{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

hw2 - See 12.5 Q 9a Solution bracket notion<.t,y>=<...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Background image of page 2
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Background image of page 4
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: See 12.5 Q 9a}: Solution: bracket notion < .t,y :>=<; —3,4 s» H {1,5 :> if, I? notion xi" + y} = —si“ + 4} + :{i‘ + 5}) Q16: the line through (2,4,5) that is parallel to {-1.23} x=2—I Solution: y=—l + 2! z=5+Tr Q52a} Given lines 13:1: 4r,_y =l—2t,z = 2+2t, £2:x=1+r,y =l—r,s=—l+4t , prove tl're}.r intersect at the point {2,U,3}. ' 41': = l + 1‘1 I —- 2!, = l H I: I. and I: is the solution of echation 2 + 2!] = —l + 4.3, so these two lines niust intersect at some point. Substituting rut: into the line equations, we can find the intersection (2,0,3) .- Proof : By solving equations { , we obtain :1 = LI" 2J2 = l . We also find c) Find a line perpendicular to Evil.” and passing through their intersection. We know the directional vector of undetermined line can be written as e 4,—2,2 } x {Ii—1,4 :>=< —6,#ln¢,—-2 :- , so that the solution satisfying above conditions a: = 2 —61' x = 2 + 3r should be y = -l4i' ; this solution is equivalent to y = '3'! z =3—21‘ z =3+I flow «'2'- Ewe-114:1 waists-i to Saline out: Emblem. Became .13” cm; /'_> Wm???" em Firemen, We lime; i5: . (All ma: )r-o Maul: Sec +' - ._ 12.6 i: i 0' a‘ero (fir—j) ZE—zriza u ' 311 the given points A(—2,1,1), B(0,2,3) and 1., Q11: Find the equa ionaofplatle that NC“, ,.0- 1}. olution: Using two vectors AB =< 2,1,2 > and AC :4 3, —1,—2 > in this plane, we find one normal direction vectCIr of’rhis plane Ex .51?— = 10 J 517 . Then, using anyone of these three points, we can find the plane equation 2y —- z —I = D . ' Ta «tut 0" Q2 We first assume that the plane equation 4x — 2y+ 72 +1: : 0. Since it pass through file origin point (0,0,0), we know r: = 0, say, the plane equation is 4x— 2y+ 72 = 0 . Q36: Proof. we know this plane passing through the three points A(a, 0 ,0), B(0, b ,,0) CG), 0 He} then one of normal direction vector of this plane 15 fl]? )1 A? =< be, ac, ab >. Combining with anyone point of A,B,C, we can get the plane equation box +acy + abz = abc, i.e., xfa+yfb+zfc =l. AIS; 'V‘bue rs chlrluer ”(Bron-f I?“ Q.” . 1'th “‘5 f“ 91-1. .m- ‘ ..-—I IKE -“ '5'“ M PA Lee 11-.. ¢:‘> ,, 2‘ r-é‘x in o ' n—O {a '0' L {:3 in? +flcj +£1.49? sol/LC 6'": 7f 4, y +_E{ if »Sec :2? gamifi‘c §Mfad€_ 994'. ‘a’ #31114 5:5 1 “-30. 7-1;; eme‘r‘an (San 19¢ 15%» (filmed Win: 7/: 32+4g“ ; 171 + {Air , 5;; i7: is 5:94»de Elk/375m yam/baled , Me 64,. {kc 15d 19%: Swfwce: M Sfiaw» by fax? lama/k1, 6135- 1123325229 TA“; flaw“ Can lac fimgfirmed Mo 3;: fig: :4? - gt; 13¢»; hype/50mm 07¢ 31 “if MEI-”)1. I I . 131%? 534606" ‘3 . '3“; QJ%. 41"}:1‘+422:Z6 , Me (24 hmyl74nweg if £550 ghw’t + 3: - VD 2” '4? F’ Sec 52.? Gyhhfifin‘c‘axz’ M Sphen‘gw/ . Q35, fphcfi :23; 75;} "5 £7WV‘L/MM 1’79 Xz-k'jrsf / 50%” ”-5 K, Cyh‘ncbra‘m supra/ace w guymfé, 52a Mada. ...
View Full Document

{[ snackBarMessage ]}