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»VECTORI

Data publicarii: 30 August, 2010

EXEMPLUL 1

Suport teoretic:

Triunghi dreptunghic, centru de greutate, teorema bisectoarei, norma unui vector.

Enunt:

In triunghiul dreptunghic ABC (Â - drept) se da: AB = 4a, AC = 3a, a > 0,

G - centrul de greutate, iar D - piciorul bisectoarei din varful C.

Sa se calculeze lungimea segmentului DG.

Raspuns:

$DG=\frac{a\cdot{\sqrt{37}}}{6}.$ DG=\frac{a\cdot{\sqrt{37}}}{6}.

Rezolvare:

Se considera reperul ortogonal cu originea in A, axele absciselor si ordonatelor

fiind [AB, respectiv [AC. Avem:

$\overrightarrow{DG}=\overrightarrow{DA}+\overrightarrow{AC}+\overrightarrow{CG}=\overrightarrow{DA}+3a\vec{j}+{\frac{2}{3}}\cdot{\overrightarrow{CE}},$ \overrightarrow{DG}=\overrightarrow{DA}+\overrightarrow{AC}+\overrightarrow{CG}=\overrightarrow{DA}+3a\vec{j}+{\frac{2}{3}}\cdot{\overrightarrow{CE}},

unde E este mijlocul catetei AB. Apoi:

$\overrightarrow{DG}=\overrightarrow{DA}+3a\vec{j}+{\frac{2}{3}}\cdot{\frac{\overrightarrow{CA}+\overrightarrow{CB}}{2}}=\overrightarrow{DA}+3a\vec{j}+\frac{-3a\vec{j}-3a\vec{j}+4a\vec{i}}{3}={\frac{1}{3}}\cdot{(3\overrightarrow{DA}+3a\vec{j}+4a\vec{i})}.\;(1)$ \overrightarrow{DG}=\overrightarrow{DA}+3a\vec{j}+{\frac{2}{3}}\cdot{\frac{\overrightarrow{CA}+\overrightarrow{CB}}{2}}=\overrightarrow{DA}+3a\vec{j}+\frac{-3a\vec{j}-3a\vec{j}+4a\vec{i}}{3}={\frac{1}{3}}\cdot{(3\overrightarrow{DA}+3a\vec{j}+4a\vec{i})}.\;(1)

Conform teoremei bisectoarei, rezulta:

${\frac{DA}{DB}=\frac{AC}{BC}}$ {\frac{DA}{DB}=\frac{AC}{BC}} $\Leftrightarrow$ \Leftrightarrow ${\frac{DA}{DA+DB}=\frac{AC}{AC+BC}}\;(proportii\;derivate);\;deci:$ {\frac{DA}{DA+DB}=\frac{AC}{AC+BC}}\;(proportii\;derivate);\;deci:

$\frac{DA}{4a}={\frac{3a}{3a+5a}}\Leftrightarrow{\cdots}\Leftrightarrow{DA=\frac{3a}{2}}.\;(2)$ \frac{DA}{4a}={\frac{3a}{3a+5a}}\Leftrightarrow{\cdots}\Leftrightarrow{DA=\frac{3a}{2}}.\;(2)

Din (1) si (2) deducem:

$\overrightarrow{DG}={\frac{1}{3}}\cdot{({-\frac{9a}{2}}\cdot{\vec{i}}+3a\vec{j}+4a\vec{i})}=\cdots=-{\frac{a}{6}}\cdot{\vec{i}}+a\vec{j}.$ \overrightarrow{DG}={\frac{1}{3}}\cdot{({-\frac{9a}{2}}\cdot{\vec{i}}+3a\vec{j}+4a\vec{i})}=\cdots=-{\frac{a}{6}}\cdot{\vec{i}}+a\vec{j}.

Lungimea segmentului DG este deci:

$||\overrightarrow{DG}||=\sqrt{(-\frac{a}{6})^2+a^2}=\cdots=\frac{a\cdot{\sqrt{37}}}{6}.$ ||\overrightarrow{DG}||=\sqrt{(-\frac{a}{6})^2+a^2}=\cdots=\frac{a\cdot{\sqrt{37}}}{6}.