Найдите cos угла между векторами

Косинус угла между векторами a, b считается как: cos fi = a*b / (|a|*|b|). Если вектор a = (xa, ya, za), b = (xb, yb, zb), то a*b = xa*xb + ya*yb + za*zb |a| = sqrt(xa^2 + ya^2 + za^2), |b| = sqrt(xb^2 + yb^2 + zb^2). 1) a = (2;-1;3), b = (2;-1;0) a*b = 2*2 + (-1)*(-1) + 3*0 = 5 |a| = sqrt(2^2 + (-1)^2 + 3^2) = sqrt(14) |b| = sqrt(2^2 + (-1)^2 + 0^2) = sqrt(5) cos fi = 5 / (sqrt(14) * sqrt(5)) = sqrt(5/14) = sqrt(70)/14 2) a = (2;2;1), b = (4;1;-7) a*b = 2*4 + 2*1 + 1*(-7) = 3 |a| = sqrt(2^2+2^2+1^2)=3 |b| = sqrt(4^2 + 1^2 + (-7)^2) = sqrt(66) cos fi = 3 / (3 * sqrt(66)) = sqrt(1/66) = sqrt(66)/66 3) a = (2;2;-1), b = (-3;6;-6) a*b = 2*(-3) + 2*6 + (-1)*(-6) = 12 |a| = sqrt(2^2 + 2^2 + (-1)^2) = 3 |b| = sqrt((-3)^2 + 6^2 + (-6)^2) = 9 cos fi = 12 / (3*9) = 4/9 4) a = (2;-1;3), b = (4;1;-7) a*b = 2*4 + (-1)*1 + 3*(-7) = -14 |a| = sqrt(2^2 + (-1)^2 + 3^2) = sqrt(14) |b| = sqrt(4^2 + 1^2 + (-7)^2) = sqrt(66) cos fi = (-14)/(sqrt(14)*sqrt(66)) = -sqrt(14/66)=-sqrt(7/33)=-sqrt(231)/33 5) a = (3;5;-2), b = (2;-1;0) a*b = 3*2 + 5*(-1) + (-2)*0 = 1 |a| = sqrt(3^2 + 5^2 + (-2)^2) = sqrt(38) |b| = sqrt(2^2 + (-1)^2 + 0^2) = sqrt(5) cos fi = 1/(sqrt(38)*sqrt(5)) = sqrt(190)/190