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I have no idea how to approah this problem. I have been tryng to complete it for 3 days but with no success. Any help would be greatly appreciated.

Consider a line L which goes by a point Q and is parallel to a vector v. Let P be any point and let R be its mirror image across the line L.

(a) Find a formula for R starting from P, Q and vector v. Sketch your construction.

(b) Assume that P(1, 6, 2), Q(-4, 1, 2) and vector v = <-2,-1,5>. Find R.

(a) To find the point opposite a line, let the line be at y = A. Let the point P be at (x,B).

The line is B-A units over the line. To get the point that is that far below, take the same x with A - height. That is, (x,A-(B-A)) = (x,2A-B), where x is from P, A is from Q, and B is the y height of P.

For (b), find vector PQ. Find the magnitude of PQ, and divide by the magnitude, so it has magnitude we now have a vector with magnitude 1. Find the magnitude of v, and divide v by that scalar so the magnitude of the vector is 1. Compute the x, y, and z differences, and then subtract them from the vector QL with magnitude 1.

Determine the length of QP and multiply this new vector by that amount to get the QR.

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