# Homework Help: Cartesian and vector equations

1. May 20, 2013

### Jbreezy

1. The problem statement, all variables and given/known data
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.

2. Relevant equations

For the parametric eq.

x = a + λu where u is a unit vector

3. 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

2. May 20, 2013

### CAF123

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

3. May 20, 2013

### Jbreezy

OK. Does it look correct otherwise? Thanks.

4. May 20, 2013

### Staff: Mentor

5. May 20, 2013

### Jbreezy

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

6. May 20, 2013

### CAF123

I believe so

7. May 20, 2013

### Jbreezy

Thanks, sorry for the re- post.

8. May 20, 2013

### HallsofIvy

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.

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.

9. May 20, 2013

### Jbreezy

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