Vectors Summary

1. Scalar product (dot product):

θcos|||| baba =•

Laws of dot product:

(i) abba •=• (ii) acabcabacba •+•=•+•=+• )(

(iii) 2|| aaa =• (angle between two identical vectors is 0 degrees)

(iv) aba ⇒=• 0 and b are perpendicular

Applications:

(i) Projection vector:

Length of projection L= |∧

• ADAB |

B

Projection vector AC = (∧

• ADAB ) ∧

AD

Foot of perpendicular = OC =OA+AC

Shortest distance from B to line

A C D = 2|| BC = 22|| LAB −

[OR = |∧

× ADAB | ]

L

(ii) Acute angle between two lines:

•= −

||||cos

21

211

mm

mmθ where 1m and 2m are the direction vectors of the

the two lines.

(iii) Acute angle between two planes:

•= −

||||cos

21

211

nn

nnθ where 1n and 2n are the individual normals to the two

planes respectively.

(iv) Acute angle between a line and a plane:

•−= −

||||cos

2

1

nm

nmπθ where m and n are the direction vector of the line

and normal to the plane respectively, and θ is the angle between the line and plane in question.

2. Cross product (vector product):

[ ]∧

=× nbaba θsin|||| where n is a vector that is perpendicular to both a and b .

Laws of cross product:

(i) )( abba ×−=× (ii) ( ) ( )acabcabacba ×−×−=×+×=+× )(

(iii) 0~

=× aa

Applications:

(i) a Area of triangle= ||2

1ba×

b

(ii) If four points ,A ,B C and D are coplanar, then 0|| =•× ADACAB

3. Equation of lines:

Representations:

(i)

+

=

f

e

d

c

b

a

r λ (parametric form) OR mar λ+= (condensed form)

(ii) f

cz

e

by

d

ax −=

−=

− (cartesian form)

4. Equations of planes:

Representations:

(i) 21 mmar µλ ++= (parametric form)

(ii) nanr •=• (scalar product form)

(where 21 mmn ×= , a is the position vector of a point lying on the plane. )

(iii) kczbyax =++ (Cartesian form)

(where ba, and c are the components of the normal vector to the plane)

5. Skew lines:

Two lines with equations 1mar λ+= and 2mbr µ+= are said to be skew lines

if they DO +OT intersect at a common point and 1m is +OT PARALLEL to 2m .

6. Determining if line resides in plane:

A line with equation mar λ+= is said to reside in the plane knr =• if

(i) 0=• nm (ii) kna =•

7. Shortest distance from plane to origin:

For a plane with equation knr =• , the shortest distance from the plane to the

origin is given by || n

k.

8. Distance between 2 planes:

For 2 planes with equations 1knr =• and 2knr =• , where 21 kk < , the shortest

distance between them is given by:

(i) ||

1

n

k+

||

2

n

k if 1k and 2k are of different signs

(ii) −||

2

n

k

||

1

n

k if 1k and 2k are of the same signs

9. Finding intersection between various constructs:

(i) Intersection between 2 lines:

For 2 lines with equations 1mar λ+= and 2mbr µ+= , equate them to each other

in column vector form such that 1ma λ+ = 2mb µ+ . Solve for the values of λ and µ before substituting back into either of the two line equations to derive the common

point of intersection.

(ii) Intersection between line and plane:

For a line with equation mar λ+= and a plane with equation knr =• , substitute

the line equation within that of the plane equation such that ( ) knma =•+ λ . Solve

for the value of λ and subsequently derive the common point of intersection through substitution of λ into the line equation.

(iii) intersection between 2 planes:

A. If one plane is presented in scalar product form and the other in parametric

form,

Example: 6

1

3

1

=

•r ------------------------(1)

+

+

=

0

1

1

1

3

3

1

0

1

µλr -----------(2)

⇒ 6

1

3

1

1

3

31

=

•

+

+

++

λµλµλ

613931 =++++++ λµλµλ

4412 =+ µλ

λµµλ 3113 −=→=+

Substituting this back into (2),

Equation of line of intersection is

−+

+

=

0

1

1

)31(

1

3

3

1

0

1

λλr

=

+

1

0

0

1

1

2

λ (shown)

B. If both planes are presented in Cartesian form:

Example: 9=++ zyx ----------(1)

1=+−− zyx ----------(2)

(1)+(2): 5102 =⇒= zz

Let ty = and substituting this together with 5=z into (1),

We have tx −= 4

Equation of line of intersection is

−

+

=

−

=

=

0

1

1

5

0

4

5

4

tt

t

z

y

x

r (shown)

C. If both planes are presented in scalar product forms or in parametric

forms or one is presented in scalar product form and the other in parametric

form, convert the plane equations such that their configurations matches that of

either case A or B, and solve accordingly.

D. If a common point A with position vector a is known to reside on both planes,

and the two planes have normal vectors 1n and 2n , then the common line of

intersection is simply given by ( )21 nnar ×+= λ .

(iv) Intersection between 3 planes:

Extract the components of the separate plane equations to form the augmented

matrix:

1

1

1

1

d

c

b

a

r =

• , 2

2

2

2

d

c

b

a

r =

• 3

3

3

3

d

c

b

a

r =

•

3333

2222

1111

dcba

dcba

dcba

After reducing the augmented matrix to its row reduced equivalent using the

RREF function of the graphic calculator, 3 possible scenarios arise:

A. The planes intersect at one point, ie there is a unique solution to the matrix.

Example:

−

−−

−

4124

3221

4112

4100

2010

1001

1( x ) +0( y ) + 0( z )=1, therefore x =1,

0( x ) +1( y ) +0( z )=2, therefore y =2,

0( x ) +0( y ) +1( z )=4, therefore z =4

Hence, the 3 planes intersect at the point (1,2,4).

B. The three planes do not intersect at all.

Example:

−

−−

−−

2461

5121

2221

−

−

1000

075.010

05.001

For the third row in the reduced form matrix, 0=1, giving rise to a contradiction,

hence there is no common point to the 3 planes, ie they DO +OT intersect.

C. The three planes intersect at a line.

Example:

−−

−−

−−

8461

5121

2221

−

−−

0000

75.075.010

5.35.001

From the reduced row matrix, we have

zxzx2

1

2

7

2

7

2

1+−=⇒−=− ,

zyzy4

3

4

3

4

3

4

3+=⇒=−

Let ,λ=z then r =

z

y

x

=

−

04

32

7

+ =

14

32

1

λ

+

−

4

3

2

4

1

04

32

7

λ

Therefore, the three planes intersect at the line r=

+

−

4

3

2

04

32

7

t , where ℜ∈=4

λt