Parametric vector forms of planes (1 Viewer)

[yasminee96]

I need some help converting cartesian forms of planes into vector forms.

In changing, for example, 4x-3y+6z=12, into parametric vector form I have no problem.

However, I can't do it when I have something such as:
y+6z=-1, or z=2

can someone help me out?

 

[Sy123]

I need some help converting cartesian forms of planes into vector forms.

In changing, for example, 4x-3y+6z=12, into parametric vector form I have no problem.

However, I can't do it when I have something such as:
y+6z=-1, or z=2

can someone help me out?

Just imagine those equations as:

0x + y + 6z = -1 and 0x + 0y + z = 2 and apply the same method

 

[yasminee96]

Just imagine those equations as:

0x + y + 6z = -1 and 0x + 0y + z = 2 and apply the same method

I tried that and keep getting wrong answers

 

[Sy123]

I tried that and keep getting wrong answers

(I only learned parametric representation today and wasn't really listening so sorry if this is wrong)

 

[yasminee96]

(I only learned parametric representation today and wasn't really listening so sorry if this is wrong)

This makes a lot of sense. I'll try it now!

I've been using a different method where i let y=lambda and z=mu and then solve for x and sub that in then sorta just expand, which is why I was having difficulties here.
But I shall try this and let you know if it works! Thanks!

 

[yasminee96]

Braintic, this is what I've been doing

 

[Sy123]

Not up to matrices yet :/

> USYD has best maths department
> Slower than UNSW
> lecturer spends 30 min on numerical examples

gg

 

[yasminee96]

Not up to matrices yet :/

> USYD has best maths department
> Slower than UNSW
> lecturer spends 30 min on numerical examples

gg

We havent done matrices yet either!
That's just the way i like to write my vectors lol gg

 

[Sy123]

oops

Did you get the answer using my suggested method?

 

[braintic]

For z=2:

x and y can be anything, so you need linear combinations of the vectors (1,0,0) and (0,1,0).
To ensure that z=2, just add (0,0,2).

So the answer is (0,0,2) + c(1,0,0) + d(0,1,0) .

This gives all vectors of the form (c,d,2), which is precisely the plane z=2.



For y+6z=-1, first treat it as a 2-dimensional object in the y-z plane.

Let z=c:
y=1-6c
so using 2-D (y,z) vectors, this line is (1-6c, c) = (1,0) + c(-6,1)
When we introduce x, we have to allow it to be anything, so we need linear combinations of (1,0,0).
The other vectors also need to be extended to 3-D, by including an x-component.

So the plane is (0,1,0) + c(0,-6,1) + d(1,0,0)

(which I believe is a better parametrization)

 

[Squar3root]

Lol is everyone pumped/spewing over the algebra 1131/1141 test this week?

 

[integral95]

Lol is everyone pumped/spewing over the algebra 1131/1141 test this week?

I reckon there would be more people spewing over algebra than calculus