# Thread: To find the parametric equation of a line in Euclidian space knowing the following.

1. ## To find the parametric equation of a line in Euclidian space knowing the following.

If known that a line "m" passes through the point P (2,-1,3) in 3d vector space

and knowing it is parallel to the plane pi (x-y+z=0)

and cuts the line l { (1-2g)i + (3+2g)j +(2-g)k } (where "g" is just a prameter, i don't have "lambda" on this keyboard )

how would you determine the equation of the line "m"?

Help very appreciated thanks!!

2. Find the plane $\pi_1$ parallel to $\pi$ and contains $P$ . Find the intersection point $Q$ between $\pi_1$ and the line $l$ . The solution is the line passing trough $P$ and $Q$ .

Fernando Revilla

3. Here's how I would do it. Any line through (2, -1, 3) can be written in the form x= At+ 2, y= Bt- 1, z= Ct+ 3 where <A, B, C> is a vector in the direction of the line. Since the line is to be parallel to the plane x- y+ z= 0, <A, B, C> must be perpendicular to the normal to that plane, <1, -1, 1>. That is, <A, B, C>.<1, -1, 1>= A- B+ C= 0. Since that is true for any multiple of the vector, we can, for example, take A= 1 and have 1- B+ C= 0 so that B= C+ 1. That is, we can write the equation of the line as x= t+ 2, y= (C+ 1)t- 1, and z= Ct+ 3 and we only need to find C. The last condition is that the line intersect the line given by x= 1- 2g, y= 3+ 2g, z= 2- g. Putting those into the equation of the line gives 1- 2g= t+ 2, 3+ 2g= (C+ 1)t- 1, and 2- g= Ct+ 3. From the first, t= -1- 2g. Putting that into the second equation gives two equations for C and g. Of course, C gives you the equation of the line you want.