Converting scalar equation to vector equation

• Jun 2nd 2008, 08:02 PM
theowne
Converting scalar equation to vector equation
If you're given a scalar equation

2x-y+3z-24=0

And you're asked to find a vector equation, how can you do this?
• Jun 2nd 2008, 08:07 PM
mr fantastic
Quote:

Originally Posted by theowne
If you're given a scalar equation

2x-y+3z-24=0

And you're asked to find a vector equation, how can you do this?

Give the whole question.
• Jun 2nd 2008, 08:11 PM
theowne
It is a plane. That's the whole question.
• Jun 2nd 2008, 08:18 PM
sean.1986
Do you mean write it as a vector?

$\displaystyle [2, -1, 3]^T = 24$
• Jun 2nd 2008, 08:19 PM
dineshdileep
Quote:

Originally Posted by theowne
If you're given a scalar equation

2x-y+3z-24=0

And you're asked to find a vector equation, how can you do this?

its a plane's equation,i think you can write it as

(2i-j+3k).(xi+yj+zk)=24
now you can normalize the vector and you will have it in the standard form.
• Jun 2nd 2008, 08:37 PM
mr fantastic
Quote:

Originally Posted by theowne
If you're given a scalar equation

2x-y+3z-24=0

And you're asked to find a vector equation, how can you do this?

There are two ways of expressing the vector equation of a plane:

1. $\displaystyle (\vec{r} - \vec{r_0}) \cdot \vec{n}$ where $\displaystyle \vec{n}$ is a vector normal to the plane and $\displaystyle \vec{r_0}$ is the position vector of a point in the plane.

2. $\displaystyle \vec{r} = \vec{r_0} + s \vec{u} + t \vec{v}$ where $\displaystyle \vec{u}$ and $\displaystyle \vec{v}$ are two non-parallel vectors which lie in the plane and $\displaystyle \vec{r_0}$ is the position vector of a point in the plane. s and t are scalar parameters that can have in real value.

So choose which form of answer you want to give. Then extract the required vectors from the given Cartesian equation and substitute.
• Jun 2nd 2008, 08:40 PM
theowne
Yes, I know the vector form, how do I extract the direction vectors from the scalar form? All I can see from it is a normal vector, but that could be for any number of direction vectors can't it
• Jun 2nd 2008, 08:41 PM
Chris L T521
Quote:

Originally Posted by theowne
If you're given a scalar equation

2x-y+3z-24=0

And you're asked to find a vector equation, how can you do this?

Are they asking for the equation of the normal vector to the plane??
• Jun 2nd 2008, 08:44 PM
mr fantastic
Quote:

Originally Posted by theowne
Yes, I know the vector form, how do I extract the direction vectors from the scalar form? All I can see from it is a normal vector, but that could be for any number of direction vectors can't it

A vector normal to the plane ax + by + cz = d is n = ai + bj + ck. You should know this.

You should be able to find a point in the plane and hence get the position vector of this point (=> the position vector $\displaystyle \vec{r_0}$ of a point in the plane).

And, if you want two non-parallel vectors lying in the plane, yuo should be able to find three non-collinear points A, B and C lying in the plane and then construct the vectors u = AB and v = AC, say ....
• Jun 2nd 2008, 09:21 PM
theowne
Quote:

A vector normal to the plane ax + by + cz = d is n = ai + bj + ck. You should know this.
Yes, which is why I said "All I can see from it is a normal vector". Regardless, thanks for the end of the post.