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alextraza22
@alextraza22
November 2021
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Решите, пожалуйста. Задача с векторами.
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Kазак
Зададим векторам конкретные значения, удовлетворяющие условию
a(2;0)
b(5cos(120°); 5sin(120°)) = (-5/2; 5√3/2)
p = xa + 17b = x(2;0) + 17(-5/2; 5√3/2) = (2x; 0) + (-85/2; 85√3/2)
p = (2x - 85/2; 85√3/2)
q = 3a - b = 3(2; 0) - (-5/2; 5√3/2) = (6; 0) + (5/2; -5√3/2) = (6 + 5/2; 0 - 5√3/2) = (12/2 + 5/2; - 5√3/2)
q = (17/2; - 5√3/2)
Для ортогональных векторов скалярное произведение равно 0
p·q = 0
(2x - 85/2)(17/2) + (85√3/2)(- 5√3/2) = 0
17x - 85*17/4 - 85*3*5/4 = 0
умножим на 4
4*17x - 85*17 - 85*3*5 = 0
разделим на 17
4x - 85 - 5*3*5 = 0
4x - 85 - 75 = 0
4x - 160 = 0
x = 40
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Answers & Comments
a(2;0)
b(5cos(120°); 5sin(120°)) = (-5/2; 5√3/2)
p = xa + 17b = x(2;0) + 17(-5/2; 5√3/2) = (2x; 0) + (-85/2; 85√3/2)
p = (2x - 85/2; 85√3/2)
q = 3a - b = 3(2; 0) - (-5/2; 5√3/2) = (6; 0) + (5/2; -5√3/2) = (6 + 5/2; 0 - 5√3/2) = (12/2 + 5/2; - 5√3/2)
q = (17/2; - 5√3/2)
Для ортогональных векторов скалярное произведение равно 0
p·q = 0
(2x - 85/2)(17/2) + (85√3/2)(- 5√3/2) = 0
17x - 85*17/4 - 85*3*5/4 = 0
умножим на 4
4*17x - 85*17 - 85*3*5 = 0
разделим на 17
4x - 85 - 5*3*5 = 0
4x - 85 - 75 = 0
4x - 160 = 0
x = 40