Topic review (newest first)
- 2005-10-29 20:54:38
"Prove that if vector C is perpendicular to vectors A and B, then vector C is still perpendicular to vector mA + nB where m and n are scalars."
I made a mistake with the wording.
We know so far that A.C = 0 and B.C = 0. Therefore, to prove the above statement....
C.(mA + nB) = C.mA + C.nB
Using scalar properties, C.mA + C.nB = m*(A.C) + n*(B.C)
m*(A.C) + n*(B.C) = m*0 + n*0 = 0
Is it really that easy?
- 2005-10-28 21:43:58
Well, I like the way we all conduct ourselves here. Sometimes a bit rough amongst friends, but that is just how friends are. Perhaps some of that "spilled over" into maths_buff's topic in a bad way, and that is something to be careful of in the future.
And I think we do a great job helping people, keeping spammers out, and generally making this a really nice place.
The way we have reacted so strongly here is an indication of our good nature.
If we want to discuss this more, how about over in "Dark Discussions" ... ?
- 2005-10-28 19:44:34
Right. So you meant me to correct you if you were wrong about the rules, not about the spelling of senior. Well, you weren't.
No Personal Attacks or Put-Downs. This is a type of bullying, and just makes you look insecure.
This is not a place to be mean to others and these posts will not be tolerated. Light banter or constructive criticism can be allowed if it is polite and friendly. Remember, other people have feelings too. "Those who give respect shall receive it."
It's alright to deviate though. That's just a natural part of conversation, señor.
- 2005-10-28 19:32:13
Senor, correct me if I am wrong.
(And it is senior.)
That was intentional.
Used as a courtesy title before the surname, full name, or professional title of a man in a Spanish-speaking area.
Used as a form of polite address for a man in a Spanish-speaking area.
- 2005-10-28 19:20:41
Yes, I'm sorry too. I think insomnia just wanted to contribute something to the conversation and because he didn't know about vectors, he resorted to an insult. Consider them gone. (And it is senior.)
Anyway, haven't you pretty much got it already?
a.b = |a||b| cos θ, and if the dot product is 0, then the vector is perpendicular.
Take the angle between A and C to give A.C = |A||C| cos 90, which is 0, as you've said.
Multiplying A by m just gives Am.C = |Am||C| cos 90, which is still 0, so they are still perpendicular.
The same reasoning could be applied with B and C.
- 2005-10-28 13:33:39
I apologize. I am not supposed to comment on Mathsyperson's and Insomnia's replies and believe me, we are polite and have a strict code of conduct in this forum and no deviation is tolerated.
Insomnia, the new moderator : Please note.
Mathsyperson, my senior in this forum : Senor, correct me if I am wrong.
- 2005-10-28 08:43:57
I don't personally see what the fuss is all about, nor do I comprehend the reasonings behind the snide remarks. I know that a scalar only changes the magnitude, but, as MathsIsFun pointed out, in itself that is not a proof.
I was asking a question - I didn't expect to be ridiculed (i.e. Insomnia and Person). Nice to see that internet etiquette is at its best in this forum.
Thank you for your reply, MathsIsFun. As for the other two, keep your bragging for the locker room.
- 2005-10-27 22:08:20
This isn't a proof, but intuitively it does not matter how much longer A and B grow, C will always be perpendicular.
A scalar does not change direction of a vector.
(Different of course if a vector is the resultant of two other vectors, and those vectors change length)
- 2005-10-27 10:50:06
"Prove that if vector C is perpendicular to vectors A and B, then vector C is still perpendicular to vectors A and B when A and B are multiplied by scalars m and n respectively."
If we have perpendicular vectors on a 2D plane, then V.W = 0 when the dot product of vectors V and W is found.
It can also be seen from the dot product formula that V.W = ( V ) ( W) cos 90 and cos 90 = 0.
But how to prove they are perpendicular after being multiplied by scalars remains the question.