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== 点の存在範囲2 ==

【基本】
 平面上の相異なる3点A(aw),B(bw),C(cw)が同一直線上にないとき,
pw=saw+tbw+ucw
s+t+u=1・・・(1)
は平面全体に対応します。

pw=saw+tbw+ucw
s+t+u=1, s≧0,t≧0,u≧0・・・(2)
は△ABCの内部及び周上に対応します。
[解説]
(1)←
pw=saw+tbw+ucw(s+t+u=1)はuを消去して変形すると
pw=saw+tbw+(1−s−t)cw
    =cw+s(awcw)+t(bwcw)
    =OC+sCA+tCB

まず,OCにより原点から点Cに進み,次にsCA+tCBで平面全体を表すことができます。

(2)←
pw=saw+tbw+ucw(s+t+u=1, s≧0,t≧0,u≧0)
   =OC+sCA+tCB( s≧0,t≧0,u=1−s−t≧0)
   =OC+sCA+tCB( s≧0,t≧0,s+t≦1)
は△ABCの内部及び周上となります。


※条件s+t+u=1の役割
[解説1]
2つのベクトルだけで平面上のすべての点を表現することができるので,3つのベクトルを使うと,同一ベクトルの表し方が何通りも生じます。

s+t+u=1の条件がなければ,右図のような点Pについて
1aw+0bw+1cw (s+t+u=2)
2aw+1bw+0cw (s+t+u=3)
0aw+(−1)bw+2cw (s+t+u=1)

など,何通りも表し方ができます。

これを防ぎ,点と式を1対1に対応させるためには,s+t+u=1のような条件式を1つ加える必要があります.
【ここまでのポイント】
p=sa+tb+uc
s+t+u=1
は,1つの文字uを消去することにより
p=c+sCA+tCB
と読むことができます.これにより,まずCまで行って,次にCA,CBで作られる網目を目盛りとした位置を表すことになります.
- - - -
もちろん,1つの文字sを消去することにより
p=a+tAB+uAC
と読むこともでき,これにより,まずAまで行って,次にAB,ACで作られる網目を目盛りとした位置を表すと考えることもできます.
- - - -
tを消去しても同様にできますが,とりあえず1つは確実にできるようにしましょう.

※条件s+t+u=1の役割
[解説2]
2点A(a),B(b)を結ぶ線分ABm:nに内分する点Pの位置ベクトルをpとおくと
p=na+mbm+n(m,n>0)
が成り立ちます.(これは教科書などに書かれている基本公式です)
この内分公式を2段階に分けて,次のように組み立てます.
まず,2点A(a),B(b)を結ぶ線分ABk:lに内分する点Dの位置ベクトルをdとおくと
d=la+kbk+l(k,l>0)…(1)
次に,2点C(c),D(d)を結ぶ線分CD(k+l):mに内分する点Pの位置ベクトルをpとおくと
p=mc+(k+l)d(k+l)+m(k,l,m>0)…(2)
(1)を(2)に代入すると
p=mc+(k+l)la+kbk+l(k+l)+m
=mc+la+kb(k+l)+m=la+kb+mck+l+m
=lk+l+ma+kk+l+mb+mk+l+mc
となるから
s=lk+l+m,t=kk+l+m,u=mk+l+m
とおくと
p=sa+tb+uc(s,t,u>0,s+t+u=1)
が成り立つことになります.
(逆に)
p=0.2a+0.3b+0.5c(0.2+0.3+0.5=1)
のような式が与えられた場合
p=(0.3+0.2)0.2a+0.3b0.3+0.2+0.5c0.2+0.3+0.5
と変形できるので(0.3+0.2で割って,掛けるところが変形のポイント)
AB0.3:0.2に内分する点をDとすると,
CD(0.3+0.2):0.5に内分する点がPとなる.
このように,s,t,u>0, s+t+u=1となる実数s,t,uを用いて
p=sa+tb+uc
で表される点Pはつねに△ABCの内部にあることになります.
※ABでAから遠い方の比率がs,Bから遠い方の比率がt,CDでCから遠い方の比率がuになります

【不等式で表される領域を考えるには】
p=sa+tb+ucs+t+u=1)すなわちp=c+sCA+tCBで表される点はs, t, s+tの値に応じて次の図形上にあることになります.


○上の図で,s=−1,0,1,2,...となる場所を順に見て行くと,p=sa+tb+ucs+t+u=1, s≧0)で表される点は次の黄色で示した領域にあることが分かります.


○上の図で,t=−1,0,1,2,...となる場所を順に見て行くと,p=sa+tb+ucs+t+u=1, t≧0)で表される点は次の水色で示した領域にあることが分かります.


○上の図で,s+t=−1,0,1,2,...となる場所を順に見て行くと,p=sa+tb+ucs+t+u=1, s+t≦1)で表される点は次の緑色で示した領域にあることが分かります.


○これらの共通部分を考えることにより,次のような問題が解けます.
【例1】
平面上の相異なる3点A(a),B(b),C(c)が同一直線上にないとき,
p=sa+tb+uc
s+t+u=1, s≧0, t≧0, s+t≧1
で表される領域を答えてください.
(解答)
次の桃色で示した領域(境界線を含む)

【例2】
平面上の相異なる3点A(a),B(b),C(c)が同一直線上にないとき,
p=sa+tb+uc
s+t+u=1, s≦0, t≧0, s+t≦1
で表される領域を答えてください.
(解答)
次の桃色で示した領域(境界線を含む)


《問題》
 平面上で同一直線上にない3点A(aw),B(bw),C(cw)について,次の位置ベクトルpwで示される点Pが存在する範囲(黄色で表示)を右から選びなさい.ただし,図において境界線はすべて含まれるものとします。
○問題を一つクリックし,続けて対応する選択肢をクリックすると消えます.間違えば消えません.(難しいときは,計算用紙でuを消去した式に直すとよい。)
【式】
 
【選択肢の図】
【式】

【選択肢の図】
【式】

【選択肢の図】

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