# Konsultimet Mat avancuar 20 qershor. Ushtrime te zgjidhura dhe udhezime.

1.Jepet trekendeshi me kulme A(2;1;1), B(3;-2;2) dhe C(0;3;-1). Te gjendet gjatesia e mesores BM.

zgjidhje

Gjejme kordinatat e pikes M mesi i AC.  $x_{M}=\frac{x_{A}+x_{C}}{2}=\frac{2+0}{2}=1$,   $y_{M}=\frac{y_{A}+y_{C}}{2}=\frac{1+3}{2}=2$$z_{M}=\frac{z_{A}+z_{C}}{2}=\frac{1+(-1)}{2}=0$  .  M(1;2;0).   Gjejme vektorin $\overrightarrow{BM}=\left&space;(&space;\begin{matrix}&space;1-3\\&space;2+2\\&space;0-2\end{matrix}&space;\right&space;)=\left&space;(&space;\begin{matrix}&space;-2\\&space;4\\&space;-2\end{matrix}&space;\right&space;)$  , gjejme gjatesine e tij.  $\left&space;|&space;\overrightarrow{BM}&space;\right&space;|=\sqrt{(-2)^{2}+4^{2}+(-2)^{2}}=\sqrt{24}=2\sqrt{6}$  .

2.Jepen vektoret  $\vec{u}$ dhe $\vec{v}$ , te tille qe $\left&space;|&space;\vec{u}-\vec{v}&space;\right&space;|^{2}=\left&space;|&space;\vec{u}&space;\right&space;|^{2}+\left&space;|&space;\vec{v}&space;\right&space;|^{2}$  . Te vertetohet se keta vektore jane pingule.

zgjidhje

$\left&space;|&space;\vec{u}-\vec{v}&space;\right&space;|^{2}={\left&space;(&space;\vec{u}-\vec{v}&space;\right&space;)^{2}}={\vec{u}^{2}-2\cdot&space;\vec{u}\cdot&space;\vec{v}+\vec{v}^{2}}$    sjell ${\vec{u}^{2}-2\cdot&space;\vec{u}\cdot&space;\vec{v}+\vec{v}^{2}}=\vec{u}^{2}+\vec{v}^{2}$   katrori numerik i vektorit eshte i barabarte me katrorin e gjatesise se tij.  Prandaj kemi  $-2\cdot&space;\vec{u}\cdot&space;\vec{v}=0\Rightarrow&space;\vec{u}\cdot&space;\vec{v}=0\Rightarrow&space;\vec{u}\perp&space;\vec{v}$

3.Jepet $\left&space;|&space;\vec{a}&space;\right&space;|=2$  , $\left&space;|&space;\vec{b}&space;\right&space;|=5$  dhe  $\vec{a}\cdot&space;\vec{b}=-5$  . Gjeni   $\left&space;|\vec{a}X&space;\vec{b}&space;\right&space;|$   (gjatesine e vektorit prodhim vektorial te tyre)

zgjidhje

$2\cdot&space;5\cdot&space;cos\alpha&space;=-5\Rightarrow&space;cos\alpha&space;=-\frac{1}{2}\Rightarrow&space;sin\alpha&space;=\sqrt{1-\left&space;(&space;-\frac{1}{2}&space;\right&space;)^{2}}=\sqrt{\frac{3}{4}}=\frac{\sqrt{3}}{2}$      atehere  $\left&space;|&space;\vec{a}X\vec{b}&space;\right&space;|=2\cdot&space;5\cdot&space;\frac{\sqrt{3}}{2}=5\sqrt{3}$

4.Te gjendet siperfaqja e trekendeshi me kulme  A(3;4;-1), B(2;0;3)  dhe C(-3;5;4)

zgjidhje

**Siperfaqja e paralelogramit te ndertuar mbi dy vektore eshte e barabarte me gjatesine e vektorit prodhim vektorial te tyre.  VO Nese vektoret jane bashkevijore nuk formohet paralelogram ne kete rast prodhimi vektorial eshte i barabarte me vektorin zero. Ndryshe KNM qe dy vektore te jene bashkevijore eshte qe prodhimi vektorial i tyre te jete vektori zero.

Ne rastin tone siperfaqja e trekendeshitdo te jete gjysma e siperfaqes se paralelogramit. Gjejme vektoret $\overrightarrow{AB}=\left&space;(&space;\begin{matrix}&space;-1\\&space;-4\\&space;4\end{matrix}&space;\right&space;)$  $\overrightarrow{AC}=\left&space;(&space;\begin{matrix}&space;-6\\&space;1\\&space;5\end{matrix}&space;\right&space;)$

**Prodhimi vektorial i dy vektoreve kur jane dhene kordinatat e tyre gjendet duke llogaritur percaktorin ne te cilin ne shtyllen e pare vendosen vektoret njesi ndersa ne dy shtyllat tjera perkatesisht kordinatat e dy vektoreve.

$\overrightarrow{AB}X\overrightarrow{AC}&space;=\left&space;|&space;\begin{matrix}&space;\vec{i}&space;&&space;-1&space;&&space;-6\\&space;\vec{j}&&space;-4&space;&&space;1\\&space;\vec{k}&&space;4&space;&&space;5&space;\end{matrix}&space;\right&space;|=-20\vec{i}-24\vec{j}-\vec{k}-24\vec{k}+5\vec{j}-4\vec{i}=-24\vec{i}-19\vec{j}-25\vec{k}$

$S=\frac{1}{2}\left&space;|&space;\overrightarrow{AB}X\overrightarrow{AC}&space;\right&space;|=\frac{1}{2}\sqrt{(-24)^{2}+(-19)^{2}+(-25)^{2}}=\frac{1}{2}\sqrt{1562}$