向量a=（x1，y1），向量b=（x2，y2），且向量b平行于（共线），求a与b坐标关系 为什么向量a=（x1,y1）,向量b=（x2,y2）,则向量...

\u5411\u91cfa\uff08x1\uff0cy1\uff09\uff0c\u5411\u91cfb\uff08x2 \uff0cy2\uff09\uff0c\u82e5a\u5e73\u884c\u4e8eb\uff0c\u80fd\u63a8\u51fa\u6765\u4ec0\u4e48

x1/x2=y1/y2

a//b
\u5982\u679ca,b\u5747\u4e0d\u4e3a0 \u90a3\u4e48\u5c31\u7b49\u4ef7\u4e8ex1/y1=x2/y2
\u4ea6\u5373 x1y2-x2y1=0
\u7531\u4e8e0\u5411\u91cf\u5e73\u884c\u4e8e\u4efb\u610f\u5411\u91cf \u5982\u679ca,b\u4e2d\u6709\u4e00\u4e2a\u662f0\u5411\u91cf
\u9664\u5f0f\u53ef\u80fd\u51fa\u73b0\u5206\u6bcd\u4e3a0\u7684\u60c5\u51b5\uff0c\u4f46x1y2-x2y1=0\u59cb\u7ec8\u662f\u6ee1\u8db3\u7684

\u7efc\u4e0a a//b\u7b49\u4ef7\u4e8ex1y2-x2y1=0

解：说两种常用解法吧。（方法一）如果向量a//b，那么向量a和向量b的对应坐标成比例，即存在非零实数k使得向量a = k*b = (x1，y1) = k(x2，y2) = x1 = kx2以及y1 = ky2 = x1 *(ky2) = (kx2)*y1 = kx1y2 = kx2y1 =x1y2 = x2y1 。（方法二）如果向量a//b，那么向量a和向量b的夹角a，b等于0或者π，所以向量a·b = |a|*|b|cosa，b①，因为a，b等于0或者π，所以cosa，b = 1或者-1，对①式取绝对值，可得|a·b| = |(|a|*|b|cosa，b)| = ||a|*|b|| = |a|*|b| = |x1x2 + y1y2| = √(x12 + y12)*√(x22 + y22)，两边平方可得x12x22 + y12y22 + 2x1x2y1y2= (x12 + y12)*(x22 + y22) = x12x22 + x12y22 + y12x22 +y12y22 = x12y22 – 2x1x2y1y2+ y12x22 = 0 = (x1y2 – x2y1)2 = 0 =x1y2 – x2y1 = 0 。

鍚戦噺a=( x1, y1),鍚戦噺b=(?
绛旓細|鍚戦噺a|=鈭(x1^2+y1^2)|鍚戦噺b|=鈭(x2^2+y2^2)<鍚戦噺a,鍚戦噺b >涓轰簩鍚戦噺鐨勫す瑙 2锛屽潗鏍囧舰寮忥細鍚戦噺a•鍚戦噺b= x1x2+y1y2

鍚戦噺a=(x1,y1),鍚戦噺b=(x2,y2),涓斿悜閲廱骞宠浜(鍏辩嚎),姹俛涓巄鍧愭爣鍏崇郴
绛旓細瑙ｏ細璇翠袱绉嶅父鐢ㄨВ娉曞惂銆傦紙鏂规硶涓锛夊鏋鍚戦噺a//b锛岄偅涔堝悜閲廰鍜鍚戦噺b鐨勫搴斿潗鏍囨垚姣斾緥锛屽嵆瀛樺湪闈為浂瀹炴暟k浣垮緱鍚戦噺a = k*b = (x1锛寉1) = k(x2锛寉2) = x1 = kx2浠ュ強y1 = ky2 = x1 *(ky2) = (kx2)*y1 = kx1y2 = kx2y1 =x1y2 = x2y1 銆傦紙鏂规硶浜岋級濡傛灉鍚戦噺a//b锛岄偅涔堝悜...

