Question

нужно срочно решить в классе ()

Answer 1

2
a(3; -2; 1)
b(3; 4; -2)
c(5; 4; -2)
a - 3b = (3; -2; 1) - 3(3; 4; -2) = (3; -2; 1) - (9; 12; -6) = (-6; 10; 7)
|a - 3b| = √(6²+10²+7²) = √(36+100+49) = √185
c+b = (5; 4; -2) + (3; 4; -2) = (8;8;-4)
|c+b| = √(8²+8²+4²) = √(64+64+16) = √144 = 12
|2a| + 3|b| = 2|a| + 3|b| = 2√(3²+2²+1²) + 3√(3²+4²+2²) = 2√(9+4+1) + 3√(9+16+4) = 2√13 + 3√29
3
AB = B-A = (6;-8;-2) - (5;-8;-1) = (6-5; -8+8; -2+1) = (1; 0; -1)
CD = D-C = (7; -7; -9) - (7; -5; -11) = (7-7; -7+5; -9+11) = (0; -2; 2)
|AB| = √(1²+0²+1²) = √2
|CD| = √(0²+2²+2²) = √8
AB·CD = 1*0 + 0*(-2) - 1*2 = -2
cos(AB^CD) = AB·CD/(|AB|*|CD|) = -2/(√2√8) = -2/√16 = -2/4 = -1/2
AB^CD = arccos(-1/2) = 120°