ATAR Notes: Forum

VCE Stuff => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematics => Topic started by: phagist_ on August 26, 2008, 01:17:02 pm

Title: Geometry help!
Post by: phagist_ on August 26, 2008, 01:17:02 pm: Ok not sure if this is in the right section, but anyway.

I'm having trouble with this question

Find the distance from the point P(-1,2-1) from the line whose parametric equations are;

$x=2+2t~,~y=2-t~,~z=-3-3t$

cheers!
Title: Re: Help with distance from a point to a line.
Post by: cara.mel on August 26, 2008, 01:31:42 pm: I have answered in picture format because I fail at words:
(http://img.photobucket.com/albums/v711/happypuff/line_question.jpg)

If it doesnt make sense get someone else to clarify because I'm leaving pretty soon for work and won't be on FSN for probably 24 hours :P
Title: Re: Help with distance from a point to a line.
Post by: phagist_ on August 26, 2008, 01:34:22 pm: I can do it with Pythag.. but I'm having trouble finding the normal vector to project it onto.

Thanks anyway:)
Title: Re: Help with distance from a point to a line.
Post by: phagist_ on August 26, 2008, 02:51:26 pm: ahh nvm.. figured it out.. turns out I was using an equation wrong! it was doing my head in, haha.
Title: Re: Geometry help!
Post by: phagist_ on August 27, 2008, 06:07:30 pm: I really can't seem to get my head around geometry, I don't know why.

a) Find the vector equation of the plane through the points (2,-1,0) and (-5,-3,1) that is parallel to the line joining the points (3,5,-1) and (0,3,-2)

b) Find the parametric equations of the straight line through the origin that is perpendicular to this plane, and find where it intersects the plane.

-------------------------

but I am not sure by what it means 'vector equation'...

any help would be greatly appreciated,
cheers!
Title: Re: Geometry help!
Post by: Mao on August 27, 2008, 08:07:18 pm: by vector equation, i hope they meant: $\vec{N}\cdot (R(t)-R(t_0))=0$

as for b), basically you need to find a point on that plane which perpendicularly projects a line through the origin. The line would follow the general vector equation $r(t)=r(0)+\vec{v}t$ . Since it is perpendicular to the plane, its "velocity" must be the normal vector, and since it goes through the origin, r(0)=(0,0,0). Hence, if we let the line be L, $L(t)=\vec{N}t$
Title: Re: Geometry help!
Post by: phagist_ on August 27, 2008, 08:10:36 pm: ah yep..any ideas on how to approach this?
Title: Re: Geometry help!
Post by: Mao on August 27, 2008, 08:48:23 pm: so we know that the normal vector is <4,-10,8> (i hope), hence $\hat{N}=\frac{3}{\sqrt{5}}<2,-5,4>$

hence, the cartesian equation for the plane is $\hat{N}\cdot \vec{r}=\hat{N}\cdot \vec{r_0}$

$\implies 2x-5y+4z=c$

subbing in any of those points would yield c=9

now, we also know that the equation for the line is $L(t)=\vec{N}t=<4t,-10t,8t>$

hence, subbing in the above vector for x,y,z:

$\implies 8t+50t+32t=9$

$\implies t=0.1$

$\implies \mbox{point of intersection is }\frac{2}{5}i-j+\frac{4}{5}k$
Title: Re: Geometry help!
Post by: phagist_ on August 28, 2008, 06:21:46 pm: Quote from: phagist_ on August 27, 2008, 06:07:30 pm

(2,-1,0) and (-5,-3.1) that is parallel to the line joining the points (3,5,-1) and (0,3,-2)

shit, just realised a typo should be (-5,-3,1).

and thanks a lot Mao.
Title: Re: Geometry help!
Post by: phagist_ on August 28, 2008, 06:28:57 pm: I was wondering, by vector equation.. could they also mean (x,y,z)+t(a,b,c)+s(d,e,f) ?

and if that was the case, then could use the vector joining (2,-1,0) and (-5,-3,1) and the vector joining (3,5,-1) and (0,3,-2).

To get

(2,-1,0) + t(-7,-2,1) + s(-3,2,1)

sorry if this is a dumb question, I just don't get this stuff and it's been a loooong day/

cheers
Title: Re: Geometry help!
Post by: Mao on August 30, 2008, 06:51:02 pm: mmm it could

but i'm a little skeptical on using two parameters =P