Plane equation given two points and distance to parallel line

[queiroz]

Homework Statement

Find the plane through the points P = (1, 1, −1) and Q = (2, 1, 1) and parallel to a line r
{(1, 0, 2) + t(1, 0, 2), t ∈ R} with distance 1 to the plane

Homework Equations

A=(a1,a2,a3),B(b1,b2,b3),A and B are vectors. AxB= det[i j k;a1 a2 a3;b1 b2 b3] ";" means change of line.

A.B= a1b1+a2b2+a3b3

vector projector of A in B is proj(B)A = (A . B /||B||² ) B

norm of vector projector of A in B is ||proj(B)A|| = ||(A . B /||B||² ) B||= (|A.B|/||B||²)||B||= |A.B|/||B||

The Attempt at a Solution

I have the vector PQ=Q-P=(1,0,2) and named a point M(a,b,c) which gives me the vector PM=(a-1,b-1,c+1).
PQxPM will give me a vector perpendicular to the plane (scalar multiple of vector Normal N).
N= (-2b+2,-c+2a-3,b-1)
And the distance of the line to the plane is equal to the norm of the vector projection PP1 in the vector Normal when P1 is a point in the line and P in the plane. P1(1,0,2) and P(1,1,-1)
PP1=P1-P=(1,0,2)-(1,1,-1)=(0,-1,3)

||proj(N)PP1||= |PP1 . N|/||N||=1
{|(0,-1,3).(-2b+2,-c+2a-3,b-1)|/√[(-2b+2)²+(-c+2a-3)²+(b-1)²]}=1

And I get

3b-2a+c= √[(-2b+2)²+(-c+2a-3)²+(b-1)²] an equation with three unknown numbers.
I don't know how to proceed to find a,b and c.

Thanks

 

[mighty2000]

for your post! Your approach to solving this problem is correct so far. To find the values of a, b, and c, we can use the fact that the plane is parallel to the given line r. This means that the normal vector of the plane must be perpendicular to the direction vector of the line, which is (1, 0, 2). This gives us another equation:

N . (1, 0, 2) = 0
(-2b+2,-c+2a-3,b-1) . (1, 0, 2) = 0
-2b+4+2b-1=0
3=0

This is a contradiction, which means that there is no solution for a, b, and c. This makes sense since we are trying to find a plane that is both parallel to a line and has a distance of 1 to the plane. This is not possible, as a plane parallel to a line will either be the same plane or infinitely far away from it. Therefore, there is no unique solution for this problem.