Given the vector | a | = 4 and the vector | B | = 3, the scalar product is obtained under the following conditions: 1 / vector a is perpendicular to vector B 2 / the angle between vector a and vector B is 60 ° 3 / vector a is parallel to B

Given the vector | a | = 4 and the vector | B | = 3, the scalar product is obtained under the following conditions: 1 / vector a is perpendicular to vector B 2 / the angle between vector a and vector B is 60 ° 3 / vector a is parallel to B

|a|=4,|b|=3
one
A ⊥ B, then: a · B = 0
two
=So: a · B = | a | * | B | * cos (π / 3) = 4 * 3 / 2 = 6
three
=0 or π, then: a · B = |a | * | B | * cos (0) = 12 or a · B = |a | * | * cos (π) = - 12