Cartesian and vector equations

[Jbreezy]

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

 

[CAF123]

You can solve each of the the Cartesian equations you have for λ and hence put the equations into 'symmetric' form.

 

[Jbreezy]

OK. Does it look correct otherwise? Thanks.

 

[Mark44]

Isn't this the same question you asked in this thread? https://www.physicsforums.com/showthread.php?t=690620

If so, don't start a new thread for a question that's in an existing thread.

 

[Jbreezy]

It is the same but different. I have mad mistakes in the old one. So I redid it. Is it correct?

 

[CAF123]

OK. Does it look correct otherwise? Thanks.

I believe so

 

[Jbreezy]

Thanks, sorry for the re- post.

 

[HallsofIvy]

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 \sqrt{74}, 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 \sqrt{74}. 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 \lambda. The simplest way to do that is to solve each equation for \lambda: \lambda= -\sqrt{74}(x- 2), \lambda= \sqrt{74}(y- 1)/3, and \lambda= \sqrt{74}(z- 7)/(-8). Since those are all equal to \lambda they are equal to each other:
\frac{x- 2}{-1}= \frac{y- 1}{3}= \frac{z- 7}{-8}
Notice that the \sqrt{74} terms all cancel.

 

[Jbreezy]

I don't know just used it. The unit vector I mean. THX DOOODE!