Cálculo Vetorial no Espaço

\[ \begin{aligned} \text{a) }&\overrightarrow{MN}+\overrightarrow{JO}=\overrightarrow{MN}+\overrightarrow{NH}=\overrightarrow{MH}\\ \text{b) }&\overrightarrow{PL}+\overrightarrow{AG}=\overrightarrow{KA}+\overrightarrow{AG}=\overrightarrow{KG}\\ \text{c) }&\overrightarrow{LG}+\overrightarrow{ST}=\overrightarrow{LG}+\overrightarrow{GR}=\overrightarrow{LR}\\ \text{d) }&\overrightarrow{BF}+\overrightarrow{NO}+\overrightarrow{RH}\\ &=\overrightarrow{BF}+\overrightarrow{FQ}+\overrightarrow{QT}\\ &=\overrightarrow{BT}\\ \text{e) }&\overrightarrow{KL}-\overrightarrow{NT}+\overrightarrow{MC}\\ &=\overrightarrow{ET}+\overrightarrow{TN}+\overrightarrow{MC}\\ &=\overrightarrow{EN}+\overrightarrow{NC}\\ &=\overrightarrow{QM}+\overrightarrow{MC} =\overrightarrow{QC}\\ \text{f) }&\overrightarrow{KR}-\overrightarrow{FS}-\overrightarrow{MP}=\overrightarrow{KR}+\overrightarrow{SF}+\overrightarrow{FL}\\ &=\overrightarrow{KR}+\overrightarrow{SL}\\ &=\overrightarrow{KR}+\overrightarrow{RK}\\ &=\overrightarrow{KR}-\overrightarrow{KR} =\vec0 \end{aligned} \] \[ \begin{aligned} \text{a) }&F+\overrightarrow{LN}=F+\overrightarrow{FR}=R\\ \text{b) }&R+\overrightarrow{PL}+\overrightarrow{IB}\\ &=R+\overrightarrow{PL}+\overrightarrow{LM}\\ &=R+\overrightarrow{PM} =R+\overrightarrow{RN} =N\\ \text{c) }&T+\overrightarrow{CM}-\overrightarrow{AF}\\ &=T+\overrightarrow{CM}+\overrightarrow{FA}\\ &=T+\overrightarrow{OG}+\overrightarrow{GB}\\ &=T+\overrightarrow{OB} =T+\overrightarrow{TL} =L\\ \text{d) }&A+\overrightarrow{BS}-\overrightarrow{RQ}+\overrightarrow{LN}\\ &=A+\overrightarrow{BS}+\overrightarrow{QR}+\overrightarrow{LN}\\ &=Q+\overrightarrow{QR}+\overrightarrow{LN}\\ &=R+\overrightarrow{RH} =H \end{aligned} \]

\[ \begin{aligned} \overrightarrow{BD}-2\overrightarrow{MG} &=\overrightarrow{BD}+2\overrightarrow{GM}\\ &=\overrightarrow{AC}+\overrightarrow{CM}\\ &=\overrightarrow{AM} \end{aligned} \] \[ \begin{aligned} \overrightarrow{FP}-\frac13\overrightarrow{EH} &=\overrightarrow{FP}+\frac13\overrightarrow{HE}\\ &=\overrightarrow{AK}+\overrightarrow{HG}\\ &=\overrightarrow{AK}+\overrightarrow{KJ}\\ &=\overrightarrow{AJ} \end{aligned} \] \[ \begin{aligned} \left\|\overrightarrow{FK}-\overrightarrow{GB}\right\| &=\left\|\overrightarrow{FK}+\overrightarrow{BG}\right\|\\ &=\left\|\overrightarrow{FK}+\overrightarrow{KP}\right\|\\ &=\left\|\overrightarrow{FP}\right\| \end{aligned} \] \[ \left\|\overrightarrow{FP}\right\|=\sqrt{2^2+1^2+2^2}=3 \] \[ \begin{aligned} \left\|\overrightarrow{HM}-\frac12\overrightarrow{NB}\right\| &=\left\|\overrightarrow{HM}+\frac12\overrightarrow{BN}\right\|\\ &=\left\|\overrightarrow{HM}+\overrightarrow{MQ}\right\|\\ &=\left\|\overrightarrow{HQ}\right\| \end{aligned} \] \[ \left\|\overrightarrow{HQ}\right\|=\sqrt{2^2+2^2}=\sqrt8=2\sqrt2 \]

