Back to: Further Mathematics SS2
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
1. 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
a. (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
1. The vectors AB and C are -2i+6j-3k and -2i-3j+6k respectively. Find the scalar product AB.AC
2. Find the value of the scalar A for which the pairs of vectors 5i +3j and 2i-4Aj are perpendicular.
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