Question

If the shortest distance between the lines x-1/alpha =y+1/-1=z/1 and x+