\[ \begin{aligned} \overrightarrow{AH}+\overrightarrow{GD} &=(\overrightarrow{AD}+\overrightarrow{DH})\\ &\quad+(\overrightarrow{GA}+\overrightarrow{AD})\\ &=2\overrightarrow{AD}+(\overrightarrow{DH}-\overrightarrow{HD})\\ &=2\overrightarrow{AD}+\vec0\\ &=2\overrightarrow{AD} \end{aligned} \]

c.q.m.

\[ \begin{aligned} V_{\text{sólido}} &=V_{\text{cubo}}+V_{\text{pirâmide}}\\ 33&=3^3+\frac13\times3^2 \times\left\|\overrightarrow{EV}\right\| \end{aligned} \] \[ \begin{aligned} 33&=27+3\left\|\overrightarrow{EV}\right\|\\ \Longleftrightarrow\quad \left\|\overrightarrow{EV}\right\|&=2 \end{aligned} \]

c.q.m.

Os vetores \(\overrightarrow{EV}\) e \(\overrightarrow{GB}\) são colineares.

\[ \begin{aligned} \left\|\overrightarrow{GB}\right\| &=\left\|k\overrightarrow{EV}\right\|\\ \Longleftrightarrow\quad \left\|\overrightarrow{GB}\right\| &=|k|\left\|\overrightarrow{EV}\right\|\\ \Longleftrightarrow\quad 3&=|k|\,2\\ \Longleftrightarrow\quad |k|&=\frac32 \end{aligned} \]

Como os vetores \(\overrightarrow{EV}\) e \(\overrightarrow{GB}\) tem sentidos opostos, temos que \(k<0\), isto é, \(k=-\frac32\).

\[ \begin{aligned} \overrightarrow{AC}+\overrightarrow{EG} &=(\overrightarrow{AI}+\overrightarrow{IJ}+\overrightarrow{JC})\\ &\quad+(\overrightarrow{EI}+\overrightarrow{IJ}+\overrightarrow{JG})\\ &=(\overrightarrow{AI}+\overrightarrow{EI}) +(\overrightarrow{IJ}+\overrightarrow{IJ})\\ &\quad+(\overrightarrow{JC}+\overrightarrow{JG})\\ &=(\overrightarrow{AI}-\overrightarrow{IE}) +2\overrightarrow{IJ}\\ &\quad+(\overrightarrow{JC}-\overrightarrow{GJ})\\ &=\vec0+2\overrightarrow{IJ}+\vec0=2\overrightarrow{IJ} \end{aligned} \]

c.q.m.

\[ \begin{aligned} \text{a) }&\overrightarrow{BC}+\overrightarrow{CG}=\overrightarrow{BG}\\ \text{b) }&\overrightarrow{AC}+2\overrightarrow{JG}\\ &=\overrightarrow{AC}+\overrightarrow{CG}\\ &=\overrightarrow{AG}\\ \text{c) }&\overrightarrow{AI}+\frac12(\overrightarrow{AC}+\overrightarrow{EG}) \\ &=\overrightarrow{AI}+\frac12(2\overrightarrow{IJ})\\ &=\overrightarrow{AI}+\overrightarrow{IJ}\\ &=\overrightarrow{AJ} \end{aligned} \]

\[ \begin{aligned} \text{5.1 a) }&\overrightarrow{DA}\\ \text{b) }&\overrightarrow{AB}\text{ e }\overrightarrow{BC},\text{ por exemplo}\\ \text{c) }&\overrightarrow{AB}\text{ e }\overrightarrow{BF},\text{ por exemplo}\\ \text{d) }&\overrightarrow{AB} \end{aligned} \] \[ \begin{aligned} \text{a) }&A+\overrightarrow{BC}=A+\overrightarrow{AD}=D\\ \text{b) }&A-\frac12\overrightarrow{CA}=A+\frac12\overrightarrow{AC}=O\\ \text{c) }&\overrightarrow{FO}+\overrightarrow{AO}=\overrightarrow{FO}+\overrightarrow{OC}=\overrightarrow{FC}\\ \text{d) }&\overrightarrow{CE}-(E-F)\\ &=\overrightarrow{CE}-\overrightarrow{FE}\\ &=\overrightarrow{CE}+\overrightarrow{EF} =\overrightarrow{CF} \end{aligned} \]

\[ \begin{aligned} \overrightarrow{MA} &=\frac12\overrightarrow{CA}\\ &=-\frac12\overrightarrow{AC}\\ &=-\frac12\left(\frac23\overrightarrow{AB}\right)\\ &=-\frac13\overrightarrow{AB} \end{aligned} \]

c.q.m.

