vectors in R^3?

Question

the vectors (1,2,2) and (2,-2,1) in R^3 determine a plane P through the origin

True or false, the vector (3,0,3) lies in P
True or false, vector (2,1,-2) lies in P

im sure i can figure this out myself I just dont know the process on how to check if they belong in P.
if somone can explain the process then i should be good

they dont = 0...

thanks i get it now

Answer 1

1) Two directional vectors are not enough to define a plane. You also need a point. The given vectors have magnitude and direction but are not specifically located by any point. So a plane thru the origin could contain those vectors, but so could a plane that does not go thru the origin.

2) Is the vector w = <3, 0, 3> coplanar with
u = <1, 2, 2> and v = <2, -2, 1>?

If w can be written as a linear combination of u and v, then it is coplanar with them.

w = au + bv

Solve for a and b.

3 = a + 2b
0 = 2a - 2b
3 = 2a + b

From the second equation.

0 = 2a - 2b
2a = 2b
a = b

Plug into the first equation to solve for a.

3 = a + 2b
3 = a + 2a
3 = 3a
a = 1

a = b
b = 1

We used the first two equations to solve for a and b. Check to make sure it also fits the third equation. It does.

w = u + v

So w is coplanar with u and v.
_________________

3) Is the vector w = <2, 1, -2> coplanar with
u = <1, 2, 2> and v = <2, -2, 1>?

If w can be written as a linear combination of u and v, then it is coplanar with them.

w = au + bv

Solve for a and b.

2 = a + 2b
1 = 2a - 2b
-2 = 2a + b

Add the first two equations together.

3 = 3a
a = 1

Plug into the first equation to solve for b.

2 = a + 2b
2 = 1 + 2b
1 = 2b
b = 1/2

We used the first two equations to solve for a and b. Check to make sure it also fits the third equation. It does not.

So w is NOT coplanar with u and v. It cannot lie in the same plane.
_________________

Answer 2

The answer to the first question is true since (3, 0, 3) is the sum of the two given vectors. The sum of vectors in a plane is in the plane.
For the second part, you have to find the eq of the plane.
take the first vect crossed into the second which is
6i + 3j - 6k or we can divide by 3 and use 2i + j - 2k
The equation of the plane is a(x - xo) + b . . .
where a, b, c are the components of a vect perpend to the plane, which is the one we found, and xo, yo, zo is a point on the plane - the origin in this case. So we get
2x + y - 2z = 0.
So the vector given (2, 1, -2) doesn't lie in P it's perp. to P.

Answer 3

To check if three vectors A, B and C are on the same plane, you only need to check if (A x B)∙C = 0, the volume of a rectangular formed by the three vectors.

Answer 4

enable's see if linear mixtures of those vectors can create 3 needless to say autonomous vectors. 7u + 2v - 6w = a million 0 0 = i w - u = 0 a million 0 = j - 3u - v + 3w = 0 0 a million = ok The vectors i, j, and ok are linear mixtures of u, v, w and clearly autonomous and span R³. consequently u, v, and w additionally span R³ and could be seen a base.

Answer 5

Ya ya, check if they are coplaner!