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I have been having difficulty with this problem. Any help would be 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.

Questioner:Billy

Country:Quebec, Canada

Category:Advanced Math

Private:No

Subject:Linear algebra poit to a line

Question:distance between a point

I have been having difficulty with this problem. Any help would be 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.

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Sorry about the delay. This was a busy weekend.

Try the 'vector projection'.

Call the vector from Q to P = vector QP.

V is the vector along L and is <-2,-1,5>.

Now the vector projection of QP onto V is given by:

(see en.wikipedia.org/wiki/Vector_projection )

QP dot V

Proj = ---------- V

|V|^2

Now QP = <5, 5, 0 >

QP dot V = <5, 5, 0 > dot < -2,-1,5> = -10 - 5 + 0 = - 15

| V | ^2 = 4 + 1 + 25 = 30

- 15

Proj = ---------- <-2, -1, 5 > = <-1, -1/2, 5/2 >

30

OK. Now you can finish up:

1. Add that projection to Q. That gives you the point where PR crosses L. Call that point A.

2. Take the vector PA (from P to A)

3. Double it.

4. Add that to P and you have R.

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