\[ \left\|\overrightarrow{AE}\right\|=2\left\|\vec u\right\|=2\times\frac52=5 \]

\[ \begin{aligned} \overrightarrow{BC} &=C-B\\ &=(3,4,-8)-(-1,2,-2)\\ &=(4,2,-6) \end{aligned} \] \[ \begin{aligned} B-2\overrightarrow{BC} &=(-1,2,-2)-2(4,2,-6)\\ &=(-1,2,-2)-(8,4,-12)\\ &=(-9,-2,10) \end{aligned} \] \[ \begin{aligned} \overrightarrow{AB} &=B-A\\ &=(-1,2,-2)-(-4,3,2)\\ &=(3,-1,-4) \end{aligned} \] \[ \begin{gathered} \overrightarrow{AB}+2(-2\overrightarrow{BC})\\ =\overrightarrow{AB}-4\overrightarrow{BC}\\ =(3,-1,-4)-4(4,2,-6)\\ =(-13,-9,20) \end{gathered} \] \[ \begin{gathered} 2\overrightarrow{AB}-\frac32\overrightarrow{BC}\\ =2(3,-1,-4)-\frac32(4,2,-6)\\ =(6,-2,-8)-(6,3,-9)\\ =(0,-5,1) \end{gathered} \] \[ \begin{aligned} \overrightarrow{CA} &=C-A\\ &=(3,4,-8)-(-4,3,2)\\ &=(7,1,-10) \end{aligned} \] \[ \begin{gathered} 2\overrightarrow{AB}+2\overrightarrow{BC} +\overrightarrow{CA}\\ =(6,-2,-8)+(8,4,-12)\\ \quad +(7,1,-10)\\ =(21,3,-30) \end{gathered} \]

Como \(A\), \(B\) e \(C\) então os vetores \(\overrightarrow{AB}\) e \(\overrightarrow{AC}\), são colineares.

\[ \begin{aligned} \overrightarrow{AC} &=C-A\\ &=(b+1,-4,-9)-(a,-1,-3)\\ &=(b-a+1,-3,-6) \end{aligned} \] \[ \begin{aligned} \overrightarrow{AB} &=B-A\\ &=(0,2,b)-(a,-1,-3)\\ &=(-a,3,b+3) \end{aligned} \] \[ \begin{aligned} \frac{b-a+1}{-a} &=\frac{-3}{3} =\frac{-6}{b+3}\\ \Longleftrightarrow\quad \frac{b-a+1}{-a} &=-1\ \wedge\ -1=\frac{-6}{b+3} \end{aligned} \] \[ \Longleftrightarrow \begin{aligned} b-a+1&=a\\ {}\wedge\quad -b-3&=-6\\ {}\wedge\quad (a&\ne0\ \wedge\ b\ne-3) \end{aligned} \] \[ \Longleftrightarrow \begin{aligned} 3+1&=2a\\ {}\wedge\quad b&=3\\ {}\wedge\quad (a&\ne0\ \wedge\ b\ne-3) \end{aligned} \] \[ \Longleftrightarrow \begin{aligned} a&=2\\ {}\wedge\quad b&=3\\ {}\wedge\quad (a&\ne0\ \wedge\ b\ne-3) \end{aligned} \]

\[ \begin{aligned} \frac{-5}{-2}&=\frac{5}{2} =\frac{2}{\frac{8}{10}}\\ \Longleftrightarrow\quad \frac52&=\frac52=\frac{20}{8}\\ \Longleftrightarrow\quad \frac52&=\frac52=\frac52 \end{aligned} \]

Logo, \(\vec u\) e \(\vec v\) são colineares.

