Tag: vectores

Um vector perpendicular a outros dois 0

Um vector perpendicular a outros dois

Geometria Analítica: Infinito 11 A - Parte 1 Pág. 182 Ex. 34

Enunciado

Num referencial ortonormado do espaço, indique um vector que seja perpendicular a $\vec{u}(1,4,7)$  e a $\vec{v}(2,-1,5)$ .
Observe que qualquer outro vector nas mesmas condições é colinear com ele.

Resolução >> Resolução

var parameters = { "id": "ggbApplet", "width":256, "height":192, "showMenuBar":false, "showAlgebraInput":false, "showToolBar":false, "customToolBar":"0 39 | 1 501 67 , 5 19 , 72 | 2 15 45 , 18 65 , 7 37 | 4 3 8 9 , 13 44 , 58 , 47 | 16 51 64 , 70 | 10 34 53 11 , 24 20 22 , 21 23 | 55 56 57 , 12 | 36 46 , 38 49 50 , 71 | 30 29 54 32 31 33 | 17 26 62 73 , 14 68 | 25 52 60 61 | 40 41 42 , 27 28 35 , 6", "showToolBarHelp":false, "showResetIcon":true, "enableLabelDrags":false, "enableShiftDragZoom":false, "enableRightClick":false, "errorDialogsActive":false, "useBrowserForJS":false, "preventFocus":false, "language":"pt", // use this instead of ggbBase64 to load a material from GeoGebraTube // "material_id":12345, "ggbBase64":"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// is3D=is 3D applet using 3D view, AV=Algebra View, SV=Spreadsheet View, CV=CAS View, EV2=Graphics View 2, CP=Construction Protocol, PC=Probability Calculator, DA=Data Analysis, FI=Function Inspector, PV=Python, macro=Macro View var views = {'is3D': 0,'AV': 0,'SV': 0,'CV': 0,'EV2': 0,'CP': 0,'PC': 0,'DA': 0,'FI': 0,'PV': 0,'macro': 0}; var applet = new GGBApplet(parameters, '5.0', views); window.onload = function() {applet.inject('ggbApplet')};

Apliquemos os vectores livres $\vec{u}(1,4,7)$ e $\vec{v}(2,-1,5)$ num ponto A.…

Determine uma equação cartesiana do plano mediador do segmento [AB] 0

Determine uma equação cartesiana do plano mediador do segmento [AB]

Geometria Analítica: Infinito 11 A - Parte 1 Pág. 181 Ex. 31

Enunciado

Determine uma equação cartesiana do plano mediador do segmento [AB], sendo:

  1. $A(4,-1,2)$ e $B(2,7,0)$.
     
  2. $A(-4,1,7)$ e $B(3,2,-5)$.

Resolução >> Resolução

  1. var parameters = { "id": "ggbApplet", "width":283, "height":277, "showMenuBar":false, "showAlgebraInput":false, "showToolBar":false, "customToolBar":"0 39 | 1 501 67 , 5 19 , 72 | 2 15 45 , 18 65 , 7 37 | 4 3 8 9 , 13 44 , 58 , 47 | 16 51 64 , 70 | 10 34 53 11 , 24 20 22 , 21 23 | 55 56 57 , 12 | 36 46 , 38 49 50 , 71 | 30 29 54 32 31 33 | 17 26 62 73 , 14 68 | 25 52 60 61 | 40 41 42 , 27 28 35 , 6", "showToolBarHelp":false, "showResetIcon":true, "enableLabelDrags":false, "enableShiftDragZoom":false, "enableRightClick":false, "errorDialogsActive":false, "useBrowserForJS":false, "preventFocus":false, "language":"pt", // use this instead of ggbBase64 to load a material from GeoGebraTube // "material_id":12345, 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// is3D=is 3D applet using 3D view, AV=Algebra View, SV=Spreadsheet View, CV=CAS View, EV2=Graphics View 2, CP=Construction Protocol, PC=Probability Calculator, DA=Data Analysis, FI=Function Inspector, PV=Python, macro=Macro View var views = {'is3D': 1,'AV': 0,'SV': 0,'CV': 0,'EV2': 0,'CP': 0,'PC': 0,'DA': 0,'FI': 0,'PV': 0,'macro': 0}; var applet = new GGBApplet(parameters, '5.0', views); window.onload = function() {applet.inject('ggbApplet')}; O plano mediador so segmento [AB] é o lugar geométrico dos pontos $P(x,y,z)$ do espaço, tais que $\overrightarrow{MP}.\overrightarrow{AB}=0$, sendo M o ponto médio de [AB].
Identifique o conjunto de pontos do plano definidos pela condição 0

Identifique o conjunto de pontos do plano definidos pela condição

Geometria Analítica: Infinito 11 A - Parte 1 Pág. 181 Ex. 30

Enunciado

Sendo $A(0,9)$ e $B(-8,2)$, identifique o conjunto de pontos $P(x,y)$ do plano que verificam a condição:

  1. $\overrightarrow{AP}.\overrightarrow{BP}=0$;
     
  2. $\overrightarrow{MP}.\overrightarrow{AM}=0$, sendo M o ponto médio de [AB].

Resolução >> Resolução

  1. Tente identificar o lugar geométrico definido pela condição $\overrightarrow{AP}.\overrightarrow{BP}=0$.
    Caso não consiga, execute a animação sem activar “Mostrar lugar geométrico”.
Equação de uma recta que passa em A e é perpendicular a r 0

Equação de uma recta que passa em A e é perpendicular a r

Geometria Analítica: Infinito 11 A - Parte 1 Pág. 181 Ex. 29

Enunciado

Considere, num referencial o. n. $(O,\vec{i},\vec{j})$, a recta r de equação $(x,y)=(3,2)+k(-3,-1),k\in \mathbb{R}$ e o ponto $A(-1,4)$.

