The Vector Product

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

Find using 5

5 1 6

2 5 12 2

2 5 8 2

6,2,2

Q t

x

y

z

Q

6 1 5

2 12 10

2 8 10

PQ

��������������

22 25 10 10 225 15PQ ��������������

The shortest distance between the point and the plane is therefore 15 units