Transcript The Geometry of a Tetrahedron

The Geometry of a
Tetrahedron
Footnote 18:Section 10.4
Mark Jeng
Professor Brewer
What is a tetrahedron?
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A tetrahedron is a solid
with 4 vertices: P, Q, R,
and S.
There are also 4
triangular faces
opposite the vertices as
shown in the figure.
Problem 1
1. Let v1, v2, v3, and v4 be vectors with lengths
equal to the areas of the face opposite the
vertices P, Q, R, and S, respectively, and
direction perpendicular to the respective
faces and pointing outward. Show that:
v1 + v2 + v3 + v4 = 0
Setting up the vectors
The area of a triangle
is: ½|a x b|
|v1| = ½ |QS x SR|
|v2| = ½ |PS x PR|
|v3| = ½ |PS x PQ|
|v4| = ½ |PQ x PR|
Vector 1
|v1| = ½
= ½ |i(0) – j(0) + k(-qr)|
= ½ |<0,0,-qr>|
= ½ qr
The direction of the vector is pointing outward
from the xy-axis, so therefore the final vector
is:
v1 = <0,0,-½ qr>
Vector 2
|v2| = ½
= ½ |i(pr) – j(0) + k(0)|
= ½ |<pr,0,0>|
= ½ pr
The direction of the vector is pointing outward
from the yz-axis, so therefore the final vector
is:
v2 = <-½ pr ,0,0>
Vector 3
|v3| = ½
= ½ |i(0) – j(pq) + k(0)|
= ½ |<0,-pq,0>|
= ½ pq
The direction of the vector is pointing outward
from the xz-axis, so therefore the final vector
is:
v3 = <0,-½ pq,0>
Vector 4
|v4| = ½
= ½ |i(-pr) – j(-pq) + k(qr)|
= ½ |<-pr,pq,qr>|
= ½ √[(-pr)2 + (pq)2 + (qr)2]
The vector is pointing outward in the +x, +y, and
+z directions perpendicular to the plane PQR;
therefore the final vector is:
v4 = <½ pr,½ pq,½ qr>
Question 1 Results
Show that: v1 + v2 + v3 + v4 = 0
<0,0,-½ qr> + <-½ pr ,0,0> + <0,-½ pq,0> +
<½ pr,½ pq,½ qr> = 0
The x, y, and z components of the vectors
all cancel out
Problem 2
2. The volume V of a tetrahedron is 1/3 the
distance from a vertex to the opposite face,
times the area of that face.
(a) Find a formula for the volume of a
tetrahedron in terms of the coordinates
of its vertices P, Q, R, and S.
(b) Find the volume of the tetrahedron
whose vertices are P(1,1,1), Q(1,2,3),
R(1,1,2), and S(3,-1,2).
Formula for Tetrahedron Volume
(a) The volume of a tetrahedron is:
V = 1/3 d[(s1, s2, s3), PlanePQR] * ½ |PQ x PR|
where (s1, s2, s3) are the x, y, and z
components of point S, the vertex
opposite PlanePQR
Finding the Volume
(b) Find the volume using the formula:
V = 1/3 d[(s1, s2, s3), PlanePQR] * ½ |PQ x PR|
with points P(1,1,1), Q(1,2,3), R(1,1,2), and
S(3,-1,2).
The first step is to derive the vectors PQ and PR:
vPQ = <1-1, 2-1, 3-1> = <0,1,2>
vPR = <1-1, 1-1, 2-1> = <0,0,1>
Finding the Volume (cont.)
Using those 2 vectors, find the Area of that
face: ½ |PQ x PR|
= ½ |i(1) – j(0) – k(0)|
½
= ½ |<1,0,0>|
= ½ √(12 + 02 + 02)
=½
Finding the Volume (cont.)
Then, using those same 2 vectors and point
P(1,1,1), find the equation of the PlanePQR.
In the previous slide, the cross product of those
2 vectors was determined: <1,0,0>
The general equation of a plane is:
A(x-x0) + B(y-y0) + C(z-z0) = 0 where <A,B,C>
is <1,0,0> and <x0,y0,z0> is <1,1,1>
The equation of PlanePQR is then:1(x-1) = 0
Finding the Volume (cont.)
The formula for the distance between a point
and a plane is:
D = |Ax1 + By1 + Cz1 + D| / √(A2 + B2 + C2)
Using the equation of the plane, x-1 = 0, and
the point S(3,-1,2), the distance will be:
D = |1(3) + 0(-1) + 0(2) + (-1)| / √(1 + 0 + 0)
=2
Finding the Volume (cont.)
Now, going back to the volume of a tetrahedron
formula:
V = 1/3 d[(s1, s2, s3), PlanePQR] * ½ |PQ x PR|
Plug in all the components
V = 1/3*(2)*(1/2)
= 1/3
Problem 3
3. Suppose the tetrahedron in the figure has a
tri-rectangular vertex S. (This means that the
3 angles at S are all right angles.) Let A, B,
and C be the areas of the 3 faces that meet
at S, and let D be the area of the opposite
face PQR. Using the result of Problem 1,
show that:
D2 = A2 + B2 + C2
3-D Pythagorean Theorem
Going back to the results of Problem 1, the
areas of the faces are:
A = ½ qr
B = ½ pr
C = ½ pq
D = ½ √[(-pr)2 + (pq)2 + (qr)2]
3-D Pythagorean Theorem (cont.)
Therefore, D2 = A2 + B2 + C2:
(½ √[(-pr)2 + (pq)2 + (qr)2])2 =
(½ qr)2 + (½ pr)2 + (½ pq)2
¼[(pr)2 + (pq)2 + (qr)2] = ¼ (qr)2 + ¼ (pr)2 + (pq)2
The 3-D Pythagorean Theorem is verified