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?

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- June 2nd 2008, 08:02 PMtheowneConverting 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? - June 2nd 2008, 08:07 PMmr fantastic
- June 2nd 2008, 08:11 PMtheowne
It is a plane. That's the whole question.

- June 2nd 2008, 08:18 PMsean.1986
Do you mean write it as a vector?

- June 2nd 2008, 08:19 PMdineshdileep
- June 2nd 2008, 08:37 PMmr fantastic
There are two ways of expressing the vector equation of a plane:

1. where is a vector normal to the plane and is the position vector of a point in the plane.

2. where and are two non-parallel vectors which lie in the plane and 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. - June 2nd 2008, 08:40 PMtheowne
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

- June 2nd 2008, 08:41 PMChris L T521
- June 2nd 2008, 08:44 PMmr fantastic
*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 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 .... - June 2nd 2008, 09:21 PMtheowneQuote:

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