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»VECTORI

Data publicarii: 30 Martie, 2011

VECTORI IN SPATIU

Expresia analitica a unui vector:

${\overrightarrow{AB}}={({x_B}-{x_A})}{\vec{i}}+{({y_B}-{y_A})}{\vec{j}}+{({z_B}-{z_A})}{\vec{k}},$ {\overrightarrow{AB}}={({x_B}-{x_A})}{\vec{i}}+{({y_B}-{y_A})}{\vec{j}}+{({z_B}-{z_A})}{\vec{k}},

unde A(xA,yA,zA) si B(xB,yB,zB).

Produsul scalar a doi vectori:

Se numeste produsul scalar al vectorilor

$\vec{a}\:si\:\vec{b}$ \vec{a}\:si\:\vec{b}

numarul real, notat

${\vec{a}}\cdot{\vec{b}},$ {\vec{a}}\cdot{\vec{b}},

definit prin:

${\vec{a}}\cdot{\vec{b}}={|\vec{a}|}\cdot{|\vec{b}|}\cdot{\cos{(\widehat{\vec{a},\vec{b}})}}.$ {\vec{a}}\cdot{\vec{b}}={|\vec{a}|}\cdot{|\vec{b}|}\cdot{\cos{(\widehat{\vec{a},\vec{b}})}}.

Daca se dau vectorii

${\vec{a}}={x_1}{\vec{i}}+{y_1}{\vec{j}}+{z_1}{\vec{k}}$ {\vec{a}}={x_1}{\vec{i}}+{y_1}{\vec{j}}+{z_1}{\vec{k}}

si:

${\vec{b}}={x_2}{\vec{i}}+{y_2}{\vec{j}}+{z_2}{\vec{k}},$ {\vec{b}}={x_2}{\vec{i}}+{y_2}{\vec{j}}+{z_2}{\vec{k}},

atunci:

${\vec{a}}\cdot{\vec{b}}={x_1}{x_2}+{y_1}{y_2}+{z_1}{z_2}$ {\vec{a}}\cdot{\vec{b}}={x_1}{x_2}+{y_1}{y_2}+{z_1}{z_2}

(produsul scalar a doi vectori este egal cu suma produselor coordonatelor celor doi

vectori).

Modulul unui vector:

$|\vec{a}|=\sqrt{x^2+y^2+z^2},$ |\vec{a}|=\sqrt{x^2+y^2+z^2},

unde x, y si z sunt coordonatele vectorului $\vec{a},$ \vec{a},

adica:

$\vec{a}=x\vec{i}+y\vec{j}+z\vec{k}.$ \vec{a}=x\vec{i}+y\vec{j}+z\vec{k}.

Unghiul dintre doi vectori:

${\vec{a}}={x_1}{\vec{i}}+{y_1}{\vec{j}}+{z_1}{\vec{k}}$ {\vec{a}}={x_1}{\vec{i}}+{y_1}{\vec{j}}+{z_1}{\vec{k}}

${\vec{b}}={x_2}{\vec{i}}+{y_2}{\vec{j}+{z_2}{\vec{k}}},$ {\vec{b}}={x_2}{\vec{i}}+{y_2}{\vec{j}+{z_2}{\vec{k}}},

care formeaza un unghi de masura φ, este dat de formula:

${cos}{\varphi}=\frac{{\vec{a}}\cdot{\vec{b}}}{|\vec{a}|\cdot|\vec{b}|}=\frac{{x_1}{x_2}+{y_1}{y_2}+{z_1}{z_2}}{{\sqrt{{x_1}^{2}+{y_1}^{2}+{z_1}^{2}}}\cdot{\sqrt{{x_2}^{2}+{y_2}^{2}+{z_2}^{2}}}}.$ {cos}{\varphi}=\frac{{\vec{a}}\cdot{\vec{b}}}{|\vec{a}|\cdot|\vec{b}|}=\frac{{x_1}{x_2}+{y_1}{y_2}+{z_1}{z_2}}{{\sqrt{{x_1}^{2}+{y_1}^{2}+{z_1}^{2}}}\cdot{\sqrt{{x_2}^{2}+{y_2}^{2}+{z_2}^{2}}}}.

Corolar:

Vectorii nenuli

${\vec{a}}={x_1}{\vec{i}}+{y_1}{\vec{j}}+{z_1}{\vec{k}}$ {\vec{a}}={x_1}{\vec{i}}+{y_1}{\vec{j}}+{z_1}{\vec{k}}

${\vec{b}}={x_2}{\vec{i}}+{y_2}{\vec{j}}+{x_2}{\vec{k}}$ {\vec{b}}={x_2}{\vec{i}}+{y_2}{\vec{j}}+{x_2}{\vec{k}}

sunt ortogonali daca si numai daca:

${\vec{a}}\cdot{\vec{b}}={x_1}{x_2}+{y_1}{y_2}+{z_1}{z_2}=0.$ {\vec{a}}\cdot{\vec{b}}={x_1}{x_2}+{y_1}{y_2}+{z_1}{z_2}=0.

Observaţie:

Din

${\vec{a}}\cdot{\vec{b}}={|\vec{a}|}\cdot{|\vec{b}|}\cdot{\cos{(\widehat{\vec{a},\vec{b}})}}$ {\vec{a}}\cdot{\vec{b}}={|\vec{a}|}\cdot{|\vec{b}|}\cdot{\cos{(\widehat{\vec{a},\vec{b}})}}

se deduce inegalitatea:

$|{\vec{a}}\cdot{\vec{b}}|\leq{|\vec{a}|}\cdot{|\vec{b}|},$ |{\vec{a}}\cdot{\vec{b}}|\leq{|\vec{a}|}\cdot{|\vec{b}|},

intrucat |cosφ| este cel mult egal cu 1, pentru orice φ € [0, π],

de unde rezulta inegalitatea Cauchy-Schwarz, in cazul particular n = 3 :

${({x_1}{x_2}+{y_1}{y_2}+{z_1}{z_2})}^{2}\leq{({x_1}^{2}+{y_1}^{2}+{z_1}^{2})}{({x_2}^{2}+{y_2}^{2}+{z_2}^{2})}.$ {({x_1}{x_2}+{y_1}{y_2}+{z_1}{z_2})}^{2}\leq{({x_1}^{2}+{y_1}^{2}+{z_1}^{2})}{({x_2}^{2}+{y_2}^{2}+{z_2}^{2})}.