Why was ∠ BAC = θ given in this problem?

[Jbreezy]

1. Does [abc] mean a⋅(b×c) or (a×b)⋅c , or does it matter?
2. What happens to the value of [abc] if you swap two of the vectors?
3. What happens to the value of [abc] it two of the vectors are the same?
4. What happens if you have a constant k times one of the vectors: [(ka)bc]?
5. What happens to its value if one of the vectors is a sum: [(a+d)bc]?
6. What does it mean geometrically about the vectors if [abc]=0?

1. It doesn't matter you can exchange the dot and cross.
2. It depends if it cyclicic or not. Cyclic rotations leave it unchanged. If it is not cyclicic I would say you would get the negative.
3. Evaluates to 0.
4. The constant just changes the value of cross procuct by k times.
5.[(a+d)bc]?
Isn't the cross product distributive over addition so you would have a cross b + d cross c
6. If [abc] = 0 the angle between the vectors is 0 or 180 and the vectors are coplanar.

 

[LCKurtz]

1. Does [abc] mean $a\cdot (b \times c)$ or $(a\times b) \cdot c$, or does it matter?
2. What happens to the value of [abc] if you swap two of the vectors?
3. What happens to the value of [abc] it two of the vectors are the same?
4. What happens if you have a constant k times one of the vectors: [(ka)bc]?
5. What happens to its value if one of the vectors is a sum: [(a+d)bc]?
6. What does it mean geometrically about the vectors if [abc]=0?

Start by answering those questions. You need to know them to work out the hint I will give you next.

1. It doesn't matter you can exchange the dot and cross.

Yes. That's why they can both be left out of the symbol [abc].

2. It depends if it cyclicic or not. Cyclic rotations leave it unchanged. If it is not cyclicic I would say you would get the negative.

I didn't ask about rotations. If you interchange two of the vectors, what does that always do to the value?

3. Evaluates to 0.

Yes.

4. The constant just changes the value of cross procuct by k times.

So you would write [(ka)bc] = k[abc].

5.[(a+d)bc]?
Isn't the cross product distributive over addition so you would have a cross b + d cross c

No. You are just guessing. This isn't a cross product, it is a triple scalar product. Work it out and see what happens.

6. If [abc] = 0 the angle between the vectors is 0 or 180 and the vectors are coplanar.

The angle between what vectors is 0 or 180 and why? Just because a box product comes out zero doesn't mean the internal cross product is zero. Do you know what the box product [abc] represents geometrically, and why it's called a box product in the first place?

 

[Jbreezy]

I didn't ask about rotations. If you interchange two of the vectors, what does that always do to the value?

3. Evaluates to 0.
Yes.

4. The constant just changes the value of cross procuct by k times.
So you would write [(ka)bc] = k[abc].

Yes!

5.[(a+d)bc]?
Isn't the cross product distributive over addition so you would have a cross b + d cross c
No. You are just guessing. This isn't a cross product, it is a triple scalar product. Work it out and see what happens.

(a+d) dot b cross (a+d) dot c
?


6. If [abc] = 0 the angle between the vectors is 0 or 180 and the vectors are coplanar.
The angle between what vectors is 0 or 180 and why? Just because a box product comes out zero doesn't mean the internal cross product is zero. Do you know what the box product [abc] represents geometrically, and why it's called a box product in the first place?

It is the volume of the parallelepiped OK yes it doesn't mean that the cross product is 0. So it just means that the volume of the parallelepiped is 0. COPLANAR!

 

[Jbreezy]

2 is it negates the value

 

[LCKurtz]

What about #5? It is important. Try looking here:
http://www.vitutor.com/alg/determinants/properties_determinants.html

 

[Jbreezy]

You talking about 7 on the link?
7. If all the elements of a line or column are formed by two addends, the above mentioned determinant decomposes in the sum of two determinants.

 

[Jbreezy]

Does property 7 apply to subtraction?

 

[LCKurtz]

You talking about 7 on the link?
7. If all the elements of a line or column are formed by two addends, the above mentioned determinant decomposes in the sum of two determinants.

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]?

Also, for my information, what grade level are you in and what math courses have you finished?

 

[Jbreezy]

HA! Grade level! I'm old don't worry about that.
I know the intelligence radiates.

Is it crap to start at the thing your trying to prove and work from that?
Or in this case 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]?

Please answer those two. You will need them.

Also, for my information, [STRIKE]what grade level are you in and[/STRIKE] what math courses have you finished?

And that.

HA! Grade level! I'm old don't worry about that.
I know the intelligence radiates.

Is it crap to start at the thing your trying to prove and work from that?

That's "you're", not "your". Once you have the box product formulas all done I will give you a hint what to do to solve the problem.

