WEEK FOUR
TOPIC:MECHANICS (VECTOR GEOMETRY)
SCALAR OR DOT PRODUCT OF TWO VECTORS
The scalar or dot product of two vectors a and b is written as a.b and pronounced as (a dot b). Therefore, a.b =|a| |b| cos dot is defined as a.b = a b cos where is the angle between vectors a and b
If a = a1 I + a2j and b = b1 I b2j
Thus a .b = (a)1bi ii + ab2j I 1 + 2 bi I h +a2 b2 j
Recall that I and j are mutually perpendicular unit vector hence
i.i = |x| cos 0 =1
i.j = |x| cos 90 =0
j.i = |x| cos 90 =0
j.i =|x| cos 0 =1
Hence, a.b =a1b1 + a2 b2
Examples
- Find the scale product of the following vectors 9i -2j + k and I – 3j -4k
Solution:
A=(9i- 2j +k) and b= (i-3j -4k)
a.b = (9i-2j +k) (i-3j-4k)
=9 (1) -2(-3) + 1(-4)=9+6-4a.b =11
2. Let a = 3i+2j, b = -4i+2j and c = i+4j, calculate a.b, a.c and a. (b+c)
Solution:
I a.b = (3i + 2j ) (-4i+2j) = 3 (-4) +2(2)
= -12+4
=-8
II a.c = (3i+2j) (I +4j)
= 3 (1) + 2 (4)
= 3+8 = 11
III a.(b+c)
Find (b+c) = -4i + 2j +i +4j
=-3i +6j - (b+c) = (3i+2j) (-3i +6j)
=3(-3) + 2(6)
= -9+12 = 3.PERPENDICULARITY OF VECTORS:
If two vectors P and q are in perpendicular directions, thus p.q =0
Example 1: show that the vectors p = 3i+ 2j and q= -2i + 3j are perpendicular.
Solution:
P:q = (3i+2j) (-2i +3j)
=3(-2) + 2(3)
=-6+ =0
Since p.q=0, then the vectors p and q are perpendicular.
2. If p= 4i + kj and q=2i – 3j are perpendicular, find the value of k, where k is a scalar..
Solution:
p.q=0
(4i+kj)(2i-3j)=0
4(2) + k(-3)=0
-3k=-8
K=8/3.EQUAL VECTORS: Vectors p ad q are equal if p is equal to q.
Example: find the value of the scalar K for which the vectors 2ki + 3j and 8i+kj
Solution:
2ki +3j = 8i +kj
Hence, 2ki =8i, 3j=
2k = 8 12=3k
K=8/2 k=12/3
K=4 k=4
EVALUATION
- The vectors AB and C are -2i+6j-3k and -2i-3j+6k respectively. Find the scalar product AB.AC
- Find the value of the scalar A for which the pairs of vectors 5i +3j and 2i-4Aj are perpendicular.
ANGLES BETWEEN TWO VECTORS
Is the angle between two vectors and from dot product where a.b=|a|b| cos . Hence, Cos
Where = Magnitude of vector a= 21 +22
|b| =Magnitude of vector b=21 +22Example:
Find the angle between the vectors pp=2i – 2j + k and q=12i +4j – 3k
Solution:
Cos =
p.q= (2i-2j+k)(12i+4j-3k)
=2(12) -2(4)_1(3)
= 24-8-3
=13
|p|=2 + (-2)2 + 12=+4+1 ==3
|q|= 2 + 42 + (-3) = 144+16+9= 169 =13
Cos
Cos =1/3. =Cos-1 (1/3)DIRECTION COSINES A VECTOR:
The direction is specified by the angles which the vector makes wit x and y axes. If we represent these angles by and respectively then,
Cos = Cos =
Example: find the direction cosine of the vector 4i + 3J – 11k
Solution:
Let a = 4i + 3J-11k
|a|= 2 + y2 +z2 = |=2 + 2 + (-11)2 =
Direction cosine, Cos Cos =Cos=EVALUATION
- Find the angle between the vectors 2i + 3j +6k and 3i+4j+12k
- Find the direction cosine of vector a = 10i- j+2k
EVALUATION: find the projection of the vector a on the vector b if a=5i-4j+2k and b=6i – j +3k
GENERAL REVISION EVALUATION
- Given that a=4i – 2j +k, b=2i – j +3k and c=5i +2k find (i) (a+b)c (ii) a=c+b.c
- If a = 4i – 2j +k, b=6i +5j find (i) the unit vector in direction of b. (ii) the projection of a on b (iii) the unit ve4xtor in the direction of a (iv) the projection of b on a.
READING ASSIGNMENT: Read vector Geometry, Further Mathematics project II page 236-240
WEEKEND ASSIGNMENT
- If = 3i + 4j and b=gi +2k are perpendicular, what is the value of g? A.-4 B.3 C.-8/3
- Find the value of the scalar k for which the vectors ki + 8j and 3i + are equal. A. 3 B.6 C.9
- Find the projection of the vector a on the vector if a = 4i + 6j and b=3i-2j. A.-3\ 52 B.5\ 13 C. 0
- Calculate the angle between a = -4i +2j and b =I -3j. A.450 B.600 C.1350
- Find the scalar product of vectors – 2i-3j and 4i +5j? A.-23 B.23 C.7
THEORY
1a) Given that a = 4i – 5j + 2k and b = -7i + 3j – 6k find the scalar product of a and b (b) find the direction cosine 2a + 3b
2 ) Find the angle between p = 6i + 2j – 4k and q = 9i + 5j