Problem 18. The endpoints are   (4,3) - 5(-1,-2) = (9,13)   and   (4,3) - 2(-1,-2) = (6,7)   respectively. Therefore, the direction indicator for the segment is   (6,7) - (9,13) = (-3,-6),   and the segment is described as   {(9,13) + t(-3,-6): 0 t 1}.
 

Problem 26. The direction indicator of the ray is   (-4,-1) - (5,3) = (-9,-4).    The order of the subtraction is important. Finally, to get the unit direction indicator, as requested in the problem, we must multiply by the reciprocal of the length in order to get a vector of length 1. Thus, the answer is (-9,-4).
 

Problem 30. Let the line be given parametrically as   {P + sU : s },   and let the given points P₀, P₁, ..., P_n correspond to the parameter values s₀, s₁, ..., s_n respectively.

We're given that P_j is between P_j-1 and P_j-1 for j = 1, ..., n-1. What this means for the parameter values is that for each   j   we have either   s_j-1 < s_j < s_j+1   or   s_j-1 > s_j > s_j+1. What we have to prove is that all of these inequalities line up coherently. More specifically, in the case where   s₀ < s1 < s₂   we have to prove that we have the following chain of inequalities:

{And strictly speaking, we also would have to prove that the opposite chain of inequalities holds in the case where   s₀ > s1 > s₂.   But that proof would be so similar to the first case that it can safely be omitted.}

So, we will prove by induction on   p,   starting with p = 2, that we have the following chain of inequalities:

The initial step, namely p = 2, simply says that   s₀ < s1 < s₂.   This is true by assumption -- i.e., since we have decided to focus our discussion on the first case.

Thus, we proceed to the inductive step. Informally speaking, we have to show that if the chain of inequalities has been checked at the p^th stage, then the corresponding chain at the (p+1)^th stage will follow from this and the other given information. More formally, we assume that the chain of inequalities:

Well, taking  j = p  we have either:

• s_p-1 < s_p < s_p+1   or
• s_p-1 > s_p > s_p+1.

Accordingly, the second alternative cannot hold. Therefore the first alternative holds. Since the first alternative includes the inequality   s_p < s_p+1,   we can combine it with the inductive hypothesis to obtain:

Problem 40. Let's call the endpoints   P₀ = (x₁,x₂)   and   P₃ = (y₁,y₂).   One of the two trisection points is then:
      P₁ = (x₁,x₂) + ¹/₃((y₁,y₂) - (x₁,x₂)) = . . . = ²/₃(x₁,x₂) + ¹/₃(y₁,y₂).
{One or two steps were omitted from the calculation.}
If we do a similar calculation, starting with the coefficient   ²/₃   instead, then we find the other trisection point, namely:
      P₂ = ¹/₃(x₁,x₂) + ²/₃(y₁,y₂).
Now, these solutions look fairly "self-evident", but for completeness we should check that all 3 segments have the same length, namely that:
      ||P_i - P_i-1|| = ¹/₃((y₁ - x₁)² + (y₂ - x₂)²)^1/2   for   i = 1, 2, 3.  
{The computations involved in doing this may be mildly tedious, but they are reasonably straightforward compared to some other things we've done lately.}