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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.

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