
3x3 determinants: An example
Copy 1st 2 columns
1 6 2
3 1 3
4 5 2
1 6 2 1 6
3 1 3 3 1
4 5 2 4 5
 sum of

 forward
 diagonal

 products






determinant =  2  72  30    36  15  8  40  59  19
 sum of

backward


 diagonal

 products






i
j
u  v  u1 u2
v1 v2
k
u3 
v3
i
j
u1 u2
v1 v2
k i j
u3 u1 u2
v3 v1 v2
 iu2 v3  ju3v1  ku1v2  ku2 v1  iu3v2  ju1v3
u  v   u2v3  u3v2  i   u1v3  u3v1  j   u1v2  u2v1  k
u  v  u2v3  u3v2 , u3v1  u1v3 , u1v2  u2v1
Let u  1, 2,1 and v  3,1, 2 Find u  v.
i j k
u  v  1 2 1 
3 1 2
i j k i j
1 2 1 1 2
3 1 2 3 1
u  v   4  1 i   3  2  j  1  6  k
u  v  3,5,7
Geometric Properties of the cross product:
Let u and v be nonzero vectors and let  be the angle between u and v.
1. u  v is orthogonal to both u and v.
2. u  v  u v sin 
3. u  v  0 if and only if u and v are scalar
multiples of each other.
4. u  v  area of the parallelogram
determined by u and v.
5. 12 u  v  area of the triangle having
u and v as adjacent sides.
u
u
h  u sin 
v
u
v
v
Problem: Compute the area of the triangle two of whose sides are given by the vectors
u  1, 0, 2 and v  1,3, 2
i j k
i j
1 0 2 1 0
1 3 2 1 3
i j k
u v  1 0 2
1 3 2
 (0  6)i  (2  2) j  (3  0)k
 6i  3k
 6, 0,3
| u  v | = 36  9  45
area =
1
45
2
Algebraic Properties of the cross product:
Let u, v, and w be vectors and let c be a scalar.
1. u  v    v  u 
2. u   v  w   u  v  u  w
3. c  u  v    cu   v  u   cv 
4. 0  v  v  0  0
5. (cv)  v  0
6. u   v  w    u  v   w
7. u   v  w    u  w  v   u  v  w
Volume of the parallelepiped determined by the vectors a, b, and c.
Area of the base  b  c
Height  compbca  a cos 

Volume  b  c a cos 
this stands
Volume  a   b  c   for absolute
value
a   b  c  is called the scalar triple product
The vectors are in the same plane  coplanar  if the scalar triple product is 0.
Problem: Compute the volume of the parallelepiped spanned by the 3 vectors
u  1, 0, 2 , v  1,3, 2 and w  1,3, 4
Solution:
From slide 9:
u v 
6, 0,3  1,3, 4
6, 0,3
 6  12  6
Quicker:
i
j
(u  v )  w  u1 u2
v1 v2
i
j
 u1 u2
v1 v2
k i j
u3 u1 u2
v3 v1 v2
1 3 4
(u  v )  w  1 0 2
1 3 2
k
u3   w1 , w2 , w3 
v3
  w1 , w2 , w3  =
Volume = 6
 0 6 12 6 6  0  6
w1
u1
v1
w2
u2
v2
w3
u3
v3
=
Triple scalar product
(take absolute value)
In physics, the cross product is used to measure torque.
Q
Consider a force F acting on a rigid body at a point given by a position vector r.
The torque   measures the tendency of the body to rotate about the origin  point P 
  rF
  r  F  r F sin 
 is the angle between the force and position vectors 
r

