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\centerline
{
\bf Solutions to the ``QUIZ'' for Lecture 1
}

{\bf Version of Dec. 17, 2017} (Thanks to Sarah Law)

1. Show that the triangle with vertices $P=(1,0,0)$, $Q=(0,1,0)$, and
$R=(0,0,1)$ is an equilateral triangle.
\bigskip

{\bf Solution}: 
$$
dist(P,Q)=\sqrt{(1-0)^2+(0-1)^2+(0-0)^2}=\sqrt{2} \quad ,
$$
$$
dist(P,R)=\sqrt{(1-0)^2+(0-0)^2+(0-1)^2}=\sqrt{2} \quad ,
$$
$$
dist(Q,R)=\sqrt{(0-0)^2+(0-1)^2+(1-0)^2}=\sqrt{2} \quad .
$$
Since the three sides of triangle PQR all have the same
lengths, it follows that the triangle is equilateral.


2. Determine whether the following two lines ever meet.
If they do meet, where?
$$
{\bf r}_1(t)= \langle 1,0,0 \rangle + t \langle 1,2,3 \rangle  \quad ,
\quad 
{\bf r}_2(t)= \langle 0,1,0 \rangle + t \langle 2,1,3 \rangle \quad .
$$

{\bf Solution:} The very {\bf first} step is to use another symbol,
say $s$, for 
${\bf r}_2(t)= \langle 0,1,0 \rangle + t \langle 2,1,3 \rangle$,
getting
$$
{\bf r}_1(t)= \langle 1,0,0 \rangle + t \langle 1,2,3 \rangle  \quad ,
\quad 
{\bf r}_2(s)= \langle 0,1,0 \rangle + s \langle 2,1,3 \rangle \quad .
$$
Spelling it out, we have
$$
{\bf r}_1(t)= \langle 1+t,2t,3t \rangle \quad ,
\quad 
{\bf r}_2(s)= \langle 2s,1+s,3s \rangle \quad .
$$
Setting ${\bf r}_1(t)={\bf r}_2(s)$, by equating respective
components we have {\bf three} equations and {\bf two} unkowns
($s$ and $t$):
$$
1+t =2s \quad , \quad 2t=1+s \quad , \quad 3t=3s \quad .
$$
From the third equation we get $t=s$, plugging into the first gives
$1+t=2t$ so $1=t$, and so $s=1$, and plugging into the second equation
we get $2=2$ which is correct, so we found a solution and this
means that these two lines {\bf meet}. 

To find {\bf where} they meet 
you plug-in $t=1$ into 
$$
{\bf r}_1(t)= \langle 1+t,2t,3t \rangle
$$
getting 
$$
{\bf r}_1(1)= \langle 1+1,2\cdot 1,3\cdot 1 \rangle=\langle 2,2 ,3 \rangle
\quad .
$$
To be on the safe side you should plug-in $s=1$ into
$$
{\bf r}_2(s)= \langle 2s,1+s,3s \rangle \quad ,
$$
 getting
$$
{\bf r}_2(1)= \langle 2,2,3 \rangle \quad ,
$$
which is indeed the same. We are almost done! 

But beware! They do not meet at $\langle 2,2,3 \rangle$, this
is nonsense, since they must meet at a point. The {\bf final} step
is to convert the vector $\langle 2,2,3 \rangle$ into the point
$(2,2,3)$ (which is the tip of the
arrow of the vector (if it starts at the origin)).

{\bf Ans.:} The two lines do intersect each other, and the
intersection point is $(2,2,3)$.

{\bf Comments:} To my emarassment I did it this kind of problem in the lecture.
People who followed my "method" got the right answer by luck.

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