\[ \begin{aligned} \overrightarrow{BC} &=C-B\\ &=(1,-1,1)-(-1,-1,2)\\ &=(2,0,-1) \end{aligned} \] \[ \begin{aligned} P+\overrightarrow{BC} &=(0,1,1)+(2,0,-1)\\ &=(2,1,0) \end{aligned} \]

c.q.m.

\[ \begin{aligned} \overrightarrow{AB} &=B-A\\ &=(1,-3,2)-(-2,-1,4)\\ &=(3,-2,-2) \end{aligned} \] \[ 2\vec v+\vec w=\overrightarrow{AB}+\vec u \Longleftrightarrow \vec w=\overrightarrow{AB}+\vec u-2\vec v \] \[ \begin{aligned} \vec w &=(3,-2,-2)+(-1,4,0)\\ &\quad-2\left(-2,3,\frac12\right)\\ &=(2,2,-2)-(-4,6,1)\\ &=(6,-4,-3) \end{aligned} \] \[ \frac12\vec x+\vec u=-2\vec v \Longleftrightarrow \vec x=-2\vec u-4\vec v \] \[ \begin{aligned} \vec x &=-2(-1,4,0)-4\left(-2,3,\frac12\right)\\ &=(2,-8,0)-(-8,12,2)\\ &=(10,-20,-2) \end{aligned} \] \[ \begin{aligned} \frac32\vec v &=2\vec y-\frac12\vec u+2\overrightarrow{BA}\\ \Longleftrightarrow\quad \vec y&=\frac34\vec v+\frac14\vec u+\overrightarrow{AB} \end{aligned} \] \[ \begin{aligned} \vec y &=\frac34\left(-2,3,\frac12\right)\\ &\quad+\frac14(-1,4,0)+(3,-2,-2) \end{aligned} \] \[ \begin{aligned} \vec y &=\left(-\frac32,\frac94,\frac38\right) +\left(-\frac14,1,0\right)\\ &\quad +(3,-2,-2)\\ &=\left(\frac54,\frac54,-\frac{13}{8}\right) \end{aligned} \] \[ \begin{aligned} 2\vec e_1+\vec u &=\frac12\vec t+\vec e_2-3\vec v\\ \Longleftrightarrow\quad \vec t&=4\vec e_1+2\vec u-2\vec e_2+6\vec v \end{aligned} \] \[ \begin{aligned} \vec t &=4(1,0,0)+2(-1,4,0)-2(0,1,0)\\ &\quad+6\left(-2,3,\frac12\right) \end{aligned} \] \[ \begin{aligned} \vec t &=(4,0,0)+(-2,8,0)-(0,2,0)\\ &\quad+(-12,18,3)\\ &=(-10,24,3) \end{aligned} \]

\[ V_{\text{cubo}}=8\Longleftrightarrow a=2 \] \[ A(2,0,0),\quad B(2,2,0)\quad\text{e}\quad G(0,2,2) \]

Seja \(P(x,y,z)\).

\[ \begin{aligned} \overrightarrow{BP} &=P-B\\ &=(x,y,z)-(2,2,0)\\ &=(x-2,y-2,z) \end{aligned} \] \[ \begin{aligned} \overrightarrow{AG} &=G-A\\ &=(0,2,2)-(2,0,0)\\ &=(-2,2,2) \end{aligned} \] \[ \begin{aligned} \overrightarrow{BP} &=\frac34\overrightarrow{AG}\\ \Longleftrightarrow\quad (x-2,y-2,z)&=\frac34(-2,2,2) \end{aligned} \] \[ \Longleftrightarrow \begin{cases} x-2=-\frac32\\ y-2=\frac32\\ z=\frac32 \end{cases} \Longleftrightarrow \begin{cases} x=\frac12\\ y=\frac72\\ z=\frac32 \end{cases} \]

Portanto as coordenadas de \(P\) são \(\left(\frac12,\frac72,\frac32\right)\).

Seja \(Q(x,y,z)\) um ponto.

