Cartesian and vector equations

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Homework Statement


The position vectors of two points a and b on a line are < 2,1,7> and < 1,4,-1>
Find the parametric vector equation for any point on the line.
Find the linear Cartesian equation of the line using x, y, z as coordinates.

Homework Equations



For the parametric eq.

x = a + λu where u is a unit vector

The Attempt at a Solution



b -a = < -1, 3, -8 > , call it u
Norm of b -a = √(-1)^2 + (3)^2 + -(-8)^2
unit vector u = <-1/√74, 3/ √74 , -8 /√74>

x = < 2, 1, 7 > + λ<-1/√74, 3/ √74 , -8 /√74>


For the Cartesian equation I have.

x = 2 + λ(-1/√74)

y = 1 + λ(3/ √74)

z = 7 + λ(-8 /√74)


How is she looking?
I have a question if it says linear Cartesian equation does that mean I have to eliminate lambda and write it all together or it is fine to leave as I have?

Thanks
 

Answers and Replies

  • #2
CAF123
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You can solve each of the the Cartesian equations you have for λ and hence put the equations into 'symmetric' form.
 
  • #3
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OK. Does it look correct otherwise? Thanks.
 
  • #5
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It is the same but different. I have mad mistakes in the old one. So I redid it. Is it correct?
 
  • #6
CAF123
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OK. Does it look correct otherwise? Thanks.
I believe so
 
  • #7
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Thanks, sorry for the re- post.
 
  • #8
HallsofIvy
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Homework Helper
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Homework Statement


The position vectors of two points a and b on a line are < 2,1,7> and < 1,4,-1>
Find the parametric vector equation for any point on the line.
Find the linear Cartesian equation of the line using x, y, z as coordinates.

Homework Equations



For the parametric eq.

x = a + λu where u is a unit vector

The Attempt at a Solution



b -a = < -1, 3, -8 > , call it u
Norm of b -a = √(-1)^2 + (3)^2 + -(-8)^2
unit vector u = <-1/√74, 3/ √74 , -8 /√74>

x = < 2, 1, 7 > + λ<-1/√74, 3/ √74 , -8 /√74>


For the Cartesian equation I have.

x = 2 + λ(-1/√74)

y = 1 + λ(3/ √74)

z = 7 + λ(-8 /√74)
When t= 0, x= 2, y= 1, z= 7 which is (2, 1, 7), one of the two given points.
when [itex]\sqrt{74}[/itex], x= 2- 1= 1, y= 1+ 3= 4, z= 7- 8= -1 which is (1, 4, -1), the other point. Since all the calculations are linear, this is a straight line and since it contains the two points that define the line, it is the correct line.

But I am puzzled at your use of [itex]\sqrt{74}[/itex]. You don''t have to use a unit vector. You could just as well use x= 2- t, y= 1+ 3t, z= y- 8t, avoiding the square root.




How is she looking?
I have a question if it says linear Cartesian equation does that mean I have to eliminate lambda and write it all together or it is fine to leave as I have?

Thanks
Yes, eliminate [itex]\lambda[/itex]. The simplest way to do that is to solve each equation for [itex]\lambda[/itex]: [itex]\lambda= -\sqrt{74}(x- 2)[/itex], [itex]\lambda= \sqrt{74}(y- 1)/3[/itex], and [itex]\lambda= \sqrt{74}(z- 7)/(-8)[/itex]. Since those are all equal to [itex]\lambda[/itex] they are equal to each other:
[tex]\frac{x- 2}{-1}= \frac{y- 1}{3}= \frac{z- 7}{-8}[/tex]
Notice that the [itex]\sqrt{74}[/itex] terms all cancel.
 
  • #9
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I don't know just used it. The unit vector I mean. THX DOOODE!
 

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