18.02 Fall 2008 – Problem Set 3, Part B Solutions
1.
a) The line from
E
to
P
has the direction of
EP
=
(
x
0
−
1
,y
0
,z
0
)
and goes through
(
1
,
0
,
0
)
, so it can be parametrized as
x
= 1 + (
x
0
−
1)
t,
y
=
y
0
t,
z
=
z
0
t.
It intersects the
yz
-plane when
x
= 0, i.e. 1 + (
x
0
−
1)
t
= 0, which gives
t
=
1
1
−
x
0
.
So
y
=
y
0
1
−
x
0
,
z
=
z
0
1
−
x
0
.
(We assume
x
0
<
1 because otherwise the point
P
would lie behind the observer).
b) The image on the screen of a line segment in space is contained in the intersection of
the plane containing
E
and the line segment with the
yz
-plane. This intersection is a line,
therefore the image is a line segment on the screen.
c) Using question a), (0
,
−
3
,
1) is displayed at (
−
3
,
1) in the
yz
-plane, and (
−
1
,
2
,
5) is
displayed at (1
,
5
2
). So the image on the screen is the line segment from (
−
3
,
1) to (1
,
5
2
).
d) The trajectory will again be a line segment. The velocity of the bird is
vectorv
=
−−−→
P
0
P
1
=
(−
1
,
5
,
4
)
, so at time
t
its position is (
−
t,
−
3 + 5
t,
1 + 4
t
).
By the formula in a), this is
displayed at
y
=
5
t
−
3
t
+ 1
=
5
−
3
t
1 +
1
t
, z
=
4
t
+ 1
t
+ 1
=
4 +
1
t
1 +
1
t
.
Taking the limit as
t
→ ∞
, the position on the screen approaches (5
,
4) (even though the
actual position of the bird in 3D space is further and further away).