Why was ∠ BAC = θ given in this problem?

[Jbreezy]

Yeah OK I see. That property only applies to one at a time then no?

 

[LCKurtz]

Yes, that's the one. So how would you write the answer to question 5? [(a+d)bc]= ? And what about [a(b+d)c]?

I did this for the problem starting from the assumption and working backwards you get [ (a-c)bd] + [(b-d)ac]

Do you see that this all goes back to the question I re-quoted above that you have never answered?

 

[Jbreezy]

Yes. I see. I did answer it last post. I will write out the proof tomorrow.

By the way the text had none of this!

 

[Jbreezy]

Prove that
[abd] + [bac] + [cdb] + [dca] = 0

Since c-a, b-a, d-a are coplanar the tripple scalar product of thease vectors must be 0.

(b-a) = u and (d-a) = v

So,


\begin{bmatrix} Cx-Ax &amp; Cy-Ay &amp; Cz-Az \\ Ux &amp; Uy &amp; Uz\\ Vx &amp; Vy &amp; Vz\end{bmatrix} \<br /> = \begin{bmatrix} Cx &amp; Cy &amp; Cz \\ Ux &amp; Uy &amp; Uz\\ Vx &amp; Vy &amp; Vz\end{bmatrix} +\begin{bmatrix} Ax &amp; Ay &amp; Az \\ Ux &amp; Uy &amp; Uz\\ Vx &amp; Vy &amp; Vz\end{bmatrix}<br /> and \begin{bmatrix} Ax &amp; Ay &amp; Az \\ (Bx-Ax) &amp; (By-Ay) &amp; (Bz-Az)\\ Vx &amp; Vy &amp; Vz\end{bmatrix} =\begin{bmatrix} Ax &amp; Ay &amp; Az \\ (Bx) &amp; (By) &amp; (Bz)\\ Vx &amp; Vy &amp; Vz\end{bmatrix}+\begin{bmatrix} Ax &amp; Ay &amp; Az \\ (Ax) &amp; (Ay) &amp; (Az)\\ Vx &amp; Vy &amp; Vz\end{bmatrix}
The last determinant is 0 because of the same two row. And repeat the unpacking process and you get
[abd] + [bac] + [cdb] + [dca] = 0

 

[LCKurtz]

Yes, that's the one. So how would you write the answer to question 5? [(a+d)bc]= ? And what about [a(b+d)c]?

Yes. I see. I did answer it last post.

No you didn't. And until you do, I am done with this thread. Give the answer in terms of box products.

Prove that
[abd] + [bac] + [cdb] + [dca] = 0

Since c-a, b-a, d-a are coplanar the tripple scalar product of thease vectors must be 0.

(b-a) = u and (d-a) = v

So,


\begin{bmatrix} Cx-Ax &amp; Cy-Ay &amp; Cz-Az \\ Ux &amp; Uy &amp; Uz\\ Vx &amp; Vy &amp; Vz\end{bmatrix} \<br /> = \begin{bmatrix} Cx &amp; Cy &amp; Cz \\ Ux &amp; Uy &amp; Uz\\ Vx &amp; Vy &amp; Vz\end{bmatrix} +\begin{bmatrix} Ax &amp; Ay &amp; Az \\ Ux &amp; Uy &amp; Uz\\ Vx &amp; Vy &amp; Vz\end{bmatrix}

You have a mistake in that first equation. Start with a box product and use the formula I have been trying to get from you. You can do the whole thing with box products. You don't need to write out all the determinants once you have the box product formulas. And to say that you get the right answer after working it out isn't convincing.

 

[Jbreezy]

Prove that
[abd] + [bac] + [cdb] + [dca] = 0

(b-a) = u, (d-a) = v
[(c-a)(b-a)(d-a)] = [(c-a)(u)(v)] = [cuv] - [auv] = [c(b-a)v] - [a(b-a)v]

= [cbv]-[cav] - [abv]+ [aav]

= [cb(d-a)] - [ca(d-a)] -[ab(d-a)]

= [cbd] - [cba] +[caa] -[cad] - [abd] + [aba]

= [cbd] - [cba] - [cad] - [abd] (rearrange the first term to -[cdb])

So,
-[cdb] - [cba] - [cad] - [abd] = 0

thus,

[abd] + [bac] + [dca] +[abd] = 0

 

[LCKurtz]

Prove that
[abd] + [bac] + [cdb] + [dca] = 0

(b-a) = u, (d-a) = v
0 = [(c-a)(b-a)(d-a)] = [(c-a)(u)(v)] = [cuv] - [auv] = [c(b-a)v] - [a(b-a)v]

= [cbv]-[cav] - [abv]+ [aav]

= [cb(d-a)] - [ca(d-a)] -[ab(d-a)]

= [cbd] - [cba] +[caa] -[cad] - [abd] + [aba]

= [cbd] - [cba] - [cad] - [abd] (rearrange the first term to -[cdb])

So,
-[cdb] - [cba] - [cad] - [abd] = 0

thus,

[abd] + [bac] + [dca] +[abd] = 0

Look's like you have it. See how easy that was? Don't forget you want to start with the original box product = 0 so the end result is also.

 

[Jbreezy]

Oh yeah it was just great that the note I have for this included nothing of the rules. This was such a lovely experience. Once again fail to put something together myself.