Sabemos que \(\left\|\overrightarrow{PQ}\right\|=6\) e \(\overrightarrow{PQ}=k\vec u\).

\[ \begin{aligned} \overrightarrow{PQ} &=Q-P\\ &=(x,y,z)-(1,0,-1)\\ &=(x-1,y,z+1) \end{aligned} \] \[ k\vec u=(k,-2k,k) \] \[ \begin{aligned} \left\|\overrightarrow{PQ}\right\|&=6\\ \Longleftrightarrow\quad \left\|k\vec u\right\|&=6\\ \Longleftrightarrow\quad \sqrt{k^2+(-2k)^2+k^2}&=6\\ \Longleftrightarrow\quad \sqrt{6k^2}&=6 \end{aligned} \] \[ \Longleftrightarrow \begin{aligned} 6k^2&=36\\ \Longleftrightarrow\quad k^2&=6\\ \Longleftrightarrow\quad k&=-\sqrt6\ \vee\ k=\sqrt6 \end{aligned} \]

Como \(\overrightarrow{PQ}\) e \(\vec u\) têm o mesmo sentido, \(k=\sqrt6\).

\[ k\vec u=(\sqrt6,-2\sqrt6,\sqrt6) \] \[ \overrightarrow{PQ}=k\vec u \] \[ \Longleftrightarrow \begin{cases} x-1=\sqrt6\\ y=-2\sqrt6\\ z+1=\sqrt6 \end{cases} \] \[ \Longleftrightarrow \begin{aligned} x&=1+\sqrt6\\ {}\wedge\quad y&=-2\sqrt6\\ {}\wedge\quad z&=-1+\sqrt6 \end{aligned} \] \[ Q(1+\sqrt6,-2\sqrt6,-1+\sqrt6) \]

\[ \begin{aligned} B&=A+\overrightarrow{AB}\\ &=A+\overrightarrow{DC}\\ &=A+(C-D) \end{aligned} \] \[ \begin{aligned} B&=(3,0,0)+[(4,-7,0)-(7,-3,0)]\\ &=(3,0,0)+(-3,-4,0)\\ &=(0,-4,0) \end{aligned} \]

Logo \(B\) pertence a \(Oy\).

\[ \begin{aligned} H&=D+\overrightarrow{DH}=D+\overrightarrow{BF}\\ &=(7,-3,0)+(0,0,6)\\ &=(7,-3,6) \end{aligned} \] \[ \left\|\overrightarrow{BF}\right\|=\sqrt{0^2+0^2+6^2}=6 \] \[ \begin{aligned} \left\|\overrightarrow{DC}\right\| &=\sqrt{(-3)^2+(-4)^2+0^2}\\ &=\sqrt{25}=5 \end{aligned} \] \[ \begin{aligned} A_{[ABCDEFGH]} &=\overline{DC}^{\,2}\times\overline{BF}\\ &=5^2\times6=150 \end{aligned} \]

\[ \begin{aligned} \overrightarrow{AB} &=B-A\\ &=(1,5,-6)-(3,1,-2)\\ &=(-2,4,-4) \end{aligned} \] \[ \begin{aligned} \overrightarrow{AC} &=C-A\\ &=(a,1-a,a-2)-(3,1,-2)\\ &=(a-3,-a,a) \end{aligned} \] \[ \begin{aligned} \frac{a-3}{-2} &=\frac{-a}{4}=\frac{a}{-4}\\ \Longleftrightarrow\quad -2a+6&=-a\\ \Longleftrightarrow\quad a&=6 \end{aligned} \] \[ \begin{aligned} M_{[AB]} &=\left(\frac{3+1}{2},\frac{1+5}{2},\frac{-2-6}{2}\right)\\ &=(2,3,-4) \end{aligned} \] \[ \begin{aligned} C&=M\\ \Longleftrightarrow\quad (a,1-a,a-2)&=(2,3,-4) \end{aligned} \] \[ \Longleftrightarrow \begin{cases} a=2\\ 1-a=3\\ a-2=-4 \end{cases} \] \[ \Longleftrightarrow \begin{cases} a=2\\ a=-2\\ a=-2 \end{cases} \]

condição impossível. Logo, não existe \(a\in IR\) para o qual \(C\) é ponto médio de \([AB]\).

