Здравствуйте, помогите пожалуйста…. Created by aleksoref. algebra-ru

Здравствуйте, помогите пожалуйста…...

jordanffjrf

Ответ:

$1)\ \ |\vec{a}|=3\ \ ,\ \ |\vec{b}|=4\ \ ,\ \ \angle {(\vec{a},\vec{b})}=120^\circ \\\\\vec{a}\cdot \vec{b}= |\vec{a}|\cdot |\vec{b}|\cdot cos\angle {(\vec{a},\vec{b})}=3\cdot 4\cdot cos120^\circ =12\cdot \dfrac{-1}{2}=-6\\\\\\2)\ \ \vec{a}=\{2n;-3;-6\}\ \ ,\ \ \vec{b}=\{3;-n;-3\}\\\\\vec{a}\perp \vec{b}\ \ \ \Leftrightarrow \ \ \ \vec{a}\cdot \vec{b}=0\\\\\vec{a}\cdot \vec{b}=2n\cdot 3-3\cdot (-n)-6\cdot (-3)=6n+3n+18=9n+18=0\ \ ,\ \ n=-2$

$3)\ \ \vec{a}=\{5;-2;7\}\ \ ,\ \ \vec{b}=\{7;5;2\}\\\\cos\angle {(\vec{a},\vec{b})}=\dfrac{\vec{a}\cdot \vec{b}}{|\vec{a}|\cdot |\vec{b}|}=\dfrac{35-10+14}{\sqrt{25+4+49}\cdot \sqrt{49+25+4}}=\dfrac{39}{78}=\dfrac{1}{2}\\\\\\\angle {(\vec{a},\vec{b})}=arccos\, \dfrac{1}{2}=60^\circ$

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NNNLLL54

Verified answer

Ответ:

2)))

Объяснение:

2n*3+(-3)*(-n)+(-6)*(-3)=0

6n+3n+18=0

9n+18=0

n=-2

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