12.4E: Mazoezi ya Sehemu ya 12.4

Kwa mazoezi 1-4, vectors\(\vecs{u}\) na\(\vecs{v}\) hutolewa.

Kupata bidhaa msalaba\(\vecs{u}\times\vecs{v}\) wa wadudu\(\vecs{u}\) na\(\vecs{v}\). Eleza jibu katika fomu ya sehemu.

b. mchoro wadudu\(\vecs{u}, \, \vecs{v}\), na\(\vecs{u}\times\vecs{v}\).

1)\(\quad \vecs{u}=⟨2,0,0⟩, \quad \vecs{v}=⟨2,2,0⟩\)

Jibu

\(a. \vecs{u}\times\vecs{v}=⟨0,0,4⟩;\)

\(b.\)

2)\(\quad \vecs{u}=⟨3,2,−1⟩, \quad \vecs{v}=⟨1,1,0⟩\)

3)\(\quad \vecs{u}=2\mathbf{\hat i}+3\mathbf{\hat j}, \quad \vecs{v}=\mathbf{\hat j}+2\mathbf{\hat k}\)

Jibu

\( a. \vecs{u}\times\vecs{v}=⟨6,−4,2⟩;\)

\(b.\)

4)\(\quad \vecs{u}=2\mathbf{\hat j}+3\mathbf{\hat k}, \quad \vecs{v}=3\mathbf{\hat i}+\mathbf{\hat k}\)

5) Kurahisisha\((\mathbf{\hat i}×\mathbf{\hat i}−2\mathbf{\hat i}×\mathbf{\hat j}−4\mathbf{\hat i}×\mathbf{\hat k}+3\mathbf{\hat j}×\mathbf{\hat k})×\mathbf{\hat i}.\)

Jibu

\(−2\mathbf{\hat j}−4\mathbf{\hat k}\)

6) Kurahisisha\(\mathbf{\hat j}×(\mathbf{\hat k}×\mathbf{\hat j}+2\mathbf{\hat j}×\mathbf{\hat i}−3\mathbf{\hat j}×\mathbf{\hat j}+5\mathbf{\hat i}×\mathbf{\hat k}).\)

Katika mazoezi 7-10, vectors\(\vecs{u}\) na\(\vecs{v}\) hutolewa. Kupata kitengo vector\(\vecs{w}\) katika mwelekeo wa bidhaa msalaba vector\(\vecs{u}×\vecs{v}.\) Express jibu lako kwa kutumia kiwango kitengo wadudu.

7)\(\quad \vecs{u}=⟨3,−1,2⟩, \quad \vecs{v}=⟨−2,0,1⟩\)

Jibu

\(\vecs{w}=−\frac{\sqrt{6}}{18}\mathbf{\hat i}−\frac{7\sqrt{6}}{18}\mathbf{\hat j}−\frac{\sqrt{6}}{9}\mathbf{\hat k}\)

8)\(\quad \vecs{u}=⟨2,6,1⟩, \quad \vecs{v}=⟨3,0,1⟩\)

9)\(\quad \vecs{u}=\vecd{AB}, \quad \vecs{v}=\vecd{AC},\) wapi\(A(1,0,1),\, B(1,−1,3)\), na\(C(0,0,5)\)

Jibu

\(\vecs{w}=−\frac{4\sqrt{21}}{21}\mathbf{\hat i}−\frac{2\sqrt{21}}{21}\mathbf{\hat j}−\frac{\sqrt{21}}{21}\mathbf{\hat k}\)

10)\(\quad \vecs{u}=\vecd{OP}, \quad \vecs{v}=\vecd{PQ},\) wapi\(P(−1,1,0)\) na\(Q(0,2,1)\)

11) Kuamua idadi halisi\(α\) kama hiyo\(\vecs{u}\times\vecs{v}\) na\(\mathbf{\hat i}\) ni orthogonal, wapi\(\vecs{u}=3\mathbf{\hat i}+\mathbf{\hat j}−5\mathbf{\hat k}\) na\(\vecs{v}=4\mathbf{\hat i}−2\mathbf{\hat j}+α\mathbf{\hat k}.\)

