Vectors in R3 ---- how to approach this problem

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October 10th 2010, 04:41 PM
nikcs123

Vectors in R3 ---- how to approach this problem

(a) Given a general (not necessarily rectangular) tetrahedron, let $V_{1}, V_{2}, V_{3}, V_{4}$ denote vectors whose lengths are equal to the areas of the four faces, and whose directions are perpendicular to the these faces and point outward. Show that $V_{1} + V_{2} + V_{3} + V_{4} = 0$ .

(b) Now consider a rectangular tetrahedron, i.e., one which has one vertex at which all three angles are right angles. Let D denote the area of the face opposite this rectangular vertex (i.e., the analog of a hypothenuse), and let A;B;C denote the areas of the other three faces. Prove that $D^2 = A^2 + B^2 + C^2$ .
October 10th 2010, 05:15 PM
bondesan

The approach I'm thinking for a) would be associate each edge with a vector. If you cross product two adjacent vectors, suppose A and B, you will get a vector, lets call it C, which is perpendicular to the face supported by those two vectors. You can find the area by taking |C|/2. Other edges will be a subtraction of two other known vectors, so maybe this will help to prove the asked identity.
October 10th 2010, 06:37 PM
nikcs123

Thanks bondesan.

http://nrich.maths.org/content/01/09/six6/pic6.gif

So I have tetrahedron ABCD (ignore PQSR in the picture, only decent picture i could find).

So in order to define vectors $V_{1}$ , $V_{2}$ , $V_{3}$ , $V_{4}$ , I start with one of the sides, lets use side ABC.

To find a vector normal to side ABC, I take the cross product (AB x AC) of 2 edges. That produces a vector with a direction normal and a magnitude of 2x the area of ABC (the area of a parallelogram with sides AB and AC). So then $V_{1} = \frac{AB \times AC}{2}$ .

Analogously for the other 3 sides, it can be found that:
$V_{2} = \frac{AB \times AD}{2}$ .
$V_{3} = \frac{AD \times AC}{2}$ .
$V_{4} = \frac{BC \times BD}{2}$ .

Is that correct? At this point I'm still confused as to how to prove that the sum of those 4 vectors is equal to 0.

EDIT: from here the denominators can be removed and the equation can be simplified to: (AB x AC) + (AB x AD) + (AD x AC) + (BC x BD) = 0.
October 10th 2010, 07:25 PM
nikcs123

As for part (b), I took a different tetrahedron (one with B as the "rectangular vertex") so D is the area of side ACD.

Using similar properties, I found that $D = \frac{\parallel AC \times AD\parallel}{2}$ , and then $A = \frac{\parallel AB \times AC\parallel}{2}$ , $B = \frac{\parallel AB \times AD\parallel}{2}$ , $C = \frac{\parallel BC \times BD\parallel}{2}$ .

So then from here I have to prove that:
$(\frac{\parallel AC \times AD\parallel}{2})^2 = (\frac{\parallel AB \times AC\parallel}{2})^2 + (\frac{\parallel AB \times AD\parallel}{2})^2 + (\frac{\parallel BC \times BD\parallel}{2})^2$

Simplifying, $(\parallel AC \times AD\parallel)^2 = (\parallel AB \times AC\parallel)^2 + (\parallel AB \times AD\parallel)^2 + (\parallel BC \times BD\parallel)^2$

stuck at this point
October 11th 2010, 10:17 AM
bondesan

Quote:

Originally Posted by nikcs123

http://nrich.maths.org/content/01/09/six6/pic6.gif

Is that correct? At this point I'm still confused as to how to prove that the sum of those 4 vectors is equal to 0.

EDIT: from here the denominators can be removed and the equation can be simplified to: (AB x AC) + (AB x AD) + (AD x AC) + (BC x BD) = 0.

I think you are correct. You must notice that, for instance, BC can be written in terms of other vectors: BC = AC - AB - and this will help you to simplify the whole thing, remembering some properties of the cross product ( AxB = -BxA, ...) and so on.

For b) maybe you should use ||AxB|| = |A|.|B|.sin t where t is the angle between A and B. You will have a lots of sin(pi/2) = 1.

Regards
October 11th 2010, 08:31 PM
nikcs123

Thanks again.

I've managed to completely go through and prove part a.

As for part b, I redefined my vectors as $D = \frac{\parallel AC \times AD\parallel}{2}$ , $A = \frac{\parallel AB \times BC\parallel}{2}$ , $B = \frac{\parallel AB \times BD\parallel}{2}$ , $C = \frac{\parallel BC \times BD\parallel}{2}$ .

So then:
$(\frac{\parallel AC \times AD\parallel}{2})^2 = (\frac{\parallel AB \times BC\parallel}{2})^2 + (\frac{\parallel AB \times BD\parallel}{2})^2 + (\frac{\parallel BC \times BD\parallel}{2})^2$

$(\parallel AC \times AD\parallel)^2 = (\parallel AB \times BC\parallel)^2 + (\parallel AB \times BD\parallel)^2 + (\parallel BC \times BD\parallel)^2$

Using the fact that magnitude of a cross product is magnitude of the vectors times the angle in between, I then find that:

$(\parallel AC \parallel \parallel AD \parallel sin(CAD))^2 = (\parallel AB \parallel \parallel BC \parallel sin(ABC))^2 + (\parallel AB \parallel \parallel BD \parallel sin(ABD))^2 + (\parallel BC \parallel \parallel BD \parallel sin(DBC))^2$

Where CAD, ABC, ABD, DBC are the angles between their respective vectors. Based on the properties of a rectangular tetrahedron, angles ABC, ABD, DBC are all pi/2, so the sin of those is equal to 1. Leaves me with:

$(\parallel AC \parallel \parallel AD \parallel sin(CAD))^2 = \parallel AB \parallel ^2 \parallel BC \parallel ^2 + \parallel AB \parallel ^2 \parallel BD \parallel ^2 + \parallel BC \parallel ^2 \parallel BD \parallel ^2$

Kind of stuck at this point, don't know where to take it from here.
October 11th 2010, 10:02 PM
xxp9

a) let A, B, C, D be the 4 vertex. For the face ABC, its area vector is
(AB x AC)/2, so the question is to show that 0 = \sum (AB x AC), iterating all 4 faces.
Note that AB x AC = (A-B) x (A-C) = B x C - B x A - A x C = A x B + B x C + C x A
adding all the 4 equations up, you will get a 0, since for each edge, like AB, there are two faces( e.g ABC and ABD) adjointed, each face contributes a term A x B, but with opposite signs.

b) is a consequence of a) and the Pythagoras theorem
October 11th 2010, 10:24 PM
nikcs123

Quote:

Originally Posted by xxp9

b) is a consequence of a) and the Pythagoras theorem

Makes sense, but how does one prove that?
October 11th 2010, 10:44 PM
xxp9

Quote:

Originally Posted by nikcs123

Makes sense, but how does one prove that?

from a) V1 + V2 + V3 + V4 = 0, but now V1, V2 and V3 are orthogonal now, according to the Pythagoras theorem, we have |V1+V2+V3|^2 = |V1|^2+|V2|^2+|V3|^2. DONE