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C4 vectors

Im finding some vectors quite challenging. Im not sure how i find the point on a vector line which is closest to the origin.

I know what it is perpendular to the orogin. So the vector product equals zero. However i end up forming quadratics which dont lead anywhere.

How should i approach these kinds of questions.

I am given the vector line equation and a point on the line. I am told to find the clostest point to the origin.

Thanks.
Reply 1
Original post by jamesgates1
Im finding some vectors quite challenging. Im not sure how i find the point on a vector line which is closest to the origin.

I know what it is perpendular to the orogin. So the vector product equals zero. However i end up forming quadratics which dont lead anywhere.

How should i approach these kinds of questions.

I am given the vector line equation and a point on the line. I am told to find the clostest point to the origin.

Thanks.


Well, you know the parametric equation of a point on the plane is (all the english letters are going to be given):

x=a+λbx = a + \lambda b
y=c+λdy = c + \lambda d
z=e+λfz = e + \lambda f

Call this general point XX

Then you know OX\vec{OX}. Now you need to find λ\lambda s.t OX=0\vec{OX} \cdot \ell = 0 where \ell is the direction vector of the line (all this says is make the line OX perpendicuar to your vector line). This gives you a (linear) equation solely in λ\lambda which you can then solve for and then plug this into OX\vec{OX} after which, you then calculate the magnitude of OX\vec{OX}.

I had to make this all very abstract with so many variables because you didn't give me an example to work off, sorry!
Reply 2
Original post by Zacken
Well, you know the parametric equation of a point on the plane is (all the english letters are going to be given):

x=a+λbx = a + \lambda b
y=c+λdy = c + \lambda d
z=e+λfz = e + \lambda f

Call this general point XX

Then you know OX\vec{OX}. Now you need to find λ\lambda s.t OX=0\vec{OX} \cdot \ell = 0 where \ell is the direction vector of the line (all this says is make the line OX perpendicuar to your vector line). This gives you a (linear) equation solely in λ\lambda which you can then solve for and then plug this into OX\vec{OX} after which, you then calculate the magnitude of OX\vec{OX}.

I had to make this all very abstract with so many variables because you didn't give me an example to work off, sorry!


Thanks. I understand it now!
Reply 3
Original post by jamesgates1
Thanks. I understand it now!


No problem, make sure to test it out on a few problems but it's usually a method that has served me very well.

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