Page 33 - 4263

Dot product of vectors

−→2
Thus, the scalar square of a vector, which is written as a , equals to a
square of its length:
−→2 −→ 2
a = | a | . (5.4)

− →
π
−→
−→
−→
−→
Example 5.1. Let’s have: vector c = 3 a − 2 b , | a | = 2, b = 3, φ = / .
3
− →
Find the module of vector c . ,
−→
−→
−→
−→ 2
−→
−→
2
− →2
−→2
2
Solution. According to formula (5.4): | c | = c = (3 a −2 b ) = 9 a −2·3 a ·2 b +4 b =
√
− →
9 · 4 − 2 · 3 · 2 · 2 · 3 · + 4 · 9 = 3 · 36. So, | c | = 6 3.
1
2
Let’s consider the dot product in the view of coordinates.
Theorem 5.1.
−→
Let’s have two vectors in their coordinates: a = (x ; y ; z ), b = (x ; y ; z ). A dot
−→
1
1
1
2
2
2
product of these vectors equals to the sum of multiplications of corresponding coordinates:
−→
−→
a · b = x x + y y + z z (5.5)
1 2
1 2
1 2
⋆
− → −→ −→ −→ −→ −→
−→
PROOF. Expansions of vectors a and b through base vectors i , j , k are: a = x 1 i +
−→ − → −→ −→ −→ −→
y 1 j + z 1 k , b = x 2 i + y 2 j + z 2 k . The dot product of these vectors equals:
−→ −→ −→ 2 −→ −→ −→ −→ − → −→ −→ 2
a · b = x 1 x 2 · i + x 1 y 2 · i · j + x 1 z 2 · i · k + y 1 x 2 · i · j + y 1 y 2 · j +
−→ −→ −→ −→ −→ −→ −→ 2
+y 1 z 2 · j · k + z 1 x 2 · i · k + z 1 y 2 · j · k + z 1 z 2 · k .
−→ 2 −→ 2 − → 2 −→ −→ −→ −→ −→ −→
Taking into account, that i = j = k = 1, i · j = i · k = j · k = 0, we will get:
−→
−→
a · b = x 1 x 2 + y 1 y 2 + z 1 z 2 that it was necessary to prove. 2
Corollary 1.
1. A necessary and sufficient condition of perpendicularity of two vectors is the
equality:
x 1 x 2 + y 1 y 2 + z 1 z 2 = 0. (5.6)
−→
−→
2. The angle between vectors a = (x 1 ; y 1 ; z 1 ) and b = (x 2 ; y 2 ; z 2 ) can be
calculated according to the formula:
x 1 x 2 + y 1 y 2 + z 1 z 2
cos φ = √ √ (5.7)
2 2 2 2 2 2
x + y + z · x + y + z 2
1
1
2
1
2
−→
3. The projection of vector a = (x 1 ; y 1 ; z 1 ) onto the direction of vector
−→
b = (x 2 ; y 2 ; z 2 ) can be calculated according to the formula:
− →
pr−→ a = x 1 x 2 + y 1 y 2 + z 1 z 2 (5.8)
√
b 2 2 2
x + y + z 2
2
2
These formulas obviously implies from Theorem 5.1. 2
33