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\centerline
{
\bf Solutions to the ``QUIZ'' for Lecture 3
}
\bigskip
{\bf 1.}
Find an equation of the plane that passes through the points
$(0,1,1)$, $(1,0,1)$, $(1,1,0)$.

{\bf Sol.} Let's call $P=(0,1,1)$, $Q=(1,0,1)$, $R=(1,1,0)$.

We need {\bf two} direction vectors that lie on that plane. For example
$$
{\bf PQ} = \langle 1-0,0-1,1-1 \rangle = \langle 1, -1, 0 \rangle
$$
$$
{\bf PR} = \langle 1,0,-1 \rangle
$$
To get the {\bf normal} we take the {\bf cross-product} ${\bf PQ} \times {\bf PR}$.
$$
{\bf PQ} \times {\bf PR}=
$$
$$
\left \vert \matrix{ {\bf i} & {\bf j} & {\bf k}  \cr
1 & -1& 0\cr
1 & 0 & -1 
} \right \vert \quad=
$$
$$
{\bf i} \left \vert \matrix{ 
-1 & 0\cr
0 & -1 
} \right \vert 
-{\bf j} \left \vert \matrix{ 
1 & 0\cr
1 & -1 
} \right \vert 
+{\bf k} \left \vert \matrix{ 
1 & -1 \cr
1 & 0 
} \right \vert 
$$
$$
={\bf i} ((-1) \cdot (-1)-0 \cdot 0)
-{\bf j} (1 \cdot (-1)-0 \cdot 1)
+{\bf k} (1 \cdot 0-(-1) \cdot 1)=
$$
$$
{\bf i}+{\bf j}+{\bf k} \quad.
$$
{\bf Finally} converting to the {\bf usual} notation, we get
${\bf n}=\langle 1,1,1 \rangle$.
The equation of a general plane is
$$
a(x-x_0)+b(y-y_0)+c(z-z_0)=0 \quad,
$$
where $\langle a,b,c \rangle = \langle 1,1,1 \rangle$ is the normal vector we just found,
and $(x_0,y_0,z_0)$ is {\it any} point (either $P,Q$ or $R$). Picking $P$ we get
$$
1(x-0)+1(y-1)+1(z-1)=0 \quad,
$$
that simplies to
$$
x+y+z=2 \quad .
$$

{\bf Ans.}: $x+y+z=2$.

{\bf Comments}: About \%70 got it pefectly. About \%10 were clueless. The rest
made some calculational mistakes. Quite a few people got ${\bf n}=\langle 1,1,1 \rangle $
correctly, but messed up with the last part of plugging-in $(x_0,y_0,z_0)=(0,1,1,)$, and
got either $x+y+z=0$, or $x+y+z=3$ or other wrong things. Remember that you can always
{\bf check} your answer by plugging-in the three points and see that they lie on the planne.
For $P=(0,1,1)$: $0+1+1=2$,
for $Q=(1,0,1)$, $1+0+1=2$,
for $R=(1,1,0)$: $1+1+0=2$, they all agree. People who got, for example $x+y+z=3$ could have realized
their mistake by doing this checking.


{\bf 2.} Find the intersection of the line
$$
{\bf r}(t)=\langle 1,1,0 \rangle + t \langle 0,2,4 \rangle
$$
and the plane
$$
x+y+z=14 \quad .
$$

{\bf Solution:} First {\bf spell-out} ${\bf r}(t)$:
$$
{\bf r}(t)=\langle 1,1,0 \rangle + t \langle 0,2,4 \rangle =  \langle 1,1+2t,4t \rangle \quad,
$$
and in {\bf scalar} form
$$
x=1 \quad , \quad y=1+2t \quad , \quad z=4t \quad .
$$
Now plug these expressions for $x,y,z$ in terms of the parameter $t$ into the equation of the plane
$x+y+z=14$ getting
$$
1+(1+2t)+4t=14 \quad .
$$
Simplifying, we get
$$
2+6t=14 \quad ,
$$
so
$$
6t=12 \quad ,
$$
that gives $t=2$. Having found the lucky $t$ (namely $2$) you plug it in back into
$$
x=1 \quad , \quad y=1+2t \quad , \quad z=4t \quad ,
$$
getting
$$
x=1 \quad , \quad y=1+2 \cdot 2  =5, \quad z=4 \cdot 2 =8 \quad ,
$$
So the {\bf lucky point} that belongs both to the plane and the line is the point $(1,5,8)$.

{\bf Ans.}: The intersection of the line and the plane given by the problem is the point $(1,5,8)$ .


{\bf Comments:} I did not have a chance to do this type of problems in class (and it is also not in the handout), so very few people got it.
For those who did (about \%10), well done!

Many people got confused with the other type of problems, where you had to find a plane passing
through three points, or the direction number of the intersection of two planes.

{\bf Do not confuse problems!} It is better to admit that you don't know how to do it, then just use techniques applicable to other problems.
In particular, in this problem the cross-product is completely irrelevant. Not only is it not needed, there is no way to use it!


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