向量题，求过程答案，拜托了大佬！ 求填空题答案！！谢谢大佬！！！
\u4e00\u9053\u6570\u5b66\u9898\uff0c\u6c42\u8fc7\u7a0b\uff01\u8c22\u8c22\u5927\u4f6c\uff01AB=CD=2 BC=2 \u4e09\u89d2\u5f62BCD\u662f\u76f4\u89d2\u4e09\u89d2\u5f62\uff0c\u6240\u4ee5BD=\u6839\u53f7\u4e0b\uff08CD^2+BC^2)=2\u6839\u53f72
DD'=BB'=2\u6839\u53f72=BD ,\u4e09\u89d2\u5f62B'BD\u662f\u76f4\u89d2\u4e09\u89d2\u5f62\uff0c\u6240\u4ee5DB'=\u6839\u53f7\u4e0b\uff08BD^2+BB'^2)=4
6. \u5df2\u77e5\u70b9A(4,0,5)\u548cB(7,1,3)\u5219\u4e0e\u5411\u91cfAB\u540c\u5411\u7684\u5355\u4f4d\u5411\u91cf\u4e3a?
\u5411\u91cfAB={3\uff0c1\uff0c-2}\uff0c\u2223AB\u2223=\u221a[3²+1²+(-2)²]=\u221a14.
\u6545\u4e0e\u5411\u91cfAB\u540c\u5411\u7684\u5355\u4f4d\u5411\u91cf\u4e3a{3/\u221a14\uff0c1/\u221a14\uff0c-2/\u221a14}
7. yoz\u9762\u4e0a\u7684\u629b\u7269\u7ebfy²=2px\u7ed5z\u8f74\u65cb\u8f6c\u800c\u6210\u7684\u66f2\u9762\u65b9\u7a0b\u4e3a\uff1ax²+y²=2px;
8. \u4e8c\u91cd\u79ef\u5206∬d\u03c3=\uff1f [D\uff1ax²+y²\u22661]
∬d\u03c3=∬dxdy=\u03c0(\u5373\u5706\u5fc3\u5728\u539f\u70b9\uff0c\u534a\u5f84\u4e3a1\u7684\u56edD\u7684\u9762\u79ef).
9. \u51fd\u6570z=sin(xy)\u7684\u5168\u5fae\u5206.
dz=(∂z/∂x)dx+(∂z/∂y)dy=ycos(xy)dx+xcos(xy)dy=cos(xy)(ydx+xdy).
10. z=z(x,y)\u662f\u7531\u65b9\u7a0b x²+y²+z²-4z=0\u6240\u786e\u5b9a\u7684\u9690\u51fd\u6570\uff0c\u5219∂z/∂x=?
\u8bbeF(x,y,z)=x²+y²+z²-4z=0\uff1b\u5219∂z/∂x=-(∂F/∂x)/(∂F/∂z)=-2x/(2z-4).
可以将c移到等号右边，然后两边平方，求出cos(a,b)
解答如下：
绛旓細1銆佽繃a鐨勭粓鐐逛綔骞宠浜巄鐨勭洿绾匡紝浠鐨勭粓鐐逛负鏈夊悜绾挎鐨勮捣鐐癸紝涓巄鐨勭浉鍚屾柟鍚戞埅鍙栦笌b鍚岄暱鐨勭嚎娈碉紝浠巃鐨勮捣鐐瑰埌璇ョ嚎娈电殑缁堢偣鐨勬湁鍚戠嚎娈靛氨鏄痑+b銆2銆乤+b=b+a锛屼互a,b涓洪偦杈圭敾姝ｆ柟褰紝瀵硅绾挎棦鏄痑+b锛屽張鏄痓+a銆3銆佸叡绾挎椂鍦ㄥ悓涓鐩寸嚎涓婏紝鍚屽悜鏃朵粠a鐨勭粓鐐瑰欢闀跨殑闀垮害锛屽弽鍚戞椂浠巃鐨勭粓鐐瑰悜a鐨...
绛旓細浠栦滑澶硅鏄攼瑙掞紝鍒欎粬浠殑鍐呯Н澶т簬闆 浠栦滑鐨勫唴绉槸锛(a+mb)*(ma+b)=m(a^2+b^2)+(1+m^2)a*b=5m+(1+m^2)3 3m^2+5m+3鐨 螖=-11<0 鎵浠ヤ粬鎬绘槸澶т簬闆剁殑銆傚彟澶栵紝瑕侀櫎鍘讳竴涓壒娈婃儏褰:涓よ呭钩琛岋紝鍥犱负姝ゆ椂鍐呯Н鍚屾牱澶т簬闆 骞宠鍒欙細1:m=m:1,鎵浠 m=+/- 1 缁间笂锛氳寖鍥存槸 m鈮犅...
