Длины вектора a и b равны 16 и 9 аугол

Решите задачу:

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

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

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