 

[Jbreezy]

Does property 7 apply to subtraction?

I assume so.

Is it crap to start at the thing your trying to prove and work from that?

I maybe have it done if you know the answer to those. I have to sleep will be back tomorrow I can answer those other questions.
I took some algebra, precalc, and one calc class.

Is it crap to start at the thing your trying to prove and work from that?

Don't forget this question!

 

[LCKurtz]

Does property 7 apply to subtraction?

Yes.

Is it crap to start at the thing your trying to prove and work from that?

Since you are given the result, you could use the givens to work backwards and show the formula works. Sometimes that is easier than other times. Or sometimes you can use the givens to directly show the result. This problem can be worked either way.

One thing for sure, the result isn't even true for four arbitrary position vectors. So if you haven't used the fact that the points those position vectors represent are coplanar, your proof can't be correct. I will check in tomorrow.

 

[Jbreezy]

Yeah I don;t know if it is right then. I used to given and the property 7 of determinants and ended up with 2 determinantes and one of the rows was all 0

 

[Jbreezy]

Still I don't understand how to put theis together. I want to start from the fact the position vectors are coplanar then show that the statement is true.

 

[LCKurtz]

So how would you write the answer to question 5? [(a+d)bc]= ? And what about [a(b+d)c]?

You need to answer these before you are ready to tackle the problem.

Still I don't understand how to put theis together. I want to start from the fact the position vectors are coplanar then show that the statement is true.

It isn't the position vectors that are coplanar. It is the points they point to that are coplanar. And that will be where to start.

 

[Jbreezy]

[(a+d)bc]= (a+d) dot b cross c dot ( a+d)

and

[a(b+d)c] = a dot (b cross c ) + a dot ( c cross c )
c cross c = 0
No I really mean that the points are coplanar. So I want to start to say that since they are three vectors tripple scalar product will be 0. I figure you can do ( c -a ) cross ( b-d ) dot ( ?) and move things around because of the properties and get to the final statement.

 

[Jbreezy]

Do not tell me the answer!

 

[LCKurtz]

[(a+d)bc]= (a+d) dot b cross c dot ( a+d)

and

[a(b+d)c] = a dot (b cross c ) + a dot ( c cross c )
c cross c = 0

Those aren't right. A triple scalar (box) product has one dot and one cross. Write [a(b+d)c] as a determinant, use one of the properties of determinants to break it up and write the result in the box product [abc] form. Show me the steps for one of them.

No I really mean that the points are coplanar. So I want to start to say that since they are three vectors tripple scalar product will be 0. I figure you can do ( c -a ) cross ( b-d ) dot ( ?) and move things around because of the properties and get to the final statement.

That is the idea, but you don't want to get all tangled up with dots and crosses. You want to express everything in a box product like form like [abc]. That is why you need to know the rules for manipulating box products. In particular, the formulas we listed, including correcting the above formulas and expressing the answers as box products.

You could use any three pairs such as c-a, b-a, d-a and start with the box product of them.

[Edit] I don't mean you can't work it out with the dots and crosses. You can if you are careful.

 

[Jbreezy]

I'm having a hard time to connect this up. I understand what you are telling me but. How can you have a determinant with ( c -a) (b-a) (d-a) ? How can you expand that. I thought that property was if there was one row.
[a(b+d)c] this you just have a_x, a_y ...for the top row. Then b_x + d_x , b_y + d_y...then c_x,...for the last row. Then you just expand this into two determinants. I did this for the problem starting from the assumption and working backwards you get [ (a-c)bd] + [(b-d)ac] ...So now I look at your hint and I don't understand how you can have all three in a determinant and expand it ?
c-a, b-a, d-a ? Be careful what you tell me I don't want to be told.Thanks

 

[LCKurtz]

I'm having a hard time to connect this up. I understand what you are telling me but. How can you have a determinant with ( c -a) (b-a) (d-a) ? How can you expand that. I thought that property was if there was one row.
[a(b+d)c] this you just have a_x, a_y ...for the top row. Then b_x + d_x , b_y + d_y...then c_x,...for the last row. Then you just expand this into two determinants. I did this for the problem starting from the assumption and working backwards you get [ (a-c)bd] + [(b-d)ac] ...So now I look at your hint and I don't understand how you can have all three in a determinant and expand it ?
c-a, b-a, d-a ? Be careful what you tell me I don't want to be told.Thanks

You have to do one step at a time. So if you have (c-a)(b-a)(d-a) and you only know how to handle one, call (c-a)=z and (d-a)=w and work it like you would z(b-a)w. Then put z and w back to what they were and continue like that. Don't try to combine steps.

 

[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.