Question

The shortest distance between the lines (x - 1)/0 = (y + 1)/-1 = z/1 and x + y + z + 1 = 0, 2x – y + z + 3 = 0 is : 

(1) 1/2 

(2) 1 

(3) 1/√2 

(4) 1/√3

Answer 1

(4) 1/√3

Line of intersection of planes 

x + y + z + 1 = 0 ...(1) 

2x – y + z + 3 = 0 ...(2) 

Eliminate y 

3x + 2z + 4 = 0 

x = (-2z - 4)/3 ...(3) 

Put in equation (1)

z = –3y + 1 ...(4) 

From (3) and (4) 

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

Answer 2

      x + y + z + 1 = 0

    2x - y + z + 3 = 0

------------------

    3x + 0 + 2z +4 =0

                       3x = -2z -4 

                        x =(-2z - 4) 

                             -------     ..........(i)

                                   3

     2x - y

       x + y + z + 1 = 0

     -   -    -    -  

  -----------------

       x - 2y + 0 + 2 =0

                          x = 2y - 2.      ......... ( ii )

 

 from EQ ( i ) and ( ii ) , we get

   x = 2y -2 = (-2z -4 )

                      -------

                            3

  x=(y-1) =(z+2)/ (-3)       (divide by 2)

  -   ---    ---------

  2     1              -3

 (x-1)/0=y +1 /(-1) = z/1

  x/2=(y-1)/1 =(z+2)/(-3) 

 the shortest distance between these two skew  lines is 1/√ 3