# Thread: Vectors in R3 ---- how to approach this problem

1. ## 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 $D^2 = A^2 + B^2 + C^2$.

2. 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.

3. Thanks bondesan.

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.

4. 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

5. Originally Posted by nikcs123

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

6. 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.

7. 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

8. Originally Posted by xxp9
b) is a consequence of a) and the Pythagoras theorem
Makes sense, but how does one prove that?

9. 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