Consider the two lines x1x2x3 123 t111 and x1x2x3 234 s1

Consider the two lines [x1,x2,x3] = [1,2,3] + t[1,1,1] and [x1,x2,x3] = [2,3,4] + s[1,2,1]. Given that the lines intersect,

(a) Find the parametric equation of the plane containing the two lines.

(b) Find the normal equation of the plane containing the two lines.

Solution

a) x1 = 1 + t ; x2 = 2 +t ; x3 = 3 +t

x1 = 2 +s ; x2 = 3 +2s ; x3 = 4 +s

the plane contains the two lines, their direction vectors are parallel to the plane. Hence their cross product will be a normal vector.

d1 = < 1 ,1, 1> ; d2 = <1, 2, 1 >

cross product : d1 x d2 = < -1. 0 ,1 >

We need one point from any of the two lines: ( 1, 2, 3)

Equation Of Plane : -1(x -1) +0(y -2) + 1( z-3) =0

-x +1 +z -3 =0 ----> z-x = 2

x = z -2.

Let z = t as the secondary variable. x = t-2

parametric form : ( t - 2 , t)

(b) Normal form equation :

x - z +2 =0

Sqrt( a^2 +b^2 +c^2) = sqrt(1^ +2^2) =+/- sqrt5

Normal form of equation : ( x -z +2)/+/- sqrt5

x(/+/-sqrt5) -z/(+/-sqrt5) +2/(+/-sqrt) =0