Jibu

\(α=10\)

12) Onyesha kwamba\(\vecs{u}\times\vecs{v}\) na\( 2\mathbf{\hat i}−14\mathbf{\hat j}+2\mathbf{\hat k}\) haiwezi kuwa orthogonal kwa idadi yoyote α halisi, wapi\(\vecs{u}=\mathbf{\hat i}+7\mathbf{\hat j}−\mathbf{\hat k}\) na\(\vecs{v}=α\mathbf{\hat i}+5\mathbf{\hat j}+\mathbf{\hat k}\).

13) Onyesha kwamba\(\vecs{u}\times\vecs{v}\) ni orthogonal\(\vecs{u}+\vecs{v}\) na\(\vecs{u}−\vecs{v}\), wapi\(\vecs{u}\) na\(\vecs{v}\) ni nonzero wadudu.

14) Onyesha kwamba\(\vecs{v}\times\vecs{u}\) ni orthogonal kwa\( (\vecs{u}⋅\vecs{v})(\vecs{u}+\vecs{v})+\vecs{u}\), wapi\(\vecs{u}\) na\(\vecs{v}\) ni nonzero wadudu.

15) Tumia mahesabu\( \begin{vmatrix}\mathbf{\hat i}&\mathbf{\hat j}&\mathbf{\hat k}\\1&−1&7\\2&0&3\end{vmatrix}\).

Jibu

\( −3\mathbf{\hat i}+11\mathbf{\hat j}+2\mathbf{\hat k}\)

16) Tumia mahesabu\( \begin{vmatrix}\mathbf{\hat i}&\mathbf{\hat j}&\mathbf{\hat k}\\0&3&−4\\1&6&−1\end{vmatrix}\).

Kwa mazoezi 17-18, vectors\(\vecs{u}\) na\(\vecs{v}\) hutolewa. Matumizi determinant nukuu kupata vector\(\vecs{w}\) orthogonal kwa wadudu\(\vecs{u}\) na\(\vecs{v}\).

17)\(\quad \vecs{u}=⟨−1, 0, e^t⟩, \quad \vecs{v}=⟨1, e^{−t}, 0⟩,\)\(t\) wapi idadi halisi

Jibu

\(\vecs{w}=⟨−1, e^t, −e^{−t}⟩\)

18)\(\quad \vecs{u}=⟨1, 0, x⟩, \quad \vecs{v}=⟨\frac{2}{x},1, 0⟩,\) wapi\(x\) nambari halisi isiyo ya sifuri

19) Kupata vector\( (\vecs{a}−2\vecs{b})×\vecs{c},\) wapi\( \vecs{a}=\begin{vmatrix}\mathbf{\hat i}&\mathbf{\hat j}&\mathbf{\hat k}\\2&−1&5\\0&1&8\end{vmatrix}, \vecs{b}=\begin{vmatrix}\mathbf{\hat i}&\mathbf{\hat j}&\mathbf{\hat k}\\0&1&1\\2&−1&−2\end{vmatrix},\) na\(\vecs{c}=\mathbf{\hat i}+\mathbf{\hat j}+\mathbf{\hat k}.\)

Jibu

\( −26\mathbf{\hat i}+17\mathbf{\hat j}+9\mathbf{\hat k}\)

20) Kupata vector\( \vecs{c}×(\vecs{a}+3\vecs{b}),\) wapi\( \vecs{a}=\begin{vmatrix}\mathbf{\hat i}&\mathbf{\hat j}&\mathbf{\hat k}\\5&0&9\\0&1&0\end{vmatrix}, \vecs{b}=\begin{vmatrix}\mathbf{\hat i}&\mathbf{\hat j}&\mathbf{\hat k}\\0&−1&1\\7&1&−1\end{vmatrix},\) na\(\vecs{c}=\mathbf{\hat i}−\mathbf{\hat k}.\)

