# Vectors. Did not know where to post this one.

• Jul 22nd 2012, 12:49 AM
pratique21
Vectors. Did not know where to post this one.
Okay so I read about the equation of a plane that is parallel to 2 vectors and passes through a point(say 'P').

And i am thinking, if we have 2 vectors to which the plane is parallel, why do we need another point? Or should that point be a parameter in determining the equation of a plane.

So I argue that there must be two cases:-
1) The plane that contains the 2 vectors to which it is parallel does not contain the point P - possible and absurd. Say the plane is parallel to $\frac{x}{1} = \frac{y}{0} = \frac{z}{0}$ and $\frac{x}{0} = \frac{y}{1} = \frac{z}{0}$ and I want it to pass through say (0,0,4). I cannot have such a plane.

(2) So the point has to lie on that plane. Say the vectors are ~a and ~b . So the k1a + k2b gives us the entire plane for different values og k1 and k2.

My question is that if it also has to pass through P (with ~OP=P) isn't k1a + k2b a sufficient equation ? I think so.

But my book says differently. It says that the equation will become P+k1a + k2b .

• Jul 22nd 2012, 01:58 AM
Reckoner
Re: Vectors. Did not know where to post this one.
There are infinitely many planes that are parallel to any two given vectors. Knowing a point on the plane allows us to find a unique solution.

I don't really understand your argument. In (1), you talk about a plane parallel to the lines given by

$x = t, y = 0, z = 0$

and

$x = 0, y = t, z = 0$

(at least I think that's what you mean - I changed to a parametric form to avoid the division by zero). Then you say that there is no such plane that passes through (0,0,4), but this is clearly untrue: The plane $z = 4$ meets both conditions.
• Jul 22nd 2012, 05:00 AM
pratique21
Re: Vectors. Did not know where to post this one.
I get it. Really dumb of me. I guess i misunderstood 'parallel'. I thought it meant the plane also had to contain those vectors. Ahh! Silly of me. Thanks.
Yes, thats what i meant with those lines but the argument is now kaput. :)
• Jul 22nd 2012, 05:19 AM
earboth
Re: Vectors. Did not know where to post this one.
Quote:

Originally Posted by pratique21
Okay so I read about the equation of a plane that is parallel to 2 vectors and passes through a point(say 'P').

And i am thinking, if we have 2 vectors to which the plane is parallel, why do we need another point? Or should that point be a parameter in determining the equation of a plane.

So I argue that there must be two cases:-
1) The plane that contains the 2 vectors to which it is parallel does not contain the point P - possible and absurd. Say the plane is parallel to $\frac{x}{1} = \frac{y}{0} = \frac{z}{0}$ and $\frac{x}{0} = \frac{y}{1} = \frac{z}{0}$ and I want it to pass through say (0,0,4). I cannot have such a plane.

(2) So the point has to lie on that plane. Say the vectors are ~a and ~b . So the k1a + k2b gives us the entire plane for different values og k1 and k2.

My question is that if it also has to pass through P (with ~OP=P) isn't k1a + k2b a sufficient equation ? I think so.

But my book says differently. It says that the equation will become P+k1a + k2b .

2. I've attached a sketch to illustrate what I meant: Only the greyed plane contains the point P. To reach any point on this plane you start at the origin, go to P and then a multiple of $\vec u$ and a multiple of $\vec v$.