Find the equation of the two planes that bisect the angle between the lines.....?

Question

r=<4,1,2> + t<-2,1,2> and r = <2,1,1> + s<2,0,1>. (A plane bisects the angle between two lines if one line is the reflection through the plane of the other line.)

Answer 1

These lines meet at (4,1,2) since the 3 equations for the 3 coordinates have a solution t=0, s=1. Reparameterize the lines through their intersection using unit directions: The two new lines are:

r1 = (4,1,2) + t*(-2,1,2)/3 and

r2 = (4,1,2) +s*(2,0,1)/sqr(5).

Then the distance from a point on a line to (4,1,2) is just s or t (actually its absolute value). We can visualize the plane between them as the locus of points equidistant from two points, one on the first line and the other on the second line, an equal distance from (4,1,2). Actually the first plane is obtained using s=t and the second one using s=-t. In fact the t will cancel in the following equations, so we may take t=1. First let s=t=1 to solve for a point (x,y,z) on the first plane equidistant from the two points r1 and r2:

(x - 4 + 2/3)^2 + (y - 1 - 1/3)^2 + (z - 2 -2/3)^2 =

(x - 4 - 2/sqr(5))^2 + (y - 1)^2 + (z - 2 - 1/sqr(5))^2.

Multiplying this out and cancelling common terms we get

2(x-4)*2/3 + (2/3)^2 - 2(y-1)*1/3 + (1/3)^2 - 2(z-2)*2/3 +(2/3)^2 =

-2(x-4)*2/sqr(5) + (2/sqr(5))^2 - 2(z-2)*1/sqr(5) +(1/sqr(5))^2

Simplify this to get the first plane. Repeat above with s=-1,t=1 to get the second plane.

Answer 2

It's a sphere. Say: Sqrt( (2-x)^2 + (-1-y)^2 + (-1-z)^2) = Sqrt( (1-x)^2 +(2-y)^2 + (-7-z)^2) Just solve that. Start out by squaring both sides, then expand the squared terms in the parentheses, then get everything to one side. I had this question on my calc final. It's a pain in the butt.