Telling if three vectors in the same plane? Help

[samh]

Telling if three vectors in the same plane?? Help!

Hey guys I've got a question about vectors that's confusing me. Here's what my book says:

Two vectors are called independent in R² if they are not multiples of each other. Three vectors in R³ are called independent if they are not in the same plane.

Example: (1,0,1), (0,1,0) and (1,1,1) are not independent (since (1,0,1) + (0,1,0) = (1,1,1)).

I don't understand the example... How do you tell if three vectors in R³ are in the same plane? Why does adding the first two and getting (1,1,1) show this? I'm real rusty on all this plane stuff... Any help is appreciated.

 

[neutrino]

Well, if one of them is NOT in the same plane formed by the other two, then you can't completely describe that vector in terms of the other two. The simplest example is the standard unit vectors in cartesian coordinates : i, j and k. You can never express k as c₁i+c₂j, for contants c₁ and c₂. Such vectors are known as linearly independent.

Think of it this way, when you add two vectors (or scalar multiples of them), the resulting vector goes from the "tail" of one to the "head" of the other. The "tail", "head" and the point where these two vectors meet all form a unique plane.

 

[bomba923]

I don't understand the example... How do you tell if three vectors in R3 are in the same plane? Why does adding the first two and getting (1,1,1) show this? I'm real rusty on all this plane stuff... Any help is appreciated.

Any three given vectors <x₁,y₁,z₁>, <x₂,y₂,z₂>, <x₃,y₃,z₃>
are coplanar iff <x₁,y₁,z₁> $\times$ <x₂,y₂,z₂> • <x₃,y₃,z₃> = 0

 

[HallsofIvy]

Three vectors are in the same plane (coplanar) if and only if they are "dependent". That is, there is some linear combination Av₁+ Bv₂+ Cv₃= 0 with not all of A, B, C equal to 0. Equivalently, one vector can be written as a linear combination of the other two.

Bomba923's last post "Any three given vectors <x1,y1,z1>, <x2,y2,z2>, <x3,y3,z3> are coplanar iff <x1,y1,z1>$\times$<x2,y2,z2> • <x3,y3,z3> = 0" is a good way of checking. Even more simply that calculation can be written as a single determinant:
$$\left|\begin{array}{ccc}x_1 & y_1 & z_1 \\ x_2 & y_2 & z_2 \\ x_3 & y_3 & z_3 \end{array}\right|$$

 

[robphy]

(adding to the last previous two posts...)

that HallsofIvy's determinant (and bomba923's scalar triple product) can be intrepreted as the volume enclosed by a box (parallelepiped) with sides given by your three vectors. If that volume is zero, the box is "degenerate"... and the three vectors lie on a common plane.

 

[HallsofIvy]

(adding to the last previous two posts...)

that HallsofIvy's determinant (and bomba923's scalar triple product) can be intrepreted as the volume enclosed by a box (parallelepiped) with sides given by your three vectors. If that volume is zero, the box is "degenerate"... and the three vectors lie on a common plane.

Good point!