Finding x-value for vector in a plane defined by two non-parallel vectors

[teejmd]

Homework Statement

Any two vectors that are not parallel define a plane. So p = i + j -k and q = 2i +j define a plane. For what values of x is the vector r = xi + j + k in this plane?

Homework Equations

The Attempt at a Solution

If r* was in the plane formed by p* and q*, would r* be a scalar multiple of p* and q* to be certain it was in the plane? Solving that I get the only x value being 2. I'm not sure if this was the right approach and would appreciate some help! thanks

 

[LCKurtz]

Homework Statement

Any two vectors that are not parallel define a plane. So p = i + j -k and q = 2i +j define a plane. For what values of x is the vector r = xi + j + k in this plane?

Homework Equations

The Attempt at a Solution

If r* was in the plane formed by p* and q*, would r* be a scalar multiple of p* and q* to be certain it was in the plane? Solving that I get the only x value being 2. I'm not sure if this was the right approach and would appreciate some help! thanks

Two vectors don't determine a plane, they just determine the orientation of a plane. Any plane parallel to that plane would also be parallel to the same vectors. So what you are really asking is for what x is r = xi + j + k parallel to the plane.

Hint: Any vector perpendicular to the normal vector for the plane is parallel to the plane.

Alternate hint: The parallelepiped formed by p,q, and r would have volume 0.

 

[Dazz4C]

Two vectors don't determine a plane, they just determine the orientation of a plane. Any plane parallel to that plane would also be parallel to the same vectors. So what you are really asking is for what x is r = xi + j + k parallel to the plane.

Hint: Any vector perpendicular to the normal vector for the plane is parallel to the plane.

Alternate hint: The parallelepiped formed by p,q, and r would have volume 0.

Two vectors form a plane in between them...question is asking for when r is coplanar to q * p.

Correct me if I'm wrong. Can be solved by using 0 = r*(q*p)

 

[teejmd]

Two vectors don't determine a plane, they just determine the orientation of a plane. .

Eh? That's what the actual question says and that's what I've been taught, anyone able to clarify?

Anyway what I've ended up doing is saying if r* was in the plane, it's cross product with either p* or q* (p* x r* etc) would be parallel to p* x q* (p* q* cross product). i.e. they are both normal to the plane formed by p* and q*, making them parallel => making them scalar multiples of each other. Vector parallel still confuse me.

I set [(p* x q*) x (p* x r*)] = 0 vector so that they are parallel and solved for x, getting x=3.

 

[HallsofIvy]

If you have been taught that all vectors have their "tails" at the origin, you have been taught wrong. If the problem really says "Any two vectors that are not parallel define a plane", it is wrong. (If it said "position vectors", that would be a different matter- although I dislike that phrase.)

The two vectors p = i + j -k and q = 2i +j define the planes x+ 2y- k= C for any C. If C= 0, then that plane includes the origin and so would include the "position vectors". One can determine if three vectors, u, v, and w, lie in the same plane by looking at their "triple product", $u\cdot(v\times w)$. They will be in the same plane if and only if that is 0.

(I just realized this is a physics problem, not a math problem. Ah, well, physicists tend to be unfortunately loose with mathematics terminology.)