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 , u 2 , u 3 and v v 1 , v 2 , v 3 is uv uv23  uvuv 3231,,  uvuv 1312  uv 21

check: (u v )  u uv23  uvuv 3231 ,  uvuv 1312 ,  uv 21  uuu 123 , ,

uvu231  uvu 321  uvu 312  uvu 132  uvu 123  uvu 213  0 similary: (u v )  v  0 length...... An easier way to remember the formula for the cross products is in terms of determinants:

ab 12 2x2 determinant: ad bc 4  6   2 cd 34

3x3 determinants: An example Copy 1st 2 columns 1 6 2  sum of  sum of 1 6 2 1 6  forward backward 3 1 3   3 1 3 3 1 diagonal diagonal 4 5 2  4 5 2 4 5 products products

determinant =  2  72  30 36  15  8 40 59 19 i j k i j k i j uv u1 u 2 u 3  u1 u 2 u 3 u 1 u 2

v1 v 2 v 3 v1 v 2 v 3 v 1 v 2

iuv23  j uv 31  k uv 12  k uv 21  i uv 32  j uv 13

u v uv23  uv 32 i  uv 13  uv 31 j  uv 12  uv 21  k

uv uv23  uvuv 3231,,  uvuv 1312  uv 21 Let u 1,  2,1 and v  3,1,  2 Find u  v .

i j k i j k i j uv 1  2 1  1 2 1 1 2 3 1 2 3 1 2 3 1

u v 41   i 32 j 16k

uv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. u

u 4. uv area of the parallelogram h  u sin

v determined by uv and . v 1 5. 2 uv area of the triangle having

uv and as adjacent sides. v Problem: Compute the area of the triangle two of whose sides are given by the vectors uv 1,0,2 and   1,3,  2 i j k i j i j k 1 0 2 1 0 uv1 0 2 1 3  2  1 3 1 3 2

(0  6)i  (  2  2) j  (3  0) k  63ik 

1 6,0,3 |u  v | = 36  9  45 area = 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  c u  v  u  c v 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 bc

Height compbc a  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: 6,0,3  1,3,  4 From slide 9: uv 6,0,3  6  12   6 Volume = 6 Quicker: 1 3 4 i j k (uv )  w  1 0 2 (uv )  w  u1 u 2 u 3  w 1 , w 2 , w 3  1 3 2 v1 v 2 v 3  0 6 12 6 606

i j k i j w1 w 2 w 3 u u u  u1 u 2 u 3 u 1 u 2 w1 , w 2 , w 3  = 1 2 3 = Triple scalar product

v1 v 2 v 3 v 1 v 2 v1 v 2 v 3 (take absolute value) In physics, the cross product is used to measure torque . Q

Consider a force Fr acting on a rigid body at a point given by a position vector . The torque   measures the tendency of the body to rotate about the origin  point P  rF  rF rFsin r  is the angle between the force and position vectors  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 by is the intercept

yy 0 Point slope formula:  m (xy00 , ) is on the line xx 0

y y0 y 1 y 0 (x , y ) and ( x , y ) Two point formula:  0 0 1 1 are on the line x x0 x 1 x 0 Equations of Lines and Planes A a point on the line P0 x 0 , y 0 , z 0  In order to find the equation of a line, we need : B a direction vector for the line v  abc , ,

r r0 t v vector equation of line L z Here r0 is the vector from the origin P0 P tv P x , y , z to a specific point P0 on the line P0 x 0 , y 0 , z 0  r L r is the vector from the origin to

r0 a general point P ( x , y , z ) on the line v y

v is a vector which is parallel to a vector r0 x 0,, y 0 z 0 x r  x,, y z that lies on the line v is not unique: 2 v , or  v will also do

P0 P tv t a,, b c vector equation of the line L

r r0 t v or x,,,,,, y z x0 y 0 z 0 t a b c equating components we get the parametric equations of the line L

xx0  aty,  y 0  btz ,  z 0  ct

Solving for t we get the symmetric equations of the line L

x x y  y z  z 0 0 0 a b c Problem:

Find the parametric equations of the line containing P01 5,1,3 and P  3,  2,4 . (could also choose A a point on the line P0 x 0 , y 0 , z 0  choose P0   5,1,3 P0  3, 2,4 ) B a direction vector for the line v  abc , ,

v PPPP0 1  1 0 3 5, 2 1,4  3 2, 3,1 or v  2,3, 1

x,,,,,, y z x0 y 0 z 0 t a b c 5,1,3 t 2,3,  1

The line is: x 5  2 t , y  1  3 t , z  3  t 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. Determine whether the lines LL and are parallel, skew Problem: 12 or intersecting. If they intersect, find the point of intersection.

L : x 8  2 s , y   6  4 s , z  5  s L1 : x 3  t , y  5  3 t , z   1  4 t 2

Set the x coordinate equal to each other: 3ts  8  2 , or 2st   5

Set the y coordinate equal to each other: 5 3ts   6  4 , or 4st 3   11

We get a system of equations: 2st   5 or 4st 2   10 t 1 4st 3   11 4 st  3   11 s 2

1  4ts  5  Find the point of intersection using L1 : Check to make sure that the z 1  41  5  2 values are equal for this ts and .     x 31    3 3 check y 5  3  1 4,2,3 z  1  4  1 Planes In order to find the equation of a plane, we need : this vector is called A a point on the plane P0 x 0 , y 0 , z 0  the normal vector B a vector that is orthogonal to the plane n  abc , , 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  y 0  c z  z 0   0 scalar equation of the plane r0OP 0 x 0,, y 0 z 0 ax by  cz  ax  by  cz or r OP x,, y z n  abc,, 0 0 0 ax by  cz  d P P r -,, r  x  x y  y z  z 0 0 0 0 0 linear equation of the plane Problem: Determine the equation of the plane that contains the lines LL12 and .

L : x 8  2 s , y   6  4 s , z  5  s L1 : x 3  t , y  5  3 t , z   1  4 t 2

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  abc , ,

We have two vectors in the plane: from LL12 : uv  1,3,  4 and from :  2,  4,1

i j k uv  1 3  4  3 16i 18 j 46k  13i  7 j  2 k is normal 2 4 1

n  13,7,2 13(x 3)  7( y  5)  2( z  1)  0 or 13x 7 y  2 z  72