Geometry question in 3 space.

[phagist_]

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

(a) Find the vector equation of the plane through the points (2,−1, 0) and (−5,−3, 1)
that is parallel to the line joining the points (3, 5,−1) and (0, 3,−2).
(b) Find the parametric equations of the straight line though the origin that is perpendicular
to this plane, and find where it intersects the plane.

2. Relevant equations

\hat{n} \cdot (r~-~r_{0}) = 0

3. The attempt at a solution
Ok, I found two vectors, on joining the points (2,−1, 0) and (−5,−3, 1) and another joining the points (3, 5,−1) and (0, 3,−2), seeing as they are parallel they will share the same normal vector \hat{n}.

Then using \hat{n} \cdot (r~-~r_{0}) = 0 with r being (-5,-3, 1) and r_{0} being (2,-1,0). To get the vector form of the equation plane.

but I am not exactly sure what points to use at this stage.

and I haven't attempted b as yet.

thanks!

[Defennder]

(a) Use a vector operation to get the normal vector.

(b)For this you need the normal vector from a). Note that k\mathbf{n} for some real value of k will give you a point on the plane. Think about what the line corresponding to this position vector means.

[phagist_]

ok, are my methods correct, so far?

thanks.

[phagist_]

anyone?

[HallsofIvy]

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

(a) Find the vector equation of the plane through the points (2,−1, 0) and (−5,−3, 1)
that is parallel to the line joining the points (3, 5,−1) and (0, 3,−2).
(b) Find the parametric equations of the straight line though the origin that is perpendicular
to this plane, and find where it intersects the plane.

2. Relevant equations

\hat{n} \cdot (r~-~r_{0}) = 0

3. The attempt at a solution
Ok, I found two vectors, on joining the points (2,−1, 0) and (−5,−3, 1) and another joining the points (3, 5,−1) and (0, 3,−2), seeing as they are parallel they will share the same normal vector \hat{n}.
Those two vectors are certainly NOT parallel. Why would you think they are? The vector connecting the first two points is <7, 2, -1> and, connecting the last two, <3, 2, 1>. Since one is not a multiple of the other, they are not parallel.

Then using \hat{n} \cdot (r~-~r_{0}) = 0 with r being (-5,-3, 1) and r_{0} being (2,-1,0). To get the vector form of the equation plane.

but I am not exactly sure what points to use at this stage.

and I haven't attempted b as yet.

thanks!

[phagist_]

Ok yep, I assumed that because <3, 2, 1> was parallel to the plane it would be parallel to a vector lying on that plane <7, 2, -1>.

So where exactly do I go from here?
thanks.

[Defennder]

It isn't parallel, which is a good thing since otherwise that vector operation wouldn't work! For any plane there is an infinite number of vectors which are parallel to it; just think of rotating an arrow on a board by 360 degrees, every possible vector corresponding to the arrow direction is parallel to the plane. It is precisely that it is non-parallel that you can apply the said vector operation to get the normal vector.