Dot product of two vectors
a · b = |a| · |b| cos α
Dot product - formulas
Dot product formula for plane problems
In the case of the plane problem the dot product of vectors a = {ax ; ay} and b = {bx ; by} can be found by using the following formula:
Dot product formula for spatial problems
In the case of the spatial problem the dot product of vectors a = {ax ; ay ; az} and b = {bx ; by ; bz} can be found by using the following formula:
Dot product formula for n dimensional space problems
In the case of the n dimensional space problem the dot product of vectors a = {a1 ; a2 ; ... ; an} and b = {b1 ; b2 ; ... ; bn} can be found by using the following formula:
Properties of dot product of vectors
- The dot product of a vector with itself is always greater than zero or equal to zero:
a · a ≥ 0
- The dot product of a vector with itself is zero if and only if the vector is the zero vector:
a · a = 0 <=> a = 0
- The dot product of a vector with itself is equal to the square of its magnitude:
a · a = |a|2
- The dot product operation is communicative:
a · b = b · a
- If the dot product of two not zero vectors is is zero, then these vectors are orthogonal:
a ≠ 0, b ≠ 0, a · b = 0 <=> a ┴ b
- (αa) · b = α(a · b)
- The dot product operation is distributive:
(a + b) · c = a · c + b · c
Dot product - example
Examples of calculation of the dot product of vectors for plane problems
Solution: a · b = 1 · 4 + 2 · 8 = 4 + 16 = 20.
Solution: a · b = |a| · |b| cos α = 3 · 6 · cos 60˚ = 9.
Solution:
p · q = (a + 3b) · (5a - 3b) = 5 a · a - 3 a · b + 15 b · a - 9 b · b == 5 |a|2 + 12 a · b - 9 |b|2 = 5 · 32 + 12 · 3 · 2 · cos 60˚ - 9 · 22 = 45 +36 -36 = 45.
Examples of calculation of the dot product of vectors for spatial problems
Solution: a · b = 1 · 4 + 2 · 8 + (-5) · 1 = 4 + 16 - 5 = 15.
Examples of calculation of the dot product of vectors for n dimensional space problems
Solution: a · b = 1 · 4 + 2 · 8 + (-5) · 1 + 2 · (-2) = 4 + 16 - 5 -4 = 11.
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