21) [T] Tumia bidhaa msalaba\(\vecs{u}\times\vecs{v}\) kupata angle papo hapo kati ya wadudu\(\vecs{u}\) na\(\vecs{v}\), wapi\(\vecs{u}=\mathbf{\hat i}+2\mathbf{\hat j}\) na\(\vecs{v}=\mathbf{\hat i}+\mathbf{\hat k}.\) Express jibu katika digrii mviringo kwa integer karibu.

Jibu

\( 72°\)

22) [T] Tumia bidhaa msalaba\(\vecs{u}\times\vecs{v}\) kupata angle obtuse kati ya wadudu\(\vecs{u}\) na\(\vecs{v}\), wapi\(\vecs{u}=−\mathbf{\hat i}+3\mathbf{\hat j}+\mathbf{\hat k}\) na\(\vecs{v}=\mathbf{\hat i}−2\mathbf{\hat j}.\) Express jibu katika digrii mviringo kwa integer karibu.

23) Tumia sine na cosine ya angle kati ya vectors mbili zisizo na zero\(\vecs u\) na\(\vecs v\) kuthibitisha utambulisho wa Lagrange:\(\|\vecs{u}\times\vecs{v}\|^2=\|\vecs{u}\|^2\|\vecs{v}\|^2−(\vecs{u}⋅\vecs{v})^2\).

24) Thibitisha utambulisho Lagrange\(\|\vecs{u}\times\vecs{v}\|^2=\|\vecs{u}\|^2\|\vecs{v}\|^2−(\vecs{u}⋅\vecs{v})^2\) kwa wadudu\(\vecs{u}=−\mathbf{\hat i}+\mathbf{\hat j}−2\mathbf{\hat k}\) na\(\vecs{v}=2\mathbf{\hat i}−\mathbf{\hat j}.\)

25) Nonzero wadudu\(\vecs{u}\) na\(\vecs{v}\) kuitwa collinear kama kuna nonzero scalar\(α\) vile kwamba\( \vecs{v}=α\vecs{u}\). Kuonyesha kwamba\(\vecs{u}\) na\(\vecs{v}\) ni collinear kama na tu kama\( \vecs{u}\times\vecs{v}=0.\)

26) Nonzero wadudu\(\vecs{u}\) na\(\vecs{v}\) kuitwa collinear kama kuna nonzero scalar\(α\) vile kwamba\( \vecs{v}=α\vecs{u}\). Onyesha kwamba wadudu\( \vecd{AB}\) na\(\vecd{AC}\) ni collinear, wapi\(A(4,1,0), \, B(6,5,−2),\) na\(C(5,3,−1).\)

27) Pata eneo la parallelogram na pande zilizo karibu\(\vecs{u}=⟨3,2,0⟩\) na\(\vecs{v}=⟨0,2,1⟩\).

Jibu

\(7\)

28) Pata eneo la parallelogram na pande zilizo karibu\(\vecs{u}=\mathbf{\hat i}+\mathbf{\hat j}\) na\(\vecs{v}=\mathbf{\hat i}+\mathbf{\hat k}.\)

29) Fikiria pointi\(A(3,−1,2),\, B(2,1,5),\) na\(C(1,−2,−2).\)

Pata eneo la parallelogram\(ABCD\) na pande zilizo karibu\(\vecd{AB}\) na\( \vecd{AC}\).

pata eneo la pembetatu\(ABC\).

pata umbali kutoka hatua\(A\) hadi mstari\(BC\).