绛旓細1銆乤锛3i锛峧锛2k锛宐锛漣锛2j锛峩 (-2a)路(3b)锛濓紞6(a路b)锛濓紞6[3脳1锛1脳2锛(-2)脳(-1)]锛濓紞18 2銆鍚戦噺CB锛漛锛峜锛孉C锛漜锛峚锛屽洜涓篈,B,C鍏辩嚎锛屾墍浠ュ悜閲廋B銆丄C鍏辩嚎锛屽張鐐笴鍒嗗悜閲廈A鐨勬瘮鏄伙紝鍒機B锛澪籄C锛屾墍浠锛峜锛澪(c锛峚)锛屽嵆位a锛媌锛(1锛嬑)c锛0 鐢蔽籥锛媌锛(...
绛旓細2a+b=(-3,6)a+kb= ( -3+3k, 1+4k)锛2a+b锛.锛坅+kb锛=20 (-3,6).( -3+3k, 1+4k) =20 -3(-3+3k) +6(1+4k)=20 15k+15=20 15k=5 k=1/3
绛旓細1 BA=OA-OB BM=BA/6=(OA-OB)/6 =(a-b)/6 OM=OB+BM =b+(a-b)/6 =a/6+5b/6 --- OD=OA+OB=a+b ON=2OD/3=2a/3+2b/3 --- MN=ON-OM =2a/3+2b/3-a/6-5b/6 =a/2-b/6 2 c=位a+b锛宒=a+(2位-1)d c涓巇鍚屽悜锛屽嵆锛歞=kc锛宬>0 鍗筹細a+(2位-1)d=...
绛旓細k1.(伪1+伪2+伪3)+k2.(伪1+2伪2+3伪3)+k3.(伪1+4伪2+9伪3) =0 (k1+k2+k3)伪1+(k1+2k2+4k3)伪2+(k1+3k2+9k3)伪3=0 => k1+k2+k3=0 (1)k1+2k2+4k3=0 (2)k1+3k2+9k3=0 (3)2(2)-(1)k1+3k2+7k3=0 (4)(3)-(4)2k3=0 k3=...
绛旓細妤间笂鐨勬濊矾鍩烘湰涓婃槸姝ｇ‘鐨勶紝涓嶈繃鎴戣繕鏄兂鎸夋垜鑷繁鐨勬濊矾鍋氫竴閬嶃傦紙1锛夆埖A銆丮銆丏涓夌偣鍏辩嚎锛屸埓鍚戦噺OM=渭鍚戦噺OA+(1-渭)鍚戦噺OD 鈭礏銆丮銆丆涓夌偣鍏辩嚎锛屸埓鍚戦噺OM=位鍚戦噺OC+(1-位)鍚戦噺OB 宸茬煡鍚戦噺OC=1/4鍚戦噺OA锛屽悜閲OD=1/2鍚戦噺OD锛岃仈绔嬩袱涓柟绋嬶紝鍗 位/4=渭锛1-位=(1-渭)/2锛岃В寰椢=...
绛旓細璇佹槑涓や釜鍚戦噺a,b骞宠鐨勪竴涓柟娉曟槸锛氳瘉鏄庡瓨鍦ㄤ竴涓疄鏁発锛屼娇寰梐=kb銆3a+2b=8c锛宎-5b=-3c銆傛墍浠3a+2b=8c=-8/3(-3c)=-8/3(a-5b)锛3(3a+2b)=-8(a-5b)锛17a=34b锛宎=2b銆傛墍浠涓巄骞宠銆傦紞锛嶏紞 杩涗竴姝ヨ繕鍙互寰楀埌b=c锛屾墍浠=2b=2c锛屾墍浠,b,c涓や袱骞宠銆
绛旓細璁綪(k,k,2k).鍒橮A=(1-k, 2-k, 3-2k),PB=(2-k, 1-k, 2-2k)PA.PB=(1-k)(2-k)+(2-k)(1-k)+(3-2k)(2-2k)=6k^2-16k+10=6(k-1)(k-5/3)鐭:褰搆=-(-16/12)=4/3鏃,鍙栧緱灏忓,鏈灏忓间负:-6/9=-2/3.鐐筆(4/3,4/3,8/3)
绛旓細鍙戜簡鍥剧墖锛屾渶蹇洖绛旓紝鏈涢噰绾筹紝璋㈣阿
