Vector equation for a plane

**Oiler** · January 28th 2011, 10:28 PM

Hey guys, quick question. If a plane passes through the points (a,b,c) and (d,e,f), and is parallel to the vector (3,2,1). Am i right in assuming the vector equation for the plane will be (x,y,z) = (a,b,c) + [(d,e,f) - (a,b,c)]s + (3,2,1)s. Thanks

**janvdl** · January 28th 2011, 10:44 PM

Originally Posted by Oiler

Hey guys, quick question. If a plane passes through the points (a,b,c) and (d,e,f), and is parallel to the vector (3,2,1). Am i right in assuming the vector equation for the plane will be (x,y,z) = (a,b,c) + [(d,e,f) - (a,b,c)]s + (3,2,1)s. Thanks

If it passes through points $(a,b,c)$ and $(d,e,f)$ then $t(a - d, b - e, c - f)$ is also a vector it is parallel to.

$(x,y,z) = (a,b,c) + t(a-d, b-e, c-f)$
$(x,y,z) = (a,b,c) + t(3, 2, 1)$

I suppose you could combine them into:
$2(x,y,z) = 2(a,b,c) + t(3 + a - d, 2 + b - e, 1 + c - f)$

$(x,y,z) = (a,b,c) + \frac{t(3 + a - d, 2 + b - e, 1 + c - f)}{2}$

EDIT:

But I'm not sure why you would use such an elaborate method.
$(x,y,z) = (a,b,c) + t_1 (3,2,1)$
$(x,y,z) = (d,e,f) + t_2 (3,2,1)$

You could use either one of those equations as well. If this plane passes through both points, and is parallel to a certain vector then you only need one of those points and the vector it's parallel to.

**earboth** · January 29th 2011, 01:05 AM

Originally Posted by Oiler

Hey guys, quick question. If a plane passes through the points (a,b,c) and (d,e,f), and is parallel to the vector (3,2,1). Am i right in assuming the vector equation for the plane will be (x,y,z) = (a,b,c) + [(d,e,f) - (a,b,c)]s + (3,2,1)t. Thanks

Nearly! You have to use a 2nd variable. (see my correction in red)

Originally Posted by janvdl

If it passes through points $(a,b,c)$ and $(d,e,f)$ then $t(a - d, b - e, c - f)$ is also a vector it is parallel to.

That's correct!

$(x,y,z) = (a,b,c) + t(a-d, b-e, c-f)$
$(x,y,z) = (a,b,c) + t(3, 2, 1)$

I suppose you could combine them into:
$2(x,y,z) = 2(a,b,c) + t(3 + a - d, 2 + b - e, 1 + c - f)$

$(x,y,z) = (a,b,c) + \frac{t(3 + a - d, 2 + b - e, 1 + c - f)}{2}$

EDIT:

But I'm not sure why you would use such an elaborate method.
$(x,y,z) = (a,b,c) + t_1 (3,2,1)$
$(x,y,z) = (d,e,f) + t_2 (3,2,1)$

You could use either one of those equations as well. If this plane passes through both points, and is parallel to a certain vector then you only need one of those points and the vector it's parallel to.

Maybe I don't understand your reply correctly but in general the equations you gave represent straight lines which are placed in the plane you are looking for.

**janvdl** · January 29th 2011, 02:57 AM

Yeah, sorry I misread the question and went looking for lines instead of planes.

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