Plane - Linear algebra

[QuantumJG]

Find the equation of a plane (in both cartesian and parametric form) that passes through the point (1,6,-4) and contains the line:

x = 1 + 2t, y = 2 - 3t, z = 3 - t, t element of R

My problem is how can you use the line to derive the equation of the plane?

[QuantumJG]

Find the equation of a plane (in both cartesian and parametric form) that passes through the point (1,6,-4) and contains the line:

x = 1 + 2t, y = 2 - 3t, z = 3 - t, t element of R

My problem is how can you use the line to derive the equation of the plane?

Don't worry I got it!

[Damo17]

Find the equation of a plane (in both cartesian and parametric form) that passes through the point (1,6,-4) and contains the line:

x = 1 + 2t, y = 2 - 3t, z = 3 - t, t element of R

My problem is how can you use the line to derive the equation of the plane?

Well we know that the points and are apart of the plane. We now need another point in the plane so we can find the normal. So let , and we get the point .

So now let:
a be the vector from to and b be the vector from to .
We now get: and .

The normal:
(forgive me here, these are meant to represent matrices with the first 2 numbers of each in the first column, then the last two numbers the second column.)

i component
j component
k component

so


you now use to get:
Equation of plane:

[kamil9876]

If you imagine it, the information about a line on it's own will not be sufficient information for the plane, but the information about another point(not on this line) narrows it down to one plane.

we know that the parametric form of the plane is as follows:

where

we can make (ie it's in the plane and line, you get it from t=0).

we can make since it's the direction vector of the line, so it's a direction vector of the plane too.

The final direction vector can be obtains from "connecting" a point on the line to the additional point given, (1,6,-4). This gives:



Hence the parametric equation of the line is:


Cartesian equation can be derived from here by finding a normal vector.

[QuantumJG]

If you imagine it, the information about a line on it's own will not be sufficient information for the plane, but the information about another point(not on this line) narrows it down to one plane.

we know that the parametric form of the plane is as follows:

where

we can make (ie it's in the plane and line, you get it from t=0).

we can make since it's the direction vector of the line, so it's a direction vector of the plane too.

The final direction vector can be obtains from "connecting" a point on the line to the additional point given, (1,6,-4). This gives:



Hence the parametric equation of the line is:


Cartesian equation can be derived from here by finding a normal vector.

Yep after about an hour of thinking I realised that method was the way to go.

I got r = (1,2,3) + u(2,-3,-1) + s(0,4,-7)

[kamil9876]

yeah just realised a small mistake. 6-2=4 obsviously  :idiot2: