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I was thinking that the normal vector was [8,0] and that the direction vector was [0,-8] and the position vector was [8,0].
I was wrong.
My final answer was [x,y]=[8,0]+t[0,-8]. But I am wrong.
MYsesshou, are you my savior in this dark time?
Assassin, are you able to assist me?
Actually, I'm not quite sure. Sorry !! Sorry for lag. I have been fighting my computer and this site for like an hour.
I think everyone has. Ok. Oh well.
was that the only info you're given?
Yes. Alongside the info in the textbook explaining how to do this.
I'm still searching, but can you apply something like this? http://www.blurtit.com/q1900770.html (google'd) I don't have your book... is it a calc book?
I dunno. Yes, it is a Calculus and Vectors book.
Are you able to help?
Probably not, but I've been searching
I'll rally up others just in case.
Sorry, it's just been too long for me for this to be recalled.
Yeah. Happened to me when I was helping someone else on inequalities.
@Zarkon Uh, someone told me to ask you.
why is your answer wrong?
The actual answer is [x,y]=[8,2]+t[0,1].
I've double-checked to make sure the facts are right.
the -8 isn't necessary but acceptable.... both your eq. and answer equation are acceptable...
direction vector can be any vertical vector... starting point postion vector can refer to any point on line Infinite acceptable answers.
Like, I know that [8,0] can be anything, but my answer still isn't right in terms of t[0,1].
Sorry IsTim, I've searched through my calc book and solutions manual, but I guess 8-9 years is too long to remember enough of this. Hope PaxPolaris is able to help you understand. :)
direction vector can be any vertical vector ... [0,1] and [0,-8] are both parallel to the line so both are acceptable ... any vector [0,C] is acceptable.
Wait, you're sure you're not talking about position vector?
no, we're done with that...??
I don't know, they way you describe the direction vector sounds like the way my teacher described a position vector.
Position Vector: of any 1 point on the line Direction Vector: any vector parallel to the line
But then what's the constant?
the slope of the direction vector is constant ... the length of the direction vector is irrelevant
Ok. I'll just leave this question as is for now. Thanks for the help anyways, PAx.
eg. if your direction vector is [1,2]: it can also be [2,4] , [3,6], [-5,-10]....[n,2n] preferably you'd use the simplest form i.e. [1,2] or in your case [0,1] instead of [0,-8] http://www.netcomuk.co.uk/~jenolive/vect3.html
Hey Sess, I'm good now. Thanks for staying though. I'll take a look at it Pax Polaris.
Evidently, my book wasn't good enough to help. I was reading through some of this, after p.4, that seemed that it might help some. http://mysite.science.uottawa.ca/vbozi013/mat1339/ch08.pdf PaxPolaris' site is good too :)
Yay PaxPolaris for the help !
i think i missed the original problem as stated