Jibu

a.\(5\sqrt{6};\) b.\(\frac{5\sqrt{6}}{2};\) c.\(\frac{5\sqrt{6}}{\sqrt{59}} =\frac{5\sqrt{354}}{59} \)

30) Fikiria pointi\(A(2,−3,4),\, B(0,1,2),\) na\(C(−1,2,0).\)

Pata eneo la parallelogram\(ABCD\) na pande zilizo karibu\( \vecd{AB}\) na\( \vecd{AC}\).

pata eneo la pembetatu\(ABC\).

c Pata umbali kutoka hatua\(B\) hadi mstari\(AC.\)

Katika mazoezi 31-32, vectors\(\vecs{u}, \, \vecs{v}\), na\(\vecs{w}\) hutolewa.

Pata bidhaa tatu za scalar\(\vecs{u}⋅(\vecs{v}×\vecs{w}).\)

pata kiasi cha parallelepiped na kando karibu\(\vecs{u},\,\vecs{v}\), na\(\vecs{w}\).

31)\(\quad \vecs{u}=\mathbf{\hat i}+\mathbf{\hat j}, \quad \vecs{v}=\mathbf{\hat j}+\mathbf{\hat k},\) na\(\quad \vecs{w}=\mathbf{\hat i}+\mathbf{\hat k}\)

Jibu

\( a. 2; \quad b. 2\)vitengo ³

32)\(\quad \vecs{u}=⟨−3,5,−1⟩, \quad \vecs{v}=⟨0,2,−2⟩,\) na\(\quad \vecs{w}=⟨3,1,1⟩\)

33) Tumia bidhaa tatu za scalar\(\vecs{v}⋅(\vecs{u}×\vecs{w})\) na\(\vecs{w}⋅(\vecs{u}×\vecs{v}),\) wapi\(\vecs{u}=⟨1,1,1⟩, \vecs{v}=⟨7,6,9⟩,\) na\(\vecs{w}=⟨4,2,7⟩.\)

Jibu

\(\vecs{v}⋅(\vecs{u}×\vecs{w})=−1, \quad \vecs{w}⋅(\vecs{u}×\vecs{v})=1\)

34) Tumia bidhaa tatu za scalar\(\vecs{w}⋅(\vecs{v}×\vecs{u})\) na\(\vecs{u}⋅(\vecs{w}×\vecs{v}),\) wapi\(\vecs{u}=⟨4,2,−1⟩, \vecs{v}=⟨2,5,−3⟩,\) na\(\vecs{w}=⟨9,5,−10⟩.\)

35) Kupata wadudu\(\vecs{a},\, \vecs{b}\),\(\vecs{c}\) na kwa mara tatu scalar bidhaa iliyotolewa na determinant\( \begin{vmatrix}1&2&3\\0&2&5\\8&9&2\end{vmatrix}\). Tambua bidhaa zao tatu za scalar.

Jibu

\(\vecs{a}=⟨1,2,3⟩, \quad \vecs{b}=⟨0,2,5⟩, \quad \vecs{c}=⟨8,9,2⟩; \quad \vecs{a}⋅(\vecs{b}×\vecs{c})=−9\)

36) Bidhaa tatu ya scalar ya vectors\(\vecs{a},\,\vecs{b}\), na\(\vecs{c}\) hutolewa na uamuzi\( \begin{vmatrix}0&−2&1\\0&1&4\\1&−3&7\end{vmatrix}\). Kupata vector\(\vecs{a}−\vecs{b}+\vecs{c}.\)

37) Fikiria parallelepiped na kando\( OA,OB,\) na\( OC\), wapi\( A(2,1,0),B(1,2,0),\) na\( C(0,1,α).\)

Pata idadi halisi\( α>0\) kama kiasi cha parallelepiped ni\( 3\) vitengo ^3.

b Kwa\( α=1,\) kupata urefu\(h\) kutoka vertex\(C\) ya parallelepiped. Mchoro parallelepiped.

