ATAR Notes: Forum

VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: sajib_mostofa on September 26, 2010, 08:45:43 pm

Title: Parrallel vectors
Post by: sajib_mostofa on September 26, 2010, 08:45:43 pm: Q. Given $u = 2i+2j+k$ , $v = 2j+2k$ and $w = mi+nj$ , the values for m and n for which u + w is parrellel to v are?
Title: Re: Parrallel vectors
Post by: kakar0t on September 26, 2010, 08:58:59 pm: does m=-2 and n=-1

if they do i'll scan my working

if it doesn't, then dw lol
Title: Re: Parrallel vectors
Post by: sajib_mostofa on September 26, 2010, 09:00:00 pm: Yeah it does
Title: Re: Parrallel vectors
Post by: moekamo on September 26, 2010, 09:05:29 pm: $u+w=(2+m)i+(2+n)j+k$

to be parallel to v, $c(u+w)=v$ where $c$ is a constant

so $c(2+m)i+c(2+n)j+ck=2j+2k$ ,

equating i, j, k components, $c=2$ ,

$2(2+n)=2\implies n=-1$

and $2(2+m)=0 \implies m=-2$

so answer is n=-1 and m=-2 like kakar0t said :)
Title: Re: Parrallel vectors
Post by: kakar0t on September 26, 2010, 09:06:54 pm: Sorry for the bad quality, scanner doesn't pick up pencil well.

If you need me to clarify things please ask.

(same as moekamos clear, precise explanation)
Title: Re: Parrallel vectors
Post by: sajib_mostofa on September 26, 2010, 09:11:53 pm: Thanks to both of you. But why don't I get the same answer when using dot product and making $(u+w).v = 1$ ?
Title: Re: Parrallel vectors
Post by: m@tty on September 26, 2010, 09:23:09 pm: That is only true if both vectors are unit vectors. (Or if the product of their magnitudes equals 1)

Because if they are parallel, then the angel between them is zero.

Hence,
$(u+w)\cdotv=|u+w|\cdot|v|\cos(\theta)=|u+w|\cdot|v|\cdot1=|u+w||v|$
Title: Re: Parrallel vectors
Post by: sajib_mostofa on September 26, 2010, 09:44:42 pm: ahh I was unaware of that. cheers Matty
Title: Re: Parrallel vectors
Post by: Martoman on September 26, 2010, 09:56:29 pm: Quote from: moekamo on September 26, 2010, 09:05:29 pm
$u+w=(2+m)i+(2+n)j+k$

to be parallel to v, $c(u+w)=v$ where $c$ is a constant

so $c(2+m)i+c(2+n)j+ck=2j+2k$ ,

equating i, j, k components, $c=2$ ,

$2(2+n)=2\implies n=-1$

and $2(2+m)=0 \implies m=-2$

so answer is n=-1 and m=-2 like kakar0t said :)

This is the best way.

To do it quickly and sneakily, realise that the scale factor difference in the K term is 2. That then means that all components are the same except for a factor of 2... essentially what moekamo said.
Title: Re: Parrallel vectors
Post by: sajib_mostofa on September 26, 2010, 10:03:31 pm: Quote from: Martoman on September 26, 2010, 09:56:29 pm
Quote from: moekamo on September 26, 2010, 09:05:29 pm
$u+w=(2+m)i+(2+n)j+k$

to be parallel to v, $c(u+w)=v$ where $c$ is a constant

so $c(2+m)i+c(2+n)j+ck=2j+2k$ ,

equating i, j, k components, $c=2$ ,

$2(2+n)=2\implies n=-1$

and $2(2+m)=0 \implies m=-2$

so answer is n=-1 and m=-2 like kakar0t said :)

This is the best way.

To do it quickly and sneakily, realise that the scale factor difference in the K term is 2. That then means that all components are the same except for a factor of 2... essentially what moekamo said.

So you don't recommend using dot product?
Title: Re: Parrallel vectors
Post by: Martoman on September 26, 2010, 11:32:16 pm: No because this way you get the answer in about 3 lines without considering the pitfalls matty detailed. Save you confusing yourself and just go for the simplest method.