Answer to Question #114932 in Analytic Geometry for ANJU JAYACHANDRAN

Question

Does there pass a plane through the lines (x+4)/3 =y/2 =(z−1)/3
and x/2 =(y−1)/1 =(z+1)/1?Justify.

Answer 1

Two lines in space (R3) can be in three states: parallels, intersect, skew.

If they are skew lines, they do not lie in the same plane.

Check which type they belong to:

1)parallel

Direction vector (L1=(x+4)/3 =y/2=(z−1)/3) is of the form n=(3,2,3)

Direction vector (L2=x/2 =(y−1)/1 =(z+1)/1) is of the form m=(2,1,1)

n it is not equal to a*m (a∈R) So they are not parallel.

2)intersect

if these two lines intersect then they have a common point A=(x₁, y₁, z₁)

this point satisfies both equations of lines

(x₁+4)/3 = y₁/2 =( z₁−1)/3

x₁/2 =( y₁−1)/1 =( z₁+1)/1

Find the coordinates of the point using a system of two equations

1)x₁/2 =( y₁−1)/1

2)(x₁+4)/3 = y₁/2

1) x₁=2*y₁-2

2)(2*y₁-2+4)/3 = y₁/2

(2*y₁+2)/3 = y₁/2

4*y₁+4=3*y₁

y₁=-4

x₁=2*(-4)-2=-10

substitute these coordinates in both equations and find for each z₁

1)-4/2=( z₁−1)/3

2)-10/2=z₁+1

1)z₁=-6+1=-5

2)z₁=-5-1=-6

The solutions don't match so the lines don't intersect.

So they skew and it follows that the plane cannot pass through them.