\[ \begin{aligned} \overrightarrow{AE}&=k\overrightarrow{AB},\quad k>0\\ \text{e}\quad \left\|\overrightarrow{AE}\right\|&=15 \end{aligned} \] \[ \left\|\overrightarrow{AB}\right\| =\sqrt{(-2)^2+4^2+(-4)^2}=6 \] \[ \begin{aligned} \left\|\overrightarrow{AE}\right\|&=15\\ \Longleftrightarrow\quad \left\|k\overrightarrow{AB}\right\|&=15\\ \Longleftrightarrow\quad |k|\left\|\overrightarrow{AB}\right\|&=15\\ \Longleftrightarrow\quad |k|6&=15\\ \Longleftrightarrow\quad k&=\frac{15}{6}=\frac52 \end{aligned} \] \[ \overrightarrow{AE}=\frac52(-2,4,-4)=(-5,10,-10) \] \[ \begin{aligned} E&=A+\overrightarrow{AE}\\ &=(3,1,-2)+(-5,10,-10)\\ &=(-2,11,-12) \end{aligned} \]

\(E\) é o ponto médio de \([AC]\).

\[ \begin{aligned} E&\left(\frac{6-2}{2},\frac{2+4}{2},\frac{1+3}{2}\right)\\ &=(2,3,2) \end{aligned} \]

c.q.m.

\[ \begin{aligned} \overrightarrow{EF} &=F-E\\ &=(4,7,6)-(2,3,2)\\ &=(2,4,4) \end{aligned} \] \[ \left\|\overrightarrow{EF}\right\| =\sqrt{2^2+4^2+4^2}=6 \] \[ \begin{aligned} \left\|\overrightarrow{AC}\right\|^2 &=\left\|\overrightarrow{AB}\right\|^2 +\left\|\overrightarrow{BC}\right\|^2\\ \Longleftrightarrow\quad \left\|\overrightarrow{AC}\right\|^2 &=2\left\|\overrightarrow{AB}\right\|^2 \end{aligned} \] \[ \left\|\overrightarrow{AB}\right\| =\frac12\left\|\overrightarrow{AC}\right\| \] \[ \begin{aligned} \overrightarrow{AC} &=C-A\\ &=(-2,4,3)-(6,2,1)\\ &=(-8,2,2) \end{aligned} \] \[ \left\|\overrightarrow{AC}\right\| =\sqrt{(-8)^2+2^2+2^2} =\sqrt{72} \] \[ \left\|\overrightarrow{AB}\right\|^2 =\frac12(\sqrt{72})^2 \] \[ \Longleftrightarrow \left\|\overrightarrow{AB}\right\|^2=36 \Longleftrightarrow \left\|\overrightarrow{AB}\right\|=6 \] \[ \begin{aligned} V&=108\\ \Longleftrightarrow\quad \frac13\left\|\overrightarrow{AB}\right\|^2 \times\left\|\overrightarrow{EV}\right\|&=108 \end{aligned} \] \[ \Longleftrightarrow \begin{aligned} \frac13\times36 \times\left\|\overrightarrow{EV}\right\|&=108\\ \Longleftrightarrow\quad \left\|\overrightarrow{EV}\right\|&=\frac{108}{12}\\ \Longleftrightarrow\quad \left\|\overrightarrow{EV}\right\|&=9 \end{aligned} \] \[ V=E+\overrightarrow{EV} \] \[ \begin{aligned} \overrightarrow{EV}&=k\overrightarrow{EF},\quad k>0\\ \text{e}\quad \left\|\overrightarrow{EV}\right\| &=\left\|k\overrightarrow{EF}\right\|=9 \end{aligned} \] \[ \begin{aligned} \left\|k\overrightarrow{EF}\right\|&=9\\ \Longleftrightarrow\quad |k|\left\|\overrightarrow{EF}\right\|&=9\\ \Longleftrightarrow\quad |k|6&=9\\ \Longleftrightarrow\quad k&=\frac32 \end{aligned} \] \[ \overrightarrow{EV}=\frac32(2,4,4)=(3,6,6) \] \[ \begin{aligned} V&=E+\overrightarrow{EV}\\ &=(2,3,2)+(3,6,6)\\ &=(5,9,8) \end{aligned} \] \[ \begin{aligned} \overrightarrow{BE} &=E-B\\ &=(2,3,2)-(2,6,-1)\\ &=(0,-3,3) \end{aligned} \] \[ \begin{aligned} D&=E+\overrightarrow{ED}\\ &=E+\overrightarrow{BE}\\ &=(2,3,2)+(0,-3,3)\\ &=(2,0,5) \end{aligned} \]