Thread: Vectors and Planes

The problem is:

In the equation of the plane: Ax+By+Cz+D=0 spanned in R^3 by a=(1,3,0) and b=(0,1,2) find the parameters A,B,C and D.

I don`t have any clue about how to solve this problem. Does anybody have a good suggestion?

My second problem is:

Let "a", "b" and "c" be vectors in R^3. If "a" belongs to the plane spanned by "b" and "c", can these three vectors be linearly independent?

My answer to this would be, no they are always linearly dependent because "a" can be written in any case as a combination of "b" and "c" - is my answer correct?

Originally Posted by infernalmich

The problem is:
In the equation of the plane: Ax+By+Cz+D=0 spanned in R^3 by a=(1,3,0) and b=(0,1,2) find the parameters A,B,C and D.

My second problem is:
Let "a", "b" and "c" be vectors in R^3. If "a" belongs to the plane spanned by "b" and "c", can these three vectors be linearly independent?
My answer to this would be, no they are always linearly dependent because "a" can be written in any case as a combination of "b" and "c" - is my answer correct?

Your second answer is correct.

In the first you should have been given a point on the plane. Were you?

Thank you for the help.

Sadly, no point on the plane has been given. I posted above all the information which was given about the problem.

need some help in this please!!
is given the function {y=sin2x/x for x different from 0} or {y=2a for x=o}. find "a" if this function is continuous everywhere.
please help

Originally Posted by besimdervishi

need some help in this please!!
is given the function {y=sin2x/x for x different from 0} or {y=2a for x=o}. find "a" if this function is continuous everywhere.
please help

1. If you have a new question please start a new thread.
2. Your question belongs to Pre-Calculus.

3. Here are the steps you should do:

Determine $\displaystyle \displaystyle{y_0=\lim_{x \to 0}\left(\frac{\sin(2x)}{x} \right)}$. Use $\displaystyle \displaystyle{\sin(2x) = 2\sin(x) \cos(x)}$

Use the value of the limit to determine a: $\displaystyle y_0 = 2a$ Solve for a.

Keep in mind that this composed function is not differentiable at x = 0.

I still don't know how to answer the first question:

In the equation of the plane: Ax+By+Cz+D=0 spanned in R^3 by a=(1,3,0) and b=(0,1,2) find the parameters A,B,C and D.

Maybe there exists a theorem or something which I don't know. The only thing I can see is that here the three vectors are dependent.
The vectors are dependent so there has to be a way to write a third vector as a linear combination of the other two.
x₁=1 x₂=0 x₃=?
y₁=3 y₂=1 y₃=?
z₁=0 z₂=2 z₃=?
But I don't know how to continue!

Since the problem says "spanned" by, the plane must be a subspace of $\displaystyle R^3$ and so must contain (0, 0, 0), the "0 vector".

The first thing we can do is divide through by, say, A and write the equation as x+ B'y+ C'z+ D'= 0. Setting x= y= z= 0 gives D'= 0. Set x=
1, y= 2, z= 0 and then x= 0, y= 1, z= 2 to get two equations for B' and C'.

Originally Posted by infernalmich

I still don't know how to answer the first question:

In the equation of the plane: Ax+By+Cz+D=0 spanned in R^3 by a=(1,3,0) and b=(0,1,2) find the parameters A,B,C and D.

Maybe there exists a theorem or something which I don't know. The only thing I can see is that here the three vectors are dependent.
The vectors are dependent so there has to be a way to write a third vector as a linear combination of the other two.

Those vectors are not dependent. There are independent.
The equation depends on knowing a point on the plane.
If $\displaystyle \left( {{x_0},{y_0},{z_0}} \right)$ is a point on the plane, then one solution is:
$\displaystyle A=6,~B=-2,~C=1,~\&~D = 6{x_0} - 2{y_0} + {z_0}$.

Ah, now I understand a little bit better, thanks.

The vectors are not dependend because the R3-space contains at least three independent vectors, is this true?

But I don`t really understand how you got the values for A B C and D? Could you explain it a little bit more in detail please?

Originally Posted by infernalmich

Ah, now I understand a little bit better, thanks.
The vectors are not dependend because the R3-space contains at least three independent vectors, is this true?
But I don`t really understand how you got the values for A B C and D? Could you explain it a little bit more in detail please?

Those values come from $\displaystyle a\times b$.

sorry today I am slow on the uptake!

What is the reasoning behind this calculation?

As HallsofIvy suggested we have to add the (0,0,0).

So we have a plan spanned by three vectors:
(1,3,0)
(0,1,2)
(0,0,0)

now depending on this three vectors we have to find values for the parameters in this equation:
Ax+By+Cz+D=0

Why is the cross product of the vectors the result of the equation?

Originally Posted by infernalmich

now depending on this three vectors we have to find values for the parameters in this equation:
Ax+By+Cz+D=0
Why is the cross product of the vectors the result of the equation?

What do you know about vector geometry?
It seem to be very little, or else you would not have asked that question!

I definetly don't know enough about vector geometry yet. Therefore I would be very thankful if someone could explain this to me.

Originally Posted by infernalmich

sorry today I am slow on the uptake!

What is the reasoning behind this calculation?

As HallsofIvy suggested we have to add the (0,0,0).

So we have a plan spanned by three vectors:
(1,3,0)
(0,1,2)
(0,0,0)

now depending on this three vectors we have to find values for the parameters in this equation:
Ax+By+Cz+D=0

Why is the cross product of the vectors the result of the equation?

It isn't necessary to use the cross-product:

1. To define a plane in R³ you need to have a point (here: (0, 0, 0)) and two independent directions which are here $\displaystyle \vec u = \langle 1,3,0 \rangle$ and $\displaystyle \vec v = \langle 0, 1, 2 \rangle$

2. The equation of the plane is:

$\displaystyle \vec r = \langle x,y,z \rangle =\langle 0,0,0 \rangle + s \cdot \langle 1,3,0 \rangle + t \cdot \langle 0,1,2 \rangle$

or

$\displaystyle \left|\begin{array}{rcl}x&=&s \\ y &=& 3s+t \\ z &=& 2t \end{array}\right.~\implies~s = x~\wedge~t = \frac12 z$

3. Plug in the terms for s and t into the 2nd equation:

$\displaystyle y = 3x+\frac12z~\implies~ 0=6x-2y+z$

Thank you! How do I get the value of D? Is it simply 0?