(1) c=a+tb=(1,2)+t(1,−1)=(1+t,2−t)c = a + tb = (1,2) + t(1,-1) = (1+t, 2-t)c=a+tb=(1,2)+t(1,−1)=(1+t,2−t) (2) ∣c∣=3=(1+t)2+(2−t)2=1+2t+t2+4−4t+t2=5−2t+2t2|c| = 3 = \sqrt{(1+t)^2 + (2-t)^2} = \sqrt{1+2t+t^2+4-4t+t^2} = \sqrt{5-2t+2t^2}∣c∣=3=(1+t)2+(2−t)2=1+2t+t2+4−4t+t2=5−2t+2t2 9=5−2t+2t22t2−2t−4=2(t2−t−2)=2(t−2)(t+1)=0t=2,−19 = 5-2t+2t^2 \qquad 2t^2-2t-4 = 2(t^2-t-2) = 2(t-2)(t+1) = 0 \qquad t = 2,-19=5−2t+2t22t2−2t−4=2(t2−t−2)=2(t−2)(t+1)=0t=2,−1 (3) c∥dc \parallel dc∥d なので c=kdc = kdc=kd となる {1+t=k4①2−t=−k②\begin{cases} 1+t = k4 & ① \\ 2-t = -k & ② \end{cases}{1+t=k42−t=−k①② ①+②から k=1k = 1k=1 これから t=3t = 3t=3