Jibu

\( a. \, α=1; \quad b. \, h=1\)kitengo,

38) Fikiria pointi\( A(α,0,0),B(0,β,0),\) na\( C(0,0,γ)\), kwa\( α, β\), na idadi\( γ\) nzuri halisi.

a Kuamua kiasi cha parallelepiped na pande karibu\( \vecd{OA}, \vecd{OB},\) na\( \vecd{OC}\).

pata kiasi cha tetrahedron na vipeo\( O,A,B,\) na\( C\). (Kidokezo: Kiasi cha tetrahedron ni\( 1/6\) cha kiasi cha parallelepiped.)

c Kupata umbali kutoka asili ya ndege kuamua\( A,B,\) na\( C\). Mchoro parallelepiped na tetrahedron.

39) Hebu\( u,v,\) na\( w\) uwe na vectors tatu-dimensional na\(c\) uwe namba halisi. Thibitisha mali zifuatazo za bidhaa za msalaba.

a.\(\vecs u×\vecs u=\vecs 0\)

b.\(\vecs u×(\vecs v+\vecs w)=(\vecs u×\vecs v)+(\vecs u×\vecs w)\)

c.\( c(\vecs u×\vecs v)=(c\vecs u)×\vecs v=\vecs u×(c\vecs v)\)

d.\( \vecs u⋅(\vecs u×\vecs v)=\vecs 0\)

40) Onyesha kwamba vectors\(\vecs u=⟨1,0,−8⟩,\,\vecs v=⟨0,1,6⟩\), na\(\vecs w=⟨−1,9,3⟩\) kukidhi mali zifuatazo za bidhaa msalaba.

a.\(\vecs u×\vecs u=\vecs 0\)

b.\(\vecs u×(\vecs v+\vecs w)=(\vecs u×\vecs v)+(\vecs u×\vecs w)\)

c.\( c(\vecs u×\vecs v)=(c\vecs u)×\vecs v=\vecs u×(c\vecs v)\)

d.\(\vecs u⋅(\vecs u×\vecs v)=\vecs 0\)

41) Nonzero wadudu\(\vecs u,\,\vecs v\), na\(\vecs w\) inasemekana kuwa tegemezi linearly kama moja ya wadudu ni mchanganyiko linear ya wengine wawili. Kwa mfano, kuna mbili nonzero namba halisi\( α\) na\( β\) vile kwamba\(\vecs w=α\vecs u+β\vecs v\). Vinginevyo, vectors huitwa linearly kujitegemea. Kuonyesha kwamba\(\vecs u,\vecs v\), na\(\vecs w\) inaweza kuwekwa kwenye ndege moja kama na tu kama ni linear tegemezi.

42) Fikiria wadudu\(\vecs u=⟨1,4,−7⟩,\,\vecs v=⟨2,−1,4⟩,\,\vecs w=⟨0,−9,18⟩\), na\(\vecs p=⟨0,−9,17⟩.\)

a. kuonyesha kwamba\(\vecs u,\,\vecs v\), na\(\vecs w\) inaweza kuwekwa kwenye ndege moja kwa kutumia bidhaa zao mara tatu scalar

b Onyesha kwamba\(\vecs u,\,\vecs v\), na\(\vecs w\) inaweza kuwekwa kwenye ndege moja kwa kutumia ufafanuzi kwamba kuna namba mbili zisizo za sifuri halisi\( α\) na\( β\) vile vile\( w=αu+βv.\)

Onyesha kwamba\(\vecs u,\,\vecs v\), na\(\vecs p\) ni linearly huru - yaani, hakuna wadudu ni mchanganyiko linear ya wengine wawili.

43) Fikiria pointi\( A(0,0,2), B(1,0,2), C(1,1,2),\) na\( D(0,1,2).\) Je, vectors\( \vecd{AB}, \vecd{AC},\) na tegemezi\( \vecd{AD}\) linearly (yaani, moja ya wadudu ni mchanganyiko linear ya wengine wawili)?

