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 Post subject: Mirror image of plane
PostPosted: Tue, 3 Jun 2003 03:32:54 UTC 
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Find the coordinates of the mirror image of A(3, 1, 2) when reflected in the plane x+2y+z = 1

:lol:


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PostPosted: Tue, 3 Jun 2003 04:05:33 UTC 
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Joined: Mon, 19 May 2003 19:55:19 UTC
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Hello, Way Below Zero!

The plane x + 2y + z = 1 has the normal vector <1,2,1>.

We want the line through A(3,1,2) perpendicular to the plane.
We have point A(3,1,2) and direction <1,2,1>,
so the parametric equations are:
x = 3 + t, y = 1 + 2t + z = 2 + t

Find the intersection of this line and the plane:
(3 + t) + 2(1 + 2t) + (2 + t) = 1
Hence, t = -1.
They intersect at B(2,-1,1).

Here's where I used some intuition instead of formulas.

Going from A(3,1,2) to B(2,-1,1), we note the changes in the coordinates:
x decreases by 1, y decreases by 2, z decreases by 1.

If we continue in the same manner to point C, we get C(1,-3,0).

Therefore, the image of A is (1,-3,0).


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