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Question
The lines l1 passes through the point A whose position vector is i + 3j -4k and is parallel to the vector i + 2j + k. The lines l2 passes through the point B whose position vector is i + 2k and is parallel to the vector 3i + 4j + k. The point P and Q are on l1 and l2 respectively, and are such that PQ is perpendicular to both l1 and l2.
a)   Find the position vectors of P and Q.
b)   If the point (1,1,4) is the reflection of P in the plane J, find the equation of J in the form r.n=d

Sorry about the delay, but I just got to the question.

The variables to be used are a, b, and c.  To get them, a is to be used in the direction of l1,
b is to be used in the direction of l2, and c is to be used in the direction of the line connecting P (x1,y1,z1) and Q (x2,y2,z2).

a) The equation for l1 (which will use a as the variable) is (1,3,-4) + a(1,2,1).
The equation for l2 (which will use b as the variable) is (1,0,2) + b(3,4,1).

P is on l1 at (x1, y1, z1) and Q is on l2 at (x2, y2, z2).
It is given that the line PQ is perpendicular to l1 and l2.
The vector PQ is then all points (x1,y,z1) + c(x2-x1,y2-y1,z2-z1) where 0<=c<=1.

To be perpendicular to a line, the dot product of the directions needs to be zero.  Since the line we are trying to find is given as perpendicular to l1 and l2, this gives the equations
1(x2-x1) + 2(y2-y1) +(z2-z1) = 0 and 3(x2-x1) + 4(y2-y1) + (z2-z1) = 0.

For (x1,y1,z1) to be on the 1st line, that gives x1=1+a, y1=3+2a, and z1=-4+a.
For (x2,y2,z2) to be on the 2nd line, that gives x2=1+3b, y2=4b, and z2=2+b.

Now the length of the line connecting these vectors is perpendicular to both vectors, and that means it is the line of minimum length that does this.  Getting the equation of the length always gives a positive value, and squaring both sides still gives positive values.

The equation is then L²(PQ) = (3b-b)² - (4b-2a-3)² + (6+b-a)².
This reduces (do it yourself to make sure) to L²(PQ) = 6a² + 26b² - 24ab - 12b + 45.

Taking the derivative with respect to a gives d(L²)/da = 12a - 24b.
Setting that to 0 gives a = 2b.

Taking the derivative with respect to b gives d(L²)/db = 52b - 24a - 12.
Since a = 2b, that is 52b - 48b - 12 = 0.  This reduces to 4b - 12 = 0.
This means b = 3.  Since a = 2b, a = 6.

Using the equation of the lines and putting in a and b gives the points.
That means P is (1+6,3+2*6,-4+6) = (7, 15, 2) and Q is (1+3*3, 4*3, 2+3) = (10,12,5).

b) For (1,1,4) to be the reflection of P in plane J, the plane must pass through the midpoint of (1,1,4) and P = (7,15,2).  This means the plane has the point ((1+7)/2, (1+15)/2, (4+02)/2).
That is the point D is (4, 8, 3).  A perpendicular vector is given by P-D, and is
(7-4, 15-8, 2-3) = (3, 7,-1).  The equation of the plane would then be (1+3h, 1+7h, 4-h)
where h is any real number.

A note on the side: since the point is (3, 7, -1), that makes me think of 371.
A cute property of this number is the sum of the digits cubed is 371.
That is, 3^3 = 27, 7^3 = 343, and 1^3 = 1, and 27 + 343 + 1 = 371.

Questioner's Rating
 Rating(1-10) Knowledgeability = 10 Clarity of Response = 10 Politeness = 10 Comment Thank you Scott. Have a great day.

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#### Scott A Wilson

##### Expertise

I can answer any question in general math, arithetic, discret math, algebra, box problems, geometry, filling a tank with water, trigonometry, pre-calculus, linear algebra, complex mathematics, probability, statistics, and most of anything else that relates to math. I can also say that I broke 5 minutes for a mile, which is over 12 mph, but is that relevant?

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Experience in the area; I have tutored people in the above areas of mathematics for over two years in AllExperts.com. I have tutored people here and there in mathematics since before I received a BS degree back in 1984. In just two more years, I received an MS degree as well, but more on that later. I tutored at OSU in the math center for all six years I was there. Most students offering assistance were juniors, seniors, or graduate students. I was allowed to tutor as a freshman. I tutored at Mathnasium for well over a year. I worked at The Boeing Company for over 5 years. I received an MS degreee in Mathematics from Oregon State Univeristy. The classes I took were over 100 hours of upper division credits in mathematical courses such as calculus, statistics, probabilty, linear algrebra, powers, linear regression, matrices, and more. I graduated with honors in both my BS and MS degrees. Past/Present Clients: College Students at Oregon State University, various math people since college, over 7,500 people on the PC from the US and rest the world.

Publications
My master's paper was published in the OSU journal. The subject of it was Numerical Analysis used in shock waves and rarefaction fans. It dealt with discontinuities that arose over time. They were solved using the Leap Frog method. That method was used and improvements of it were shown. The improvements were by Enquist-Osher, Godunov, and Lax-Wendroff.

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Master of Science at OSU with high honors in mathematics. Bachelor of Science at OSU with high honors in mathematical sciences. This degree involved mathematics, statistics, and computer science. I also took sophmore level physics and chemistry while I was attending college. On the side I took raquetball, but that's still not relevant.

Awards and Honors
I earned high honors in both my BS degree and MS degree from Oregon State. I was in near the top in most of my classes. In several classes in mathematics, I was first. In a class of over 100 students, I was always one of the first ones to complete the test. I graduated with well over 50 credits in upper division mathematics.

Past/Present Clients
My clients have been students at OSU, people who live nearby, friends with math questions, and several people every day on the PC. I would guess that you are probably going to be one more.