  1. Determine a equação reduzida da recta s, perpendicular a r e que passa em A.
     
  2. Desenhe um quadrado de vértice A, com um lado sobre a recta s e outro sobre a recta r, e determine, analiticamente, as coordenadas dos vértices do quadrado que construiu.
Circunferência circunscrita no triângulo [ABC] 1

Circunferência circunscrita no triângulo [ABC]

Geometria Analítica: Infinito 11 A - Parte 1 Pág. 181 Ex. 28

Enunciado

Considere o triângulo [ABC], sendo $A(-5,1)$, $B(1,3)$ e $C(3,1)$.

  1. Escreva uma equação cartesiana da mediatriz do lado [AB].
     
  2. Escreva uma equação cartesiana da mediatriz do lado [BC].
     
  3. Determine as coordenadas do ponto de intersecção das medianas determinadas (circuncentro ou centro da circunferência circunscrita no triângulo).
     
  4. Escreva uma equação da circunferência circunscrita ao triângulo.
Uma circunferência e uma recta que lhe é tangente 0

Uma circunferência e uma recta que lhe é tangente

Geometria Analítica: Infinito 11 A - Parte 1 Pág. 181 Ex. 27

Enunciado

Num referencial o. n. $(O,\vec{i},\vec{j})$, considere a circunferência de equação ${{x}^{2}}+{{y}^{2}}+2x+4y+4=0$.

  1. Determine as coordenadas do centro e o raio da circunferência.
     
  2. Determine uma equação da recta tangente à circunferência no ponto $A(0,-2)$.

Resolução >> Resolução

  1. Ora, \[\begin{array}{*{35}{l}}
       {{x}^{2}}+{{y}^{2}}+2x+4y+4=0 & \Leftrightarrow  & {{(x+1)}^{2}}-1+{{(y+2)}^{2}}-4+4=0  \\
       {} & \Leftrightarrow  & {{(x+1)}^{2}}+{{(y+2)}^{2}}=1  \\
    \end{array}\]
    Logo, a circunferência tem centro $C(-1,-2)$ e raio $r=1$.
Circunferência circunscrita num triângulo 0

Circunferência circunscrita num triângulo

Geometria Analítica: Infinito 11 A - Parte 1 Pág. 181 Ex. 26

Enunciado

Considere um referencial o. n. $(O,\vec{i},\vec{j})$.

Escreva uma equação da circunferência circunscrita ao triângulo, cujos lados estão sobre as rectas de equação $y=0$, $x=0$ e $y=x+4$.

Resolução >> Resolução

Sorry, the GeoGebra Applet could not be started. Please make sure that Java 1.4.2 (or later) is installed and active in your browser (Click here to install Java now)

A circunferência circunscrita a um triângulo contém os seus vértices, pelo que o seu centro – circuncentro do triângulo – é um ponto equididtante dos seus vértices.…

Equação da recta tangente a uma circunferência 0

Equação da recta tangente a uma circunferência

Geometria Analítica: Infinito 11 A - Parte 1 Pág. 180 Ex. 25

Enunciado

  1. Verifique que $A(1,-2)$ é o ponto da circunferência C: ${{x}^{2}}+{{y}^{2}}-6x-2y-3=0$ e escreva uma equação da recta tangente a C em A.
     
  2. Determine uma equação da recta tangente à circunferência de centro $D(3,4)$ no ponto $E(1,2)$.

Resolução >> Resolução

  1. O ponto A pertence à circunferência C, pois as suas coordenadas verificam a sua equação: ${{1}^{2}}+{{(-2)}^{2}}-6\times 1-2\times (-2)-3=0\Leftrightarrow 1+4-6+4-3=0\Leftrightarrow 0=0$.
Escreva uma equação da circunferência 0

Escreva uma equação da circunferência

Geometria Analítica: Infinito 11 A - Parte 1 Pág. 180 Ex. 24

Enunciado

Sendo $A(2,1)$ e $B(-2,3)$, escreva uma equação da circunferência:

  1. de centro A e que passa no ponto B;
     
  2. de diâmetro [AB].

Resolução >> Resolução

  1. O raio da circunferência é $r=\overline{AB}=\sqrt{{{(-2-2)}^{2}}+{{(3-1)}^{2}}}=2\sqrt{5}$ e o centro é $A(2,1)$.

    Logo, uma equação dessa circunferência é ${{(x-2)}^{2}}+{{(y-1)}^{2}}=20$.
     
     

  2. O centro da circunferência é o ponto médio do segmento [AB]: $M(\frac{2-2}{2},\frac{1+3}{2})=(0,2)$.
Considere os pontos A, B e C 0

Considere os pontos A, B e C

Geometria Analítica: Infinito 11 A - Parte 1 Pág. 180 Ex. 23

Enunciado

Considere os pontos $A(5,1)$, $B(-3,2)$ e $C(3,-2)$.

  1. Escreva uma equação cartesiana da recta que contém a altura do triângulo [ABC] relativa a A.
     
  2. Calcule a área do triângulo [ABC].

Resolução >> Resolução

  1. A recta pedida passa em A e é perpendicular à recta BC.

    Como $\overrightarrow{BC}=(6,-4)$, então $\overrightarrow{r}=(2,3)$ é um vector director da recta pedida.