Determine Vector and Cartesian Equation Watch

Joe199111
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A(3,3,2) B(9,5,3) C(-1,3,4)

Determine the Vector and Cartesian equation of the lines AB and BC...

Im abit confused what the question is asking...

find a vector equation of the line going through the points A and B
and
find a vector equation of the line going through the points B and C

or something else ?

also how do you work out the Cartesian Equation/s?
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Khodu
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I think for vector you use the r= b(lamda)+a(1-lamda) formula and cartesian you use the formula (x,y,z)=a+lamda(AB)?
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dknt
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(Original post by Joe199111)
A(3,3,2) B(9,5,3) C(-1,3,4)

Determine the Vector and Cartesian equation of the lines AB and BC...

Im abit confused what the question is asking...

find a vector equation of the line going through the points A and B
and
find a vector equation of the line going through the points B and C

or something else ?

also how do you work out the Cartesian Equation/s?
Yes, they are asking for the vector equation of the line going through A, and B and then another line going through B and C.

There is a quick way of getting from the vector equation of a line to it's corresponding cartesian equations. If you haven't learnt it, try forming parametric equations, in terms of your scalar multiple of the direction vector (lambda, mu, s, t etc.), and making it the subject and then equating them.
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Joe199111
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(Original post by dknt)
Yes, they are asking for the vector equation of the line going through A, and B and then another line going through B and C.

There is a quick way of getting from the vector equation of a line to it's corresponding cartesian equations. If you haven't learnt it, try forming parametric equations, in terms of your scalar multiple of the direction vector (lambda, mu, s, t etc.), and making it the subject and then equating them.
I cant work out the Vector equation of AB how do I work out the cartesian equation ?
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dknt
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(Original post by Joe199111)
I cant work out the Vector equation of AB how do I work out the cartesian equation ?
I assume that is a typo and you can form the vector equation of AB...

If we have a vector equation in the form,

\mathbf{r} = \begin{pmatrix} a + \lambda p \\ b + \lambda q \\ c + \lambda r \end{pmatrix} , where  \lambda is your scalar multiple and a,b,c,p,q and r are constants, then we have parametric equations for x, y and z, with  \lambda as the parameter.


 x=  a + \lambda p ;  y=  b + \lambda q ;  z=  c + \lambda r ;

We can then rearrange in terms of  \lambda , giving,


 \lambda = \dfrac{x-a}{p} ;  \lambda = \dfrac{y-b}{q} ;  \lambda = \dfrac{z-c}{r} ;

and so we can equate them, eliminating the parameter, forming our catersian equations for the line.

 \dfrac{x-a}{p}  = \dfrac{y-b}{q}  = \dfrac{z-c}{r}

You should also be able to note the similarities between the vector equation of a line and its corresponding cartesian equations, letting you easily work out the cartesian equations from a vector equation and vice a versa.
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