#Attendance Quiz for Lecture 1 

#PROBLEM 1 (used maple for this problem)
> #Step 1: Setting up the function to find the distance between the verticies of the triangle.

> Dist := (p1, p2) -> sqrt((p2[1] - p1[1])^2 + (p2[2] - p1[2])^2 + (p2[3] - p1[3])^2);

> #Step 2: Setting the 3 verticies to their according coordinates.
> P := [1, 0, 0];
> Q := [0, 1, 0];
> R := [0, 0, 1];
                         P := [1, 0, 0]
                         Q := [0, 1, 0]
                         R := [0, 0, 1]

> #Step 3: Using the function and verticies defined above to find the distances between the 3 verticies of the triangle.
> Dist(P, Q);
                              2^(1/2)     
> Dist(P, R);
                              2^(1/2)   
> Dist(Q, R);
                             2^(1/2) 

> {Dist(P, Q), Dist(P, R), Dist(Q, R)};
                            {2^(1/2)}


> #All of the distances between the 3 verticies of the triangle are equal to each other and therefore, the triangle is an equilateral triangle.


#PROBLEM 2 (did not use maple, had trouble with it for this problem)

#Step 1: Assign a different variable, f, for r1(t), and keep r2(t) the same. Also, simplifying the components.

r1(f) = [1,0,0] + f[1,2,3] = [1+f, 2f, 3f]
r2(t) = [0,1,0] + t[2,1,3] = [2t, 1+t, 3t]

#Step 2: Solving out the components by setting the two lines' components equal to each other and getting 3 equations.

a)   1+f = 2t
b)   2f = 1+t
c)   3f = 3t

#Step 3: Solving out the 3 equation to get the value for s and t.

#by considering equation c, we can see that: 3f = 3t --> f = t
#so, you can consider equation a (or b, I chose a) we can see that: 1+f = 2t --> 1+t = 2t --> t = 1.
#since, t = f, the value of f is also one: f = 1.

#step 4: Find the point at where they meet by plugging in the according values into r1(f) and r2(t).
r1(f) = r1(1) = (1+1, 2(1), 3(1)) = (2, 2, 3)
r2(t) = r2(1) = (2(1), 1+1, 3(1)) = (2, 2, 3)

#The two lines r1(t) and r2(t) meet at the point (2, 2, 3).