Question

длины вектора a и b равны 16 и 9 аугол

Answer 1

$1); ; vec{a}cdot vec{b}=16cdot 9cdot cos60^circ =144cdot 0,5=72\\2); ; frac{7}{p}=frac{-5}{15}; ; to ; ; p=frac{7cdot 15}{-5} =-21\\3); ; vec{a}cdot vec{b}=-7cdot 4-14p=0; ; to ; ; p=frac{-7cdot 4}{14}=-2\\4); ; cosphi =frac{vec{a}cdot vec{b}}{|vec{a}|cdot |vec{b}|}=frac{-3-12}{sqrt{5}cdot sqrt{25}}=-frac{15}{5sqrt5} =-frac{3}{sqrt5}; ;; phi =pi -arccosfrac{3}{sqrt5}\\7); ; cosphi =frac{3}{2cdot 3sqrt3}=frac{1}{2sqrt3}= frac{sqrt3}{6}$

$8); ; vec{a}cdot vec{b}=-4-psqrt3=-16; ,; ; psqrt3=12; ,; ; p=frac{12}{sqrt3}=4sqrt3\\9); ; overline {MN}={-6;9}; ,; ; overline {MK}={3;2}\\cosM=frac{-18+18}{sqrt{36+81}cdot sqrt{9+4}}=0; ; to ; ; angle M=90^circ \\10); ; vec{a}cdot vec{b}=-8p-frac{sqrt2}{4}=-frac{65sqrt2}{4}; ,; ; 8p=frac{65sqrt2}{4}-frac{sqrt2}{4}=frac{64sqrt2}{4}=16sqrt2\\p=2sqrt2; ,; ; cosphi =frac{-65sqrt2}{4sqrt{8+frac{1}{8}}cdot sqrt{64+1}}=-frac{65sqrt2}{4sqrt{frac{65}{4}}cdot sqrt{65}}=-frac{sqrt2cdot 2sqrt2}{4}=-1$

$phi =180^circ \\11); ; overline {AB}={8;-6}\\|overline {AB}|=sqrt{8^2+6^2}=10$