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绛旓細鎵浠*b =x1*x2+y1*y2

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绛旓細绛旀锛歜=锛-x1,-y1锛夌浉浜掑钩琛岃鏄庡畠浠悓鍚戞垨鍙嶅悜,鍙坾a|=|b|,涓攂鈮燼 璇存槑瀹冧滑鍙兘鍙嶅悜涓斿鍦嗙偣鐨勮窛绂荤浉绛

鍚戦噺a绛変簬(x1,y1),鍚戦噺b=(x2,y2),鍒a鍚戦噺鍨傜洿浜b鍚戦噺鐨勫厖瑕佹潯浠舵槸 骞...
绛旓細鍏呰鏉′欢鏄細x1*x2+y1*y2=0 璇佹槑锛歛*b=|a|*|b|*cos 鍥犱负a涓巄鍨傜洿,鎵浠 =90搴 鎵浠os =0 鎵浠*b=x1*x2+y1*y2=0

鍚戦噺a=(X1,Y1),鍚戦噺b=(X2,Y2),X1Y2-X2Y1=0鏄悜閲廰鍜屽悜閲廱鍏辩嚎鐨勫厖瑕...
绛旓細鏄晩 X1Y2-X2Y1=0 鍙互鍐欎綔 x1/y1=x2/y2,姝ｆ槸鍏辩嚎鐨勫嚑浣曠洿瑙傘 闆鍚戦噺鍜屾墍鏈夊叡绾裤

涓轰粈涔堣嫢鍚戦噺a=(x1,y1),鍚戦噺b=(x2銆亂2)鍒檃鈯x1x2+y1y2=0
绛旓細鍏堣鍚戦噺涔樼Н鐨勭墿鐞嗗惈涔夛細鍚戦噺a涓鍚戦噺b鐨勭偣涔樿〃绀猴細鍚戦噺a鐨勯暱搴︼紙|a|锛変笌鍚戦噺b鍦╝涓婄殑鎶曞奖鐨勯暱搴(|b|cos蠁)鐨勪箻绉.鐢卞悜閲忎箻绉殑璁＄畻瑙勫垯,锛坸1,y1锛*锛坸2銆亂2锛=x1x2+y1y2=0杩欒鏄庡悜閲廰鐨勬ā锛堥暱搴︼級涓庡悜閲廱鍦╝涓婄殑鎶...

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绛旓細鎵璋撳钩琛鍚戦噺锛屽氨鏄案杩滀笉鐩镐氦鐨勭煝閲忥紝鎴戜滑鐪嬩綔绛夋晥鐨勭嚎褰唬鏁扮殑鏂圭▼缁 x1a1+y1a2=0 x2a1+y2a2=0 鏃犲畾瑙ｃ傞偅涔堜笂闈1,x2,y1,y2鏄父鏁帮紝a1,a2鏄彉閲忥紝閭ｄ箞鏈夌煩闃电殑琛屽垪寮=0(鍥犱负琛屽垪寮!=0鐨勬椂鍊欏彧鏈夊畾瑙(0,0)锛岃繖鏄浉浜や簬鍘熺偣鑰屼笉鏄钩琛)|x1,y1| |x2,y2|=0.鎵浠ユ湁浜唜1y2-x2y1...

鍚戦噺a=(x1,y1)鍚戦噺b=(x2,y2)鍚戦噺ab=
绛旓細ab锛漍1X2+Y1Y2

鍚戦噺a=(x1,y1).鍚戦噺b=(x2,y2).鍚戦噺a-鍚戦噺b鐨勭粷瀵瑰肩瓑浜庝粈涔
绛旓細|鍚戦噺a-鍚戦噺b| =|(x1,y1)-(x2,y2)| =|(x1-x2,y1-y2)| =鈭歔(x1-x2)^2+(y1-y2)^2]

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