stpm term1 chapter 6 vectors

F6 Mathematics T

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Revision Notes on Chapter 6 : Vectors (Term 1)  Name : ______________________________ _________________________ _____

Date : __________________

6.1 : VECTORS IN 2 & 3-D (A) : Unit Vectors & Position Vectors

 x  y 1). Position vector of a point  A( x, y , z )  r  OA  xi  yj  zk      z    2). Length of OA 

r



x2  y 2  z2  

3). Unit vector in the direction of OA (with length of 1 unit)



OA

r 



r 

OA

(B) : Algebraic Properties of Vectors

  x2  x1     AB  OB  OA  y2  y1     z  z    2 1

1). If point  A  ( x1 , y1 , z1 ) and B  ( x2 , y2 , z 2 ) ,

2). Distance between point  A( x1 , y1, z 1 )  and  B( x2 , y2 , z2 ) = OC  

3).

 a   b

λ 

A

C

μ

a

   

 x1  x2 

2



 y1  y2 

2



2

 z1  z 2 

B b

O 4). At x-axis, y = 0, z = 0 ; At y-axis, x = 0, z = 0 ; At z-axis, x = 0, y = 0 (C) : Scalar Product ( Dot Product ) of 2 Vectors

a.b 

1).

a

b

cos  

;

 cos   

a b

a

+ve or –  or – ve ve scalar

Ө

b

 a1   b1     a.b  a2 . b2  a1b1  a 2b2  a 3b3    a  b   3  3

2).

i .i  j. j  k .k  1 ; i . j  j.k  k .i  0

3). 4).

a.b  b.a

5).

a.(b  c)  a.b  a. c

6). If

a.b

k

;

if a .b  0  a  b and vice versa

example : a . b  a . is a constant, then a . k b  k ( a . b ) , for example

b b



1 b

(a .b )

ab

(D) : Vector Product ( Cross Product ) of 2 Vectors

1).

a  b=( a

2/5

b sin   ) n

b

ˆ

The direction can be determined by Right-hand-rule.

2).

a   b 

i

j

k 

a1

a2

a3

b1

b2

b3

 ( a2b3  a3b2 ) i  ( a1b3  a3b1) j  ( a1b2  a2b1) k   z

3). i  i  j  j  k  k   0 ; i  j  k , j  k  i , k  i  j ;  j  i  - k , k  j  - i , i  k  - j

k  i

i

If a  b  0 , then a // b and vice versa 4).

a  b  - b  a

5).

a  (b  c)  a  b  a  c

6).

If

k 

x

is a constant, then a  k b  k ( a  b )

7). Area of triangle ABC 

1 2

 AB  AC 

8). Area of parallelogram ABCD 

 AB  AD

(E) : Application of Dot & Cross Product of Vectors

1).

Volume of a cuboid

 a b c  ( a b sin 90) c a  b c cos 0 )  a  b ( a  b  a  b.c 2).

a

Ө

c b

a

Projections :

C 

 AC 

 AC  b

ˆ

 A

   E 

 AC . b

ˆ

b

 D

 j

i

y

6.2 : Vector Geometry

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(A) : Line (a point  A ( x1 , y1 , z 1 ) on the line and a direction vector b  ai  bj  ck   are needed)

1). Vector Equation of a line, l : r  a  t b ( for example : r  ( 3i  2 j  k )  t( 3i - j  8k )  )

  x    y      z   

 x1   y   1   z   1 



 a   b  t    c   



Vector from the origin

Position vector

Direction vector of the line

 pointing to the line

of a point on the line

(e.g.  AB  OB - OA , simplified by factorization, where t  is just a constant )

2). Cartesian Equation of a line :  x - x1 y - y1



a

b



z - z1

x 1

( for example :

c



1

y2

1



z  3

3

)

Must know : i ). interchange both vector & Cartesian equations of a line. ii ). given 2 points, form equation. iii). show a given point is on a given line. iv). direction vector of x-axis or parallel to x -axis = i , direction vector of y-axis or parallel to y-axis =  j , direction vector of z-axis or parallel to z-axis = k  3).Shortest distance (perpendicular distance) from a point C to a line =  p



