ベクトル・空間図形～同志社・理系全学部・2020数学Ⅱ

(1) P (1-t, 2t, 0)

(2) ベクトルAP = (1-t, 2t, -1)

     ベクトルBC = (-1, 2, 0)

     AP⊥BC　⇔　-(1-t) + 4t = 0   ∴ t = 1/5, AR = 3√5/5

(3) R (4/5, 2/5, 0),  ER = (±6/5, ±3/5), ∴ E (-2/5, -1/5, 0)

(4) ベクトルED = (12/5, 11/5, 0),  ED = √265/5

　 直線ED: y-2 = 11/12 (x-2) ⇔ 12 (2t-2) = 11 (-1-t)

　∴ t = 13/35,    P (22/35, 26/35, 0)