Thread: 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?

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.

It is a plane. That's the whole question.

Do you mean write it as a vector?

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.

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. 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.

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

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??

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 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 ....

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.