1.
Consider the points
P
= (1
,
2
,
3) and
Q
= (4
,
5
,
6) in
R
3
.
[6]
(a)
Find an equation for the plane that passes through the origin and contains the points
P
and
Q
. Express your answer in the form
ax
+
by
+
cz
=
d
.
If the normal vector to the plane is
n
= (
a, b, c
), then
n
·
P
= 0 and
n
·
Q
= 0
which gives the system
a
+ 2
b
+ 3
c
= 0
4
a
+ 5
b
+ 6
c
= 0
Solving this system gives
n
∈
span
{
(1
,

2
,
1)
}
.
Therefore the equation of the plane is
x

2
y
+
z
=
k
for some constant
k
. Since the plane passes through (0
,
0
,
0) we must
have
k
= 0. Scaling by 3, we get that the equation of the plane is
x

2
y
+
z
= 0.
[2]
(b)
Find the orthogonal projection of (

1
,
0
,
8) onto the normal vector to the plane in
part (a).
Using
n
= (1
,

2
,
1) for the normal vector of the plane, setting
v
= (

1
,
0
,
8), we
use the equation
proj
n
(
v
) =
n
·
v
k
n
k
n
=
(1
,

2
,
1)
·
(

1
,
0
,
8)
k
(1
,

2
,
1)
k
n
=

1+0+8
1+4+1
n
=
7
6
(1
,

2
,
1) = (
7
6
,

7
3
,
7
6
)
[2]
(c)
Find the distance from the point (

1
,
0
,
8) to the plane in part (a).
The distance from (

1
,
0
,
8) to the plane is the length of the projection onto the normal
direction of any vector
v
which connects a point on the plane to the point (

1
,
0
,
8).
Since the plane contains the origin, we can choose
v
= (

1
,
0
,
8) and compute the
projection as in part b). The distance is therefore
k
proj
n
(
v
)
k
=
k
7
6
(1
,

2
,
1)
k
=
7
6
k
(1
,

2
,
1)
k
=
7
6
√
1 + 4 + 1 =
7
√
6
6
2
of
7