F6
F6 Mathematics Mathematics TT 1/51/5
Revision Notes on Chapter 6 : Vectors (Term 1)
Revision Notes on Chapter 6 : Vectors (Term 1)
Name : _________________________
Name : ______________________________ _____ Date : __________________Date : __________________
6.1 : VECTORS IN 2 & 3-D 6.1 : VECTORS IN 2 & 3-D
(A) : Unit Vectors & Position Vectors (A) : Unit Vectors & Position Vectors
1). Position vector of a point
1). Position vector of a point (( , ,, , ))
x x A A x x y y z z r r OA OA xi xi yj yj zk zk yy z z
2). Length of 2). Length of OOA A
r r
x x y 22
y z 22
z 22 3). Unit vector in the direction of3). Unit vector in the direction of OAOA (with length of 1 unit)(with length of 1 unit) OA OA r r r r OA OA
(B) : Algebraic Properties of Vectors (B) : Algebraic Properties of Vectors
1). If point
1). If point A A
(( ,,x x y 11 y z 11,, ) z 11) aannd d B B
(( ,,x x y 22 y z 22,,z 22) ,,)2 2 11 2 2 11 2 2 11 x x xx AB AB OB OB OA OA y y yy z z z z
2). Distance between point
2). Distance between point A x A ( , , )( , , )x y 11 y z 11 z and11 and
2 2 22 22 2 2 22 22 11 22 11 22 11 22 (( ,, ,, ) ) == B B x x y y z z x x x x y y y y z z z z 3). 3). OC OC a a bb
aa bb 4). At x-axis, y4). At x-axis, y = 0, z = 0 = 0, z = 0 ; At y-axis, x = 0, ; At y-axis, x = 0, z = 0 z = 0 ; At z-axis, x = 0, y ; At z-axis, x = 0, y = 0= 0
(C) : Scalar Product ( Dot Product ) of 2 Vectors (C) : Scalar Product ( Dot Product ) of 2 Vectors
1).
1). a a b . . b
a a bb coscos ;; coscos a ba b.. a a bb +ve or+ve or – – ve scalarve scalar aa bb
2). 2). 1 1 11 2 2 22 11 11 22 22 33 33 3 3 33 .. .. a a bb a a b b a a b b a a b b a a b b a a bb a a bb
3).3). i i ..i i j
j j ..j k
k ..k k
1 1 ; ; ..i i j j j
j ..k k k
k ..i i
0 ; 0 ; iif f . a a b . b
0 0
a a b
b aannd d vviicce e vveerrssaa 4).4). a a b . . b
b b a.. a 5).5). a a b . ( . ( b c
c ) )
a a b . . b a
a cc.. 6). If6). If kk is a constant, thenis a constant, then aa . . k k bb kk ( ( . a a . ) b b ) , , for for examplexample e : : aa . . b b aa . . bb 11 ( . ( aa . ))b b
b b bb
A A BB O O C C μμ λ λ Ө Ө(D) : Vector Product ( Cross Product ) of 2 Vectors a b
2/51). a
b = ( a b sin ) n ˆ bThe direction can be determined by Right-hand-rule. a
2). 1 2 3 2 3 3 2 1 3 3 1 1 2 2 1 1 2 3 ( ) ( ) ( ) i j k a b a a a a b a b i a b a b j a b a b k b b b
3). 0 ; , , ; - , - , -i -i j j k k i j k j k i k i j j i k k j i i k j
If a
b
0 , then a // b and vice versa 4). a b
- b a
5). a
( b c
)
a b a c6). If k is a constant, then a k b k a b
(
) 7). Area of triangle ABC 12 AB AC
8). Area of parallelogram ABCD
AB
AD(E) : Application of Dot & Cross Product of Vectors
1). Volume of a cuboid ( sin 90 ) cos 0 ( ) . a b c a b c a b c a b a b a b c
2). Projections : Ө z i y x j i k ic
b
a
C
A
ED
AC
ˆ.
AC b
ˆAC b
b
6.2 : Vector Geometry 3/5 (A) : Line (a point A x ( , , )1 y 1 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 ( : ( 3 i 2 j k ) t i j ( 3 - 8 ) )k
1 1 1 x x a y y t b z z 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, wheret is just a constant )2). Cartesian Equation of a line :
1 1 1 - - - 1 2 3 ( for example : ) 1 1 3 x x y y z z x y z a b c
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 bˆ
Must know :
i ). given 1 point & 1 line, find p. (hint: using p AC bˆ
)
ii ). given 2 skewed lines, find p. (hint: using 1 2
1 2 ˆ ˆ . , b b p AC n where n b b )
iii). given 2 parallel lines, find p. (hint: using p AC bˆ
)
iv). always use modulus sign when finding distance using dot product of vectors.
(B) : Plane (a point A x ( , , )1 y 1 z 1 on the plane and a normal vector n ai bj ck are needed)
1). Vector Equation of a plane, : r n .
d ( for example : . ( 2 - 2 r i j
k )
- 3 )1 1 1 . x a y b x a y b z c z c
Vector from the origin Normal vector to unique scalar for a specific plane Pointing to the plane the 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: n of 1 1
n 2
b1 & obtain a point from l 1)3). Shortest distance (perpendicular distance) from the origin to a plan e = p
r n. ˆ4). Shortest distance from a point ( , y , z ) x1 1 1 to the plane (ax by cz d
0) 1 1 12 2 2 ax by cz d a b c
(note: this formula is derived from p
p 2
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
1 2 1 2 . cos b b b b
2). Angle between a line & a plane = , where
.
cos sin and sin b n
b n
3). Angle between 2 planes = , where
1 2 1 2 . cos n n n n
b
n
plane
1n
2n
2
plane
1
plane
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(D) : Intersection
1). To find the intersecting point between lines l 1: r 1
1 1
a b
& l 2 : r 2 a 2 b2,
Steps: i ). Start with
1 1
a b = a 2
b2ii ). 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 1or 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 oft .ii ). Substitute the value oft 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 = n 1
n2ii ). 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
