Homework 2 Solutions
Calculus III
Summer 2014
1. Consider the four points: (−1, −4, 1), (−1, 1, 2), (1, −1, 0), (2, 1, −1).
(a) Draw the shape that has these four points as vertices.
(b) Find the surface area of this shape.
(c) Reflect the point (−1, −4, 1) across the plane defined by the points
(1, −1, 0), (2, 1, −1), (−1, 1, 2).
(d) What is the surface area of the new shape?
Solution 1. (a) Define P1 = (−1, −4, 1), P2 = (−1, 1, 2), P3 = (1, −1, 0),
P4 = (2, 1, −1).
-4
y
-2
0
P2
P1
2
1
z
P3
0
-1
-1
0
P4
1
2
x
(b) The surface area can be found by adding the areas of each of the
4 faces:
1 −−→ −−→ −−→ −−→ −−→ −−→ −−→ −−→
A = (|P1 P2 ×P1 P3 |+|P1 P3 ×P1 P4 |+|P1 P2 ×P1 P4 |+|P2 P3 ×P2 P3 |)
2
1
−−→ −−→
|P1 P2 × P1 P3 | = −−→ −−→
|P1 P3 × P1 P4 | = −−→ −−→
|P1 P2 × P1 P4 | = −−→ −−→
|P2 P3 × P3 P4 | = √
= | − 8i + j − 13k| = 168
j k √
3 −1 = | − i + j + k| = 3
5 −2 j k √
5 1 = | − 15i + j − 20k| = 459
5 −2 j
k √
−2 −2 = |6i + 6k| = 72
0 −3 i j k
0 5 1
2 3 −1
i
2
3
i
0
3
i
2
3
√
√
√
1 √
A = ( 168 + 3 + 459 + 72) ≈ 22.3
2
−−→ −−→
(c) From (b) we found that P2 P3 × P2 P4 = h6, 0, 6i, so the plane has
normal vector h1, 0, 1i. Using that P3 is a point in the plane:
0 = hx − 1, y + 1, zi · h1, 0, 1i = x − 1 + z.
So the equation of the plane is x + z = 1. The line that passes
through P1 and is parallel to the normal is
hx(t), y(t), z(t)i = h−1 + t, −4, 1 + ti
The reflected point Pr must lie on this line on the opposite side
of the plane and be the same distance from the plane as P1 :
y
2
0
-2
-4
-6
3
P2
P1 reflected
2
P1
1
P3
P4
0
-1
-1
-2
1
0
x
2
2
z
Plugging the parametrization into the plane equation, we see that
the line intersects the plane when x(t) + z(t) = −1 + t + 1 + t =
1 ⇒ t = 21 , so the reflected point is at t = 1:
Pr = hx(1), y(1), z(1)i = h0, −4, 2i.
2. Consider a room ten units long in the x, y, and z directions. Specifically,
the walls of the room are the four planes x = 0, x = 10, y = 0, and
y = 10, and the floor and ceiling are z = 0 and z = 10, respectively.
A flat triangular mirror is mounted in one of the corners of the ceiling.
The corners of the mirror are at locations (10, 9, 10), (10, 10, 9), and
(9, 10, 10). You are sitting at the location (5, 0, 0) playing with your
new green laser pointer.
If you aim your laser pointer directly at the corner of the room with
coordinates (10, 10, 10), determine the coordinates where the beam will
hit the walls, or floor, of the room. (Hint: an incoming ray of light, and
the surface normal where the ray hits the surface, form a plane. The
reflected ray is in the same plane. Also, the angle between the incoming
ray and the normal is the same as the angle between the normal and
the reflected ray.)
Solution 2. Define hx(t), y(t), z(t)i and hu(t), v(t), w(t)i to be the
parametric equations of the Laser pointer line before and after striking the mirror respectively. (10, 10, 10) − (5, 0, 0) = h5, 10, 10i ⇒ L =
h1, 2, 2i is the vector which describes the direction of the Laser pointer
going towards the mirror, so we have that
hx(t), y(t), z(t)i = h5 + t, 2t, 2ti.
(10, 10, 9)−(10, 9, 10) = h0, 1, −1i and (10, 10, 9)−(9, 10, 10) = h1, 0, −1i
are two vectors in the mirror’s plane. h0, 1, −1i×h1, 0, −1i = h−1, −1, −1i =
n is the normal vector of the mirror. Using that (10, 10, 9) is a point
on the mirror:
0 = hx − 10, y − 10, z − 9i · h−1, −1, −1i = −x − y − z + 19.
The equation of the mirror is x + y + z = 29. Substituting the first
.
laser line equation: −x(t) − y(t) − z(t) = 5 + t + 2t + 2t = 29 ⇒ t = 24
5
The laser hits the mirror at
3
24
24
24
49 48 48
x
,y
,z
=
, ,
.
5
5
5
5 5 5
Now we need the direction the laser reflects off the mirror. Define R to
be the vector in the direction of the reflected laser. Consider a zoom
in of the corner of the room:
The vector N = −L·n
n is the projection of −L onto n. Define the
|n|2
vector P = −L − N such that R = N − P. Putting this together,
R = L + 2N, so that
h1, 2, 2i · h−1, −1, −1i
h−1, −1, −1i
3
7 4 4
= h− , − , − i
3 3 3
R = h1, 2, 2i − 2
Since we only care about the direction of the vector, let’s redefine R =
h−7, −4, −4i. We can use R and the point the laser hits the mirror to
parametrize the reflected path:
hu(t), v(t), w(t)i = h
49
48
48
− 7t,
− 4t,
− 4ti
5
5
5
4
To see whether the laser hits the floor, the x-axis wall, or the y-axis
wall, set each component to 0 and solve for the other coordinates:
Floor:
12
12
−
4t
=
0
⇒
t
=
.
But
x
= −7 which is
Set w(t) = 0 ⇒ 48
5
5
5
outside of the room, so the laser does not hit the floor.
x-axis wall:
Set v(t) = 0 ⇒ t =
along the x-axis.
12
5
⇒u
12
5
= −7. The Laser does not hit the wall
y-axis wall:
Set u(t) = 0 ⇒ t =
7
5
⇒v
7
5
=w
7
5
= 4.
The laser hits the wall along the y-axis at (0, 4, 4)!
5