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The Geometry of a Tetrahedron

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The Geometry of a Tetrahedron Footnote 18:Section 10.4 Mark Jeng Professor Brewer What is a tetrahedron? A tetrahedron is a solid with 4 vertices: P, Q, R, and S. – PowerPoint PPT presentation

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Title: The Geometry of a Tetrahedron


1
The Geometry of a Tetrahedron
  • Footnote 18Section 10.4
  • Mark Jeng
  • Professor Brewer

2
What is a tetrahedron?
  • 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.

3
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

4
Setting up the vectors
  • The area of a triangle is ½a x b
  • v1 ½ QS x SR
  • v2 ½ PS x PR
  • v3 ½ PS x PQ
  • v4 ½ PQ x PR

5
Vector 1
  • ½ i(0) j(0) k(-qr)
  • v1 ½ ½ lt0,0,-qrgt
  • ½ qr
  • The direction of the vector is pointing outward
    from the xy-axis, so therefore the final vector
    is
  • v1 lt0,0,-½ qrgt

6
Vector 2
  • ½ i(pr) j(0) k(0)
  • v2 ½ ½ ltpr,0,0gt
  • ½ pr
  • The direction of the vector is pointing outward
    from the yz-axis, so therefore the final vector
    is
  • v2 lt-½ pr ,0,0gt

7
Vector 3
  • ½ i(0) j(pq) k(0)
  • v3 ½ ½ lt0,-pq,0gt
  • ½ pq
  • The direction of the vector is pointing outward
    from the xz-axis, so therefore the final vector
    is
  • v3 lt0,-½ pq,0gt

8
Vector 4
  • ½ i(-pr) j(-pq) k(qr)
  • v4 ½ ½ lt-pr,pq,qrgt
  • ½ v(-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 lt½ pr,½ pq,½ qrgt

9
Question 1 Results
  • Show that v1 v2 v3 v4 0
  • lt0,0,-½ qrgt lt-½ pr ,0,0gt lt0,-½ pq,0gt lt½
    pr,½ pq,½ qrgt 0
  • The x, y, and z components of the vectors all
    cancel out

10
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).

11
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

12
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 lt1-1, 2-1, 3-1gt lt0,1,2gt
  • vPR lt1-1, 1-1, 2-1gt lt0,0,1gt

13
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 lt1,0,0gt
  • ½ v(12 02 02)
  • ½

14
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 lt1,0,0gt
  • The general equation of a plane is
  • A(x-x0) B(y-y0) C(z-z0) 0 where ltA,B,Cgt is
    lt1,0,0gt and ltx0,y0,z0gt is lt1,1,1gt
  • The equation of PlanePQR is then1(x-1) 0

15
Finding the Volume (cont.)
  • The formula for the distance between a point and
    a plane is
  • D Ax1 By1 Cz1 D / v(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) / v(1 0 0)
  • 2

16
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

17
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

18
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 v(-pr)2 (pq)2 (qr)2

19
3-D Pythagorean Theorem (cont.)
  • Therefore, D2 A2 B2 C2
  • (½ v(-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
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