the vector product. remember that the scalar product is a number, not a vector using the vector...
TRANSCRIPT
1 1
2 2
3 3
Let and
a b
a b
a b
a b
1 1 2 2 3 3The scalar product of a b a b a b a b
Remember that the scalar product is a number, not a vector
Using the vector product to multiply two vectors, will leavethe answer in form of a vector.
1 1
2 2 2 3 3 2 1 3 3 1 1 2 2 1
3 3
a b
a b a b a b a b a b a b a b
a b
a b i j k
The good news is that you will not have to rememberthis formula
1 1
2 2
3 3
Let and
find
a b
a b
a b
a b
a b
Rearrange the vectors as follows
1 2 3
1 2 3
a a a
b b b
i j k
1 1
2 2 2 3 3 2 1 3 3 1 1 2 2 1
3 3
a b
a b a b a b a b a b a b a b
a b
a b i j k
The vector product
Find the vector product of the vectors
2 3 and 2 4 a i j k b i j k
1 2 3
2 1 4
i j k
2 4 1 3 1 4 2 3 1 1 2 2
5 2 3
a b i j k
i j k
3 2 , 2 and 4 3 4
find
a i j k b i j k c i j k
a b c
1 1 2
4 3 4
b c
i j k
4 6 4 8 3 4
10 12
b c i j k
i j k
3 2 1
10 12 1
a b c
i j k
2 12 3 10 36 20
10 7 16
a b c i j k
i j k
One very important property of the vector product is:It will always be perpendicular to both vectors, a and b
a b a
b
A plane is parallel to each of the vectors
4 and 6 2 3
containing the point 3, 4, 7
a i k b i j k
The vector will be a normal vector to the plane n a b
4 6
0 2
1 3
n a b
4 6
0 2 4 0 1
1 3 6 2 3
2 18 8
i j k
n a b
n a b i j k
2 18 8 n a b i j k
The equation of the plane will therefore be in the form
2 18 8x y z k
Using the point 3, 4, 7
2 18 8
2 3 18 4 8 7
22
x y z k
k
k
The equation of the plane is
2 18 8 22 or
2 18 8 22 or
9 4 11 or
x y z
x y z
x y z
Find the equation of the plane which contains the points
2,1,2 , 0,2,5 and 2, 1,3A B C
The vector AC will be a normal vector to the planeAB n����������������������������
2 4
1 and 2
3 1
AB AC
����������������������������
2 1 3
4 2 1
7 10 8
AB AC
i j k
n
n i j k
����������������������������
7 10 8x y z k
7 10 8 using 2,1,2
7 2 10 1 8 2
20
7 10 8 20
x y z k A
k
k
x y z
Note that any combination of coordinates could have been usedto find the normal vector, as well as any coordinates could have been used to find a value of k.
1 1
2 2
Let be the normal to the plane and
be the normal to the plane .
The angle between the planes is then equal to the
angles between the normals of the planes.
n
n
1 2
1 2
cos n n
n n
Angle between two planes
1
2
The plane has equation 2 3 5 and
has equation 0
Calculate the acute angle between the planes.
x y z
x y z
1 1
2 2
: 2 3 5 2 3
: 0
x y z
x y z
n i j k
n i j k
1 2
2 2 21
22 22
2 3 1 4
2 3 1 14
1 1 1 3
n n
n
n
1
4cos 0.617
14 3
cos 0.617
51.9
Point of intersection of a line and a plane
The points of intersection can be found when the equationof the line is expressed in parametric form.
Find the point of intersection of the line
3 2 1
4 1 2with the plane : 2 4
x y z
x y z
Express the line in parametric form
3 2 1
4 1 24 3 2 2 1
x y z
x t y t z t
The points intersect the plane
2 4 sub in the values of , and
4 3 2 2 1
2 4 3 2 2 1 4
8 6 2 2 1 4
5 9 4
5 5
1
x y z x y z
x t y t z t
t t t
t t t
t
t
t
Sub the value of into the equation of the line
4 3 2 2 1
4 1 3 1
1 2 3
2 1 1 3
t
x t y t z t
x
y
z
The point of intersection is 1,3, 3
The angle between a line and a plane
Suppose the line L intersects the plane
Let be a vector in the direction of the line and
be a vector normal to the plane.
sin =
a n
a n
a n
Note that we are using sin
Find the size of the angle between the line
3 2 1
4 1 2and the plane : 2 4
x y z
x y z
4 2 is vector in the direction of the line
2 is a vector normal to the plane.
a i j k
n i j k
22 2
22 2
8 1 2 5
4 1 2 21
2 1 1 6
a n
a
n
1
5sin =
21 6
sin 0.445
sin 0.445
26.5
a n
a n
Line intersection of two planesIf two planes intersect, they intersect in a line. The parametricform of the line can be found as followsThe equations of two planes are 4 2 1 and 5.
Setting , we can find the equation of the line.
x y z x y z
z t
1: 4 2 1
4 2 1
4 1 2
x y z
x y t
x y t
2: 5
5
5
x y z
x y t
x y t
5
4 1 2
3 6 3 2
x y t
x y t
y t y t
2 5
2 5 7 2
x t t
x t t x t
2 7, 2,x t y t z t
Using the parametric form, the equation of the line is
Find the shortest distance from 3,1, 4 to the line
1 10 10
2 2 3
P
x y z
1 10 10
2 2 32 1
2 10
3 10
x y z
x t
y t
z t
3,1, 4P
Q
line L
Since lies on the line, 2 1,2 10, 3 10Q Q t t t 2 1 3 2 2
2 10 1 2 11
3 10 4 3 14
t t
PQ t t
t t
��������������
2
2 is a vector in the direction of
3
L
a
We laso know that
0
PQ
PQ
a
a
��������������
��������������
2 2 2
2 11 2 4 4 4 22 9 42
3 14 3
17 68 0
4
t
PQ t t t t
t
t
t
a��������������
2 4 2 6
2 4 11 3
23 4 14
PQ
��������������
22 26 3 2 49 7PQ ��������������
The shortest distance between the point and the line is therefore 7 units
1,12, 8P
plane Q
Find the shortest distance from 1,12, 8
to the plane : 2 2 6
P
x y z
The equation of in summetric form is given by
1 12 8
1 2 2
PQ
x y zt
: 1; 2 12; 2 8PQ x t y t z t
will intersect when
2 2 6
1 2 2 12 2 2 8 6
9 39 6
5
PQ
x y z
t t t
t
t