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0student0
@0student0
March 2022
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Ребят решите сначала 6 , а потом по возможности 5.
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Indentuum
5) a(-1, 2, 4); b(3, 2, -1)
(2a, b) = |2a|*|b|*cos(2a, b) = 2a_x*b_x + 2a_y*b_y + 2a_z*b_z
cos(2a, b) = (-6 + 8 - 8)/((4 + 16 + 64)*(9 + 4 + 1)) = -6/sqrt(1176) = -3/sqrt(294)
6) c(3, -2, 1); 2c(6, -4, 2)
|2c| = sqrt(36 + 16 + 4) = sqrt(56) = 2sqrt(14)
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Answers & Comments
(2a, b) = |2a|*|b|*cos(2a, b) = 2a_x*b_x + 2a_y*b_y + 2a_z*b_z
cos(2a, b) = (-6 + 8 - 8)/((4 + 16 + 64)*(9 + 4 + 1)) = -6/sqrt(1176) = -3/sqrt(294)
6) c(3, -2, 1); 2c(6, -4, 2)
|2c| = sqrt(36 + 16 + 4) = sqrt(56) = 2sqrt(14)