F
12.5
Lines and Planes
Recall how to describe lines in the plane (e.g. tangent lines to a graph):
y  mx  b
m is the slope
b is the y intercept
Point slope formula:
y  y0
m
x  x0
Two point formula:
y  y0 y1  y0

x  x0 x1  x0
( x0 , y0 ) is on the line
( x0 , y0 ) and ( x1 , y1 )
are on the line
Equations of Lines and Planes
A  a point on the line P0  x0 , y0 , z0 
In order to find the equation of a line, we need :
B a direction vector for the line v  a, b, c
r  r0  tv vector equation of line L
z
Here r0 is the vector from the origin
P0 P  tv
P0  x0 , y0 , z0 
P  x, y, z 
to a specific point P0 on the line
r
L
r0
x
v
r0  x0 , y0 , z0
r  x, y, z
P0 P  tv  t a, b, c
r is the vector from the origin to
a general point P  ( x, y, z ) on the line
y
v is a vector which is parallel to a vector
that lies on the line
v is not unique: 2v, or  v will also do
vector equation of the line L
r  r0  tv
or
x, y, z  x0 , y0 , z0  t a, b, c
equating components we get the
parametric equations of the line L
x  x0  at , y  y0  bt , z  z0  ct
Solving for t we get the symmetric equations of the line L
x  x0 y  y0 z  z0


a
b
c
Problem:
Find the parametric equations of the line containing P0   5,1,3 and P1   3, 2, 4  .
A  a point on the line P0  x0 , y0 , z0 
choose P0   5,1,3
B a direction vector for the line v  a, b, c
v  P0 P1  P1  P0  3  5, 2  1, 4  3  2, 3,1
or v  2,3, 1
x, y, z  x0 , y0 , z0  t a, b, c  5,1,3  t 2,3, 1
The line is: x  5  2t , y  1  3t , z  3  t
(could also choose
P0   3, 2, 4  )
Two lines in 3 space can interact in 3 ways:
A) Parallel Lines their direction vectors are scalar
multiples of each other
B) Intersecting Lines there is a specific t and s, so that the
lines share the same point.
C) Skew Lines their direction vectors are not parallel and
there is no values of t and s that make the
lines share the same point.
Problem:
Determine whether the lines L1 and L2 are parallel, skew
or intersecting. If they intersect, find the point of intersection.
L2 : x  8  2 s, y  6  4 s, z  5  s
L1 : x  3  t , y  5  3t , z  1  4t
Set the x coordinate equal to each other:
3  t  8  2s , or 2 s  t  5
Set the y coordinate equal to each other:
5  3t  6  4s , or 4s  3t  11
We get a system of equations:
2 s  t  5
4s  3t  11
or
1  4t  5  s
Check to make sure that the z
1  4  1  5   2 
values are equal for this t and s.
3  3 check
4s  2t  10
4s  3t  11
t  1
s  2
Find the point of intersection using L1 :
x  3   1
y  5  3  1
z  1  4  1
 4, 2,3
Planes
In order to find the equation of a plane, we need :
A  a point on the plane P0  x0 , y0 , z0 
this vector is called
B a vector that is orthogonal to the plane n  a, b, c
the normal vector
to the plane
n   r  r0   0
n  r  n  r0
vector equation of the plane
n   r  r0   0 
a  x  x0   b  y  y0   c  z  z0   0
r0  OP0  x0 , y0 , z0
r  OP  x, y, z
scalar equation of the plane
n  a, b, c
P0 P  r - r0  x  x0 , y  y0 , z  z0
ax  by  cz  ax0  by0  cz0 or
ax  by  cz  d
linear equation of the plane
Problem: Determine the equation of the plane that contains the lines L1 and L2 .
L1 : x  3  t , y  5  3t , z  1  4t
L2 : x  8  2 s, y  6  4 s, z  5  s
In order to find the equation of a plane, we need :
A  a point on the plane
can choose e.g. P0  (3,5, 1)
B  a vector that is orthogonal to the plane n  a, b, c
We have two vectors in the plane: from L1 : u  1,3, 4 and from L2 : v  2, 4,1
i
j
u  v  1 3
2 4
n  13, 7, 2
k
4 
1
 3  16  i  1  8 j   4  6 k  13i  7 j  2k
13( x  3)  7( y  5)  2( z  1)  0
or
is normal
13 x  7 y  2 z  72