# Thread: eqn of line from a plane and another lineFind the

1. ## eqn of line from a plane and another lineFind the

find the parametric eqn for the line lying on the plane x+y+z=3 and intersecting the line x=2+t , y=1-t, z=-3t perpendicularly.

I found the point of intersection to be (2,1,0).

how do i get the other point? I was thinking that if the plane would intersect the line perpendicularly, wouldnt any other point on that plane, that satisfies x+y+z = 3 be enough to form the parametric eqn that i need?

eg. B (1,1,1)? or (0,0,3). is there a unique solution?

find the parametric eqn for the line lying on the plane x+y+z=3 and intersecting the line x=2+t , y=1-t, z=-3t perpendicularly.

I found the point of intersection to be P(2,1,0). <<<<< OK

how do i get the other point? I was thinking that if the plane would intersect the line perpendicularly, wouldnt any other point on that plane, that satisfies x+y+z = 3 be enough to form the parametric eqn that i need?

eg. B (1,1,1)? or (0,0,3). is there a unique solution?
1. The given line has the direction vector $\vec u = (1,-1,-3)$

2. Let $Q(a,b,c)$ denote the point in the plane who is situated on the line perpendicular to the given line. Then you know:

$\overrightarrow{PQ} \cdot \vec u = 0~\wedge~a+b+c=3$

3. Plug in the known components:

$(2-a, 1-b, -c) \cdot (1,-1,-3)=0~\implies~-a+b+3c=-1$ and $a+b+c=3$ yields $(2+c, 1-2c, c)$

4. Choose a suitable value for c. I used c = 1. That means the Point Q is Q(3, -1, 1).

The equation of the line is
$(x,y,z) = \overrightarrow{OP}+r \cdot \overrightarrow{PQ}$

5. I attached a sketch: The given line in green, the point P in red, the point Q in yellow and the line PQ in orange.