Finding Position Vectors for Lines: Using Equations and Free Choice

[synkk]

Question: http://gyazo.com/54547a60ab590e355a4dc9a6af54650a

part (b), I've found the "u" as the question has labelled it

I have a question when trying to find a position vector that the line passes through:

we have 2x + 2y - z = 9 and x - 2y = 7

would it be OK to let z = 0 and then to solve to find y and x? or to add them and let z = 0? or any other number? I just don't understand *if* this is allowed, and if it is, then why?

 

[Mark44]

Question: http://gyazo.com/54547a60ab590e355a4dc9a6af54650a

part (b), I've found the "u" as the question has labelled it

I have a question when trying to find a position vector that the line passes through:

we have 2x + 2y - z = 9 and x - 2y = 7

would it be OK to let z = 0 and then to solve to find y and x? or to add them and let z = 0? or any other number? I just don't understand *if* this is allowed, and if it is, then why?

You should be able to find u and v directly from the equations of the planes. For example, the vector <A, B, C> is a normal to the plane whose equation is Ax + By + Cz = D.

If you solve the system of equations represented by your two planes, the solution will be the line of intersection. If you set z = 0, you'll end up with a single point that happens to be on the line of intersection. I'm not sure if that's what you want.

 

[synkk]

You should be able to find u and v directly from the equations of the planes. For example, the vector <A, B, C> is a normal to the plane whose equation is Ax + By + Cz = D.

If you solve the system of equations represented by your two planes, the solution will be the line of intersection. If you set z = 0, you'll end up with a single point that happens to be on the line of intersection. I'm not sure if that's what you want.

yup that's what I want, thanks.