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natalya128zz
@natalya128zz
August 2022
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Знайдіть кут між векторами a(-2,2√3) b(3,-√3)
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Kазак
Скалярное произведение векторов нам поможет
a·b = |a|*|b|*cos(β)
cos(β) = a·b/(|a|*|b|)
a·b = -2*3 + 2√3*(-√3) = - 6 - 2*3 = - 6 - 6 = -12
|a| = √(2² + (2√3)²) = √(4 + 4*3) = √16 = 4
|b| = √(3² + (√3)²) = √(9 + 3) = √12 = 2√3
cos(β) = - 12/(4*2√3) = - 3/(2√3) = - (√3)²/(2√3) = - √3/2
β = arccos(- √3/2) = 5/6*π = 150°
1 votes
Thanks 1
natalya128zz
Вы не могли бы объяснить как из -3/(2√3) получилось -√3/2?
Kазак
3 = (√3) * (√3)
natalya128zz
спасибо)
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Answers & Comments
a·b = |a|*|b|*cos(β)
cos(β) = a·b/(|a|*|b|)
a·b = -2*3 + 2√3*(-√3) = - 6 - 2*3 = - 6 - 6 = -12
|a| = √(2² + (2√3)²) = √(4 + 4*3) = √16 = 4
|b| = √(3² + (√3)²) = √(9 + 3) = √12 = 2√3
cos(β) = - 12/(4*2√3) = - 3/(2√3) = - (√3)²/(2√3) = - √3/2
β = arccos(- √3/2) = 5/6*π = 150°