I feel like an idiot....

ED E

Four months of holidays have killed my brain, now I cant remember any of my co-ord:(

Using the formula for the volume of a tetrahedron:

But I'm unsure as to how one subtracts xyz co-ordinates?

If somebody can explain(and make me feel like a complete tool:p), that'd be great.

Cheers,
LB.

Fremen

I'm not sure if this is what you're getting at, but for example (a1,a2,a3) - (b1,b2,b3) = (a1-b1,a2-b2,a3-b3), i.e. you subtract coordinatewise.

sponsoredwalk

Subtract xyz coordinates? This formula is asking you to compute the
cross product between (b - d) x (c - d) = e and then to take the dot
product (a - d) • e do you know what these are?

ED E

I definitely should know Dot Prod as we covered that for LC. Dunno about cross product, but that just be another memory fail.

Define your two vectors. Vector1 (x1, y1, z1) and Vector2 (x2, y2, z2).

DotProduct = (x1*x2 + y1*y2 + z1*z2)

Is that it for Dot?

( v1, v2) = [ y1*z2 - y2*z1 , z1*x2 - z2*x1 , x1*y2 - x2*y1 ]

And this for Cross?

I'll try it and see after lunch anyways.

sponsoredwalk

Ah yeah that'd make sense. I'll give you some videos to learn about the
cross product but basically it's an operation of vectors in 3-dimensional
space that produces another vector. Remember the dot product produces
a scalar, i.e. a single number as the result, well the cross product
produces another vector.
If you look at the formula you've been given you're asked to first compute
the cross product so that once you find the new vector (b - d) x (c - d) = e
you can then take the dot product and get a single number result.
The | | signs just mean to take the absolute value.

Another thing to be conscious of is that the absolute value of cross product
gives you the area of the parallelogram between the two vectors.
What this means is that whatever result (b - d) x (c - d) = e gives you if
you take the absolute value of |e|, i.e. take the magnitude of the
vector you'll get the area in between the two vectors. Think of this as
the base of a 3-dimensional box, then the 3'rd vector you take the dot
product with acts as if you're multiplying the base of the box times the
height to get the volume.

re-read what I've said after watching the videos:

https://www.youtube.com/watch?v=KDHuWxy53uM

https://www.youtube.com/watch?v=zA0fvwtvgvA

(2 videos).

Then, once you totally understand the cross product as shown in these
videos and in your book etc... read this thread to see where
the bloody thing comes from and why doing such weird things like
i x j = k etc... makes sense.

ED E

Thanks for the help.

This actually a Java exercise, so the understanding is secondary to the ability to write the code to execute it. Think I get it though.

Thanks again. Extra credit time;)

ED E

I got 5 on paper. Program says 5.16 recurring.

sponsoredwalk

I can't check it, I don't know what a, b, c or d are equal to numerically...
I'm sure they are all written in one of the following forms:

[latex] \overline{a} \ = \ a_x \ \hat{i} \ + \ a_y \ \hat{j} \ + \ a_z \ \hat{k} \ = \ < a_x,a_y,a_z > \ = \ (a_x,a_y,a_z) [/latex]

Basically if you followed the method properly I'd trust your evaluation
rather than the computer, but whether or not you did it properly is a
different story

ED E

sponsoredwalk wrote: »

I can't check it, I don't know what a, b, c or d are equal to numerically...
I'm sure they are all written in one of the following forms:

[latex] \overline{a} \ = \ a_x \ \hat{i} \ + \ a_y \ \hat{j} \ + \ a_z \ \hat{k} \ = \ < a_x,a_y,a_z > \ = \ (a_x,a_y,a_z) [/latex]

Basically if you followed the method properly I'd trust your evaluation
rather than the computer, but whether or not you did it properly is a
different story

I used a simple values so that it'd be easy to do, AKA base on the horizontal plane.
(1,0,1)(4,0,1)(3,0,3)(2,5,2)

sponsoredwalk

I'll assume this is what you mean:
a = (1,0,1) = 1i + 0j + 1k
b = (4,0,1) = 4i + 0j + 1k
c = (3,0,3) = 3i + 0j + 3k
d = (2,5,2) = 2i + 5j + 2k

b - d = (4 - 2)i + (0 - 5)j + (1 - 2)k = 2i - 5j - 1k

c - d = (3 - 2)i + (0 - 5)j + (3 - 2)k = 1i - 5j + 1k

a - d = (1 - 2)i + (0 - 5)j + (1 - 2)k = - 1i - 5j - 1k

(b - d) x (c - d) = (a determinant which I can't write in latex notation).
Basically watch the cross product video to see how to do a determinant
with i,k,j in the top row, 2 - 5 - 1, in the second row and 1 -5 1 in the
bottom row. You'll get:

(b - d) x (c - d) = -10i - 3j -5k

(a - d)•((b - d) x (c - d)) = (- 1i - 5j - 1k)•(-10i - 3j -5k)

(a - d)•((b - d) x (c - d)) = 10 + 15 + 5 = 30

[latex] = \ \frac{|30|}{6} \ = \ 5[/latex]

Unless I've committed an error the volume contained in these vectors
is 5 units³.

ED E

Yeah, that agrees with V=1/3(A)(h).


Pre division you get 30, which I agree you should, but my program is reading 31 for J(that value).

EDIT: Getting -10 -3 -6 instead of -10 -3 -5.........

EDIT2: "- x6-y5;" Should be x6*y5;

And it works. Thanks a million to you all. I have an excuse too, I was multi-tasking, something that should really be left to the opposite gender.