Jibu

Ndiyo,\( \vecd{AD}=α\vecd{AB}+β\vecd{AC},\) wapi\( α=−1\) na\( β=1.\)

44) Onyesha kwamba vectors\( \mathbf{\hat i}+\mathbf{\hat j}, \mathbf{\hat i}−\mathbf{\hat j},\) na\( \mathbf{\hat i}+\mathbf{\hat j}+\mathbf{\hat k}\) ni linearly huru - yaani, kuna idadi mbili zisizo zero halisi\(α\) na\(β\) vile kwamba\(\mathbf{\hat i}+\mathbf{\hat j}+\mathbf{\hat k}=α(\mathbf{\hat i}+\mathbf{\hat j})+β(\mathbf{\hat i}−\mathbf{\hat j}).\)

45) Hebu\(\vecs u=⟨u_1,u_2⟩\) na\(\vecs v=⟨v_1,v_2⟩\) uwe vectors mbili-dimensional. Bidhaa ya msalaba wa vectors\(\vecs u\) na\(\vecs v\) haijafafanuliwa. Hata hivyo, kama vectors ni kuonekana kama vectors tatu-dimensional\( \tilde{\vecs u}=⟨u_1,u_2,0⟩\) na\( \tilde{\vecs v}=⟨v_1,v_2,0⟩\), kwa mtiririko huo, basi, katika kesi hii, tunaweza kufafanua bidhaa msalaba wa\( \tilde{\vecs u}\) na\( \tilde{\vecs v}\). Hasa, katika nukuu ya kuamua, bidhaa ya msalaba\( \tilde{\vecs u}\) na\( \tilde{\vecs v}\) hutolewa na

\( \tilde{\vecs u}×\tilde{\vecs v}=\begin{vmatrix}\mathbf{\hat i}&\mathbf{\hat j}&\mathbf{\hat k}\\u_1&u_2&0\\v_1&v_2&0\end{vmatrix}\).

Tumia matokeo haya kukokotoa\( (\cos θ\,\mathbf{\hat i}+\sin θ\,\mathbf{\hat j})×(\sin θ\,\mathbf{\hat i}−\cos θ\,\mathbf{\hat j}),\) ambapo\( θ\) ni idadi halisi.

Jibu

\( −\mathbf{\hat k}\)

46) Fikiria pointi\( P(2,1), Q(4,2),\) na\( R(1,2).\)

Pata eneo la pembetatu\( PQR\).

b Kuamua umbali kutoka hatua\( R\) hadi mstari unaopita\( P\) na\( Q\).

47) Kuamua vector ya ukubwa\( 10\) perpendicular kwa ndege kupita kupitia x-axis na uhakika\( P(1,2,4).\)

Jibu

\( ⟨0,±4\sqrt{5},2\sqrt{5}⟩\)

48) Kuamua vector kitengo perpendicular kwa ndege kupita kupitia z -axis na uhakika\( A(3,1,−2).\)

49) Fikiria\(\vecs u\) na vectors\(\vecs v\) mbili tatu-dimensional. Ikiwa ukubwa wa vector ya bidhaa ya msalaba\(\vecs u×\vecs v\) ni\( k\) mara kubwa zaidi kuliko ukubwa wa vector\(\vecs u\), onyesha kwamba ukubwa wa\(\vecs v\) ni mkubwa kuliko au sawa\( k\), wapi\( k\) idadi ya asili.

50) [T] Kudhani kwamba ukubwa wa wadudu wawili nonzero\(\vecs u\) na\(\vecs v\) wanajulikana. Kazi\( f(θ)=‖\vecs u‖‖\vecs v‖\sin θ\) inafafanua ukubwa wa vector ya bidhaa ya msalaba\(\vecs u×\vecs v,\) ambapo\( θ∈[0,π]\) ni angle kati\(\vecs u\) na\(\vecs v\).

a Graph kazi\( f\).

pata kiwango cha chini kabisa na upeo wa kazi\( f\). Tafsiri matokeo.

c Kama\( ‖\vecs u‖=5\) na\( ‖\vecs v‖=2\), kupata angle kati\(\vecs u\) na\(\vecs v\) kama ukubwa wa bidhaa zao msalaba vector ni sawa na\( 9\).

