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## 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

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