AC



b

ˆ

Must know : i ). given 1 point & 1 line, find p. (hint: using  p



AC



ii ). given 2 skewed lines, find p. (hint: using  p 

AC . n

iii). given 2 parallel lines, find p. (hint: using  p

AC



b ) ˆ

, where n 

ˆ



ˆ

b1  b2 b1  b2

)

b ) ˆ

iv). always use modulus sign when finding distance using dot product of vectors. (B) : Plane (a point  A ( x1 , y1 , z 1 ) on the plane and a normal vector n  ai  bj  ck   are needed)

1). Vector Equation of a plane,   :

  x    y      z   

.

Vector from the origin Pointing to the plane

r.n  d

 a   b     c   

( for example : r . ( 2 i - 2 j  k  )  - 3 )



Normal vector to the plane

x1a  y1b  z1c

unique scalar for a specific plane

2). Cartesian Equation of a plane : ax  by  cz  d

where d  OA . n

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Must know : i ). interchange both vector & Cartesian equations of a plane. ii ). given 1 point on a plane & the normal vector, find the equation of plane. iii). find the intersection point between a plane and x, y or z axis. (e.g. for x-axis, y=0,z=0; substitute these into the plane’s Cartesian eq. to find x, i.e. (x,0,0)) iv). given 3 points on a plane,find the equation of the plane. v ). given 2 points on a plane & a vector on the plane or a parallel line equation, find the equation of the  plane. vi). given 1 point on a plane & 1 line equation on the plane, find the equation of plane. vii).show a given point is on a given plane. viii).show a given line is on a given plane.(hint: substitute the r   of the line into the r   of the plane and if the dot product is equal to the value of d, then the line is on the plane) ix). given 1 point, A above plane & 1 plane, find the perpendicular intersecting point, B from A to the plane. (hint: find OB  OA  t n , then substitute the co-ordinates of B into the plane equation to find t ) x ). Given a line, l 1 on a plane,

  

1

 which is perpendicular to another plane   2 , find

  

1

.

(hint: n1 of  1  n2  b1 & obtain a point from  l 1) 3). Shortest distance (perpendicular distance) from the origin to a plan e =  p 

r.n ˆ

4). Shortest distance from a point (  x1 , y1 , z1 ) to the plane ( ax  by  cz  d   0 ) 

ax1  by1  cz1  d   a 2  b2  c 2

(note: this formula is derived from  p  p2  p1  and can be used directly) (C) : Angles (All formulae in this part are derived using the dot product of 2 vectors) (if cos   is  – ve, find also the acute angle of   )

1). Angle between 2 lines =   , (can be acute or obtuse), where b1 . b2 cos    b1 b2 2). Angle between a line & a plane =   , where cos   sin 

and 

sin   

n b

b.n b

 

3). Angle between 2 planes =   , where cos   

n2

n1 . n2 n1 n2

 plane

 

n

 plane 1    

n1

 plane 2

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(D) : Intersection

1). To find the intersecting point between lines l 1: r 1  a1   b1 &  l 2 : r 2



a2



 b2 ,

Steps: i ). Start with a 1   b1  = a2   b2 ii ). Then equating the 3 coefficient of i , j , k  respectively to find the values of  and   . (note: the values of  and   must satisfy the 3 equations or otherwise the 2 lines not intersecting) iii). Substitute the values of  and   back into l 1 or l 2 to find the intersecting point. 2). To find the intersecting point between a line l, r  a  tb  and a plane  , r . n  d  , Steps: i ). Substitute the r   of the line into the r  of the plane to find the value of t . ii ). Substitute the value of t  back into the line equation to find the intersecting point. 3). To find the intersecting line between the plane  1 and  2 , Steps: i ). The direction vector of the intersecting line = n1  n2 ii ). Find a point on the intersecting line by using Cartesian equations of  1 and  2 : a). eliminate the variable z to obtain 1 equation with only x and y variables.  b). let x = any number which will cause y to be an integer. c). substitute the values of x and y into one of the planes’ Cartesian equation to find z value. d). write the intersecting line equation as r  a  tb

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