# Thread: does the point lie on the plane spanned by two vectors?

1. ## does the point lie on the plane spanned by two vectors?

Can you solve (5,8,12) = s(1,1,1)+t(1,2,3) for s and t? In other words does point (5,8,12) lie on the plane through the origin that is spanned by the vectors (1,1,1) and (1,2,3).

I'm new with vectors and not really sure where to start. I don't understand the question, how do 2 vectors compose a plane?

2. Originally Posted by superdude
Can you solve (5,8,12) = s(1,1,1)+t(1,2,3) for s and t? In other words does point (5,8,12) lie on the plane through the origin that is spanned by the vectors (1,1,1) and (1,2,3).

I'm new with vectors and not really sure where to start. I don't understand the question, how do 2 vectors compose a plane?

Open up parentheses in the right hand and compare entry-entry:

$\displaystyle (5,8,12)=s(1,1,1)+t(1,2,3) = (s+t,s+2t,s+3t) \Longleftrightarrow 5=s+t\,,\,\,8=s+2t$ $\displaystyle \,,\,\,12=s+3t$

Solve the above system and see if it is congruent (btw, it is not and thus the vector doesn't belong to the plane. Check this)

Tonio

3. how did you know how to do that? I mean what's the general rule?
Thanks for the response!

4. Originally Posted by superdude
how did you know how to do that? I mean what's the general rule?
Thanks for the response!

The most general, basic and indeed simple rule is: two vectors in $\displaystyle \mathbb{R}^n$ are equal iff their entries are equal correpondingly.

Thus, for example, $\displaystyle (a,b)=(c,d) \Longleftrightarrow a=c\,\,and\,\,b=d$

That's how we knew that, for instance, $\displaystyle 5 =s+t$

The rest is the definition of how to sum vectors and multiply them by scalars: you MUST have studied this otherwise you wouldn't receive exercises like the one you asked about.

Tonio