51) Kupata wadudu wote\(\vecs w=⟨w_1,w_2,w_3⟩\) kwamba kukidhi equation\( ⟨1,1,1⟩×\vecs w=⟨−1,−1,2⟩.\) Hint: Unapaswa kuwa na uwezo wa kuandika vipengele vyote vya\(\vecs w\) katika suala la moja ya constants\(w_1,w_2,\) au\(w_3\).

Jibu

Kuandika vipengele vyote kwa suala la mara kwa mara\(w_3\), njia moja ya kuwakilisha wadudu hawa ni:\(\vecs w=⟨w_3−1,w_3+1,w_3⟩,\) wapi\( w_3\) idadi yoyote halisi.
Kumbuka kwamba tunaweza kutumia parameter yoyote tunayotaka hapa. Tunaweza kuweka\(w_3 = a\). Kisha\(\vecs w=⟨a−1,a+1,a⟩\) ingekuwa pia kuwakilisha wadudu hawa.

52) Kutatua equation\(\vecs w×⟨1,0,−1⟩=⟨3,0,3⟩,\) ambapo\(\vecs w=⟨w_1,w_2,w_3⟩\) ni nonzero vector na ukubwa wa\( 3\).

53) [T] fundi anatumia 12-in. wrench kurejea bolt. Wrench hufanya\( 30°\) angle na usawa. Ikiwa fundi inatumia nguvu ya wima ya\( 10\) lb kwenye kushughulikia wrench, ni ukubwa gani wa wakati\( P\) (angalia takwimu zifuatazo)? Eleza jibu kwa paundi za mguu uliozunguka kwenye sehemu mbili za decimal.

Jibu

8.66 ft-lb

54) [T] mvulana inatumika breki juu ya baiskeli kwa kutumia nguvu ya chini ya 20 lb juu ya kanyagio wakati 6-katika. crank inafanya\( 40°\) angle na usawa (angalia takwimu zifuatazo). Kupata moment katika hatua\( P\). Eleza jibu lako kwa paundi za mguu-mviringo kwenye sehemu mbili za decimal.

55) [T] Pata ukubwa wa nguvu ambayo inahitaji kutumiwa hadi mwisho wa wrench ya 20-cm iko kwenye mwelekeo mzuri wa\(y\) -axis ikiwa nguvu inatumiwa katika mwelekeo\( ⟨0,1,−2⟩\) na inazalisha moment ya\( 100\) N · m kwa bolt iko katika asili.

Jibu

\(250\sqrt{5}\)N\(\approx 559\) N

56) [T] Ni nini ukubwa wa nguvu required kutumika hadi mwisho wa 1-ft wrench katika pembe ya\( 35°\) kuzalisha moment ya\( 20\) N · m?

57) [T] Vector nguvu\(\vecs F\) kaimu juu ya proton na malipo ya umeme ya\( 1.6×10^{−19}\,C\) (katika coulombs) kusonga katika uwanja magnetic\(\vecs B\) ambapo vector\(\vecs v\) kasi hutolewa na\(\vecs F=1.6×10^{−19}(\vecs v×\vecs B)\) (hapa,\(\vecs v\) ni walionyesha katika mita kwa pili,\(\vecs B\) ni katika tesla [T], na\(\vecs F\) iko katika newtons [N]). Kupata nguvu kwamba vitendo juu ya proton kwamba hatua katika\(xy\) -ndege katika kasi\(\vecs v=10^5\mathbf{\hat i}+10^5\mathbf{\hat j}\) (katika mita kwa sekunde) katika uwanja magnetic iliyotolewa na\(\vecs B=0.3\mathbf{\hat j}\).

