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Algebroidka
@Algebroidka
October 2021
1
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Спасайте. Прошу вашей помощи. Задание в прикрепленных.
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nKrynka
Решение
a = - i + j
a(-1 ; 1 ; 0)
b = i - 2j + 2k
b(1 ; - 2 ; 2)
CosФ = (x₁x₂ + y₁y₂ + z₁z₂) / [√(x₁² + y₁² + z₁²)(x₂² + y₂² + z₂²)]
cosФ = (-1 - 2 + 0) / [√(1 + 1 + 0)*√(1 + 4 + 4)] = - 3 / [√2 *√9] =
= - 3/3√2 = - 1/√2
cosФ = - √2/2
Ф = 3π/4
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Answers & Comments
a = - i + j
a(-1 ; 1 ; 0)
b = i - 2j + 2k
b(1 ; - 2 ; 2)
CosФ = (x₁x₂ + y₁y₂ + z₁z₂) / [√(x₁² + y₁² + z₁²)(x₂² + y₂² + z₂²)]
cosФ = (-1 - 2 + 0) / [√(1 + 1 + 0)*√(1 + 4 + 4)] = - 3 / [√2 *√9] =
= - 3/3√2 = - 1/√2
cosФ = - √2/2
Ф = 3π/4