Jibu

\(\vecs F=4.8×10^{−15}\,kN\)

58) [T] Vector ya nguvu\(\vecs F\) inayofanya proton yenye malipo ya umeme ya\( 1.6×10^{−19}\,C\) kuhamia kwenye uwanja wa magnetic\(\vecs B\) ambapo vector ya kasi\(\vecs v\) hutolewa na\(\vecs F=1.6×10^{−19}(\vecs v×\vecs B)\) (hapa,\(\vecs v\) inaelezwa kwa mita kwa pili,\(\vecs B\) ndani\( T\), na\(\vecs F\) ndani\( N\)). Ikiwa ukubwa wa nguvu\(\vecs F\) unaofanya protoni ni\( 5.9×10^{−17}\,N\) na proton inahamia kwa kasi ya 300 m/sec katika uwanja wa magnetic\(\vecs B\) wa ukubwa 2.4 T, pata angle kati ya vector kasi\(\vecs v\) ya proton na shamba la magnetic\(\vecs B\). Eleza jibu kwa digrii zilizozunguka kwa integer iliyo karibu.

59) [T] Fikiria\(\vecs r(t)=⟨\cos t,\,\sin t,\,2t⟩\) vector msimamo wa chembe kwa wakati\( t∈[0,30]\), ambapo vipengele vya\(\vecs r\) vinaelezwa kwa sentimita na wakati kwa sekunde. Hebu\( \vecd{OP}\) kuwa vector nafasi ya chembe baada ya\( 1\) sec.

a Kuamua kitengo vector\(\vecs B(t)\) (inayoitwa binormal kitengo vector) ambayo ina mwelekeo wa bidhaa msalaba vector\(\vecs v(t)×\vecs a(t),\) ambapo\(\vecs v(t)\) na\(\vecs a(t)\) ni instantaneous kasi vector na, kwa mtiririko huo, kuongeza kasi vector ya\( t\) chembe baada ya sekunde.

b Matumizi CAS taswira wadudu\(\vecs v(1),\,\vecs a(1)\), na\(\vecs B(1)\) kama wadudu kuanzia katika hatua\( P\) pamoja na njia ya chembe.

Jibu

a.\(\vecs B(t)=⟨\frac{2\sqrt{5}\sin t}{5},−\frac{2\sqrt{5}\cos t}{5},\frac{\sqrt{5}}{5}⟩;\)

b.

60) Jopo la jua limewekwa juu ya paa la nyumba. Jopo linaweza kuonekana kama limewekwa kwenye pointi za kuratibu (katika mita)\( A(8,0,0), B(8,18,0), C(0,18,8),\) na\( D(0,0,8)\) (angalia takwimu zifuatazo).

Pata vector\(\vecs n=\vecd{AB}×\vecd{AD}\) perpendicular kwa uso wa paneli za jua. Eleza jibu kwa kutumia vectors ya kawaida ya kitengo. Kumbuka kwamba ukubwa wa vector hii inapaswa kutupa eneo la mstatili\(ABCD\).

b Kudhani kitengo vector\(\vecs s=\frac{1}{\sqrt{3}}\mathbf{\hat i}+\frac{1}{\sqrt{3}}\mathbf{\hat j}+\frac{1}{\sqrt{3}}\mathbf{\hat k}\) pointi kuelekea jua wakati fulani wa siku na mtiririko wa nishati ya jua ni\(\vecs F=900\vecs s\) (katika watts kwa mita ya mraba [\( W/m^2\)]). Pata kiasi kilichotabiriwa cha nguvu za umeme jopo linaweza kuzalisha, ambalo hutolewa na bidhaa ya dot ya vectors\(\vecs F\) na\(\vecs n\) (iliyoelezwa kwa watts).

c Kuamua angle ya mwinuko wa jua juu ya jopo la jua. Eleza jibu kwa digrii zilizozunguka kwa nambari nzima iliyo karibu. (Kidokezo: Pembe kati ya vectors\(\vecs n\)\(\vecs s\) na angle ya mwinuko ni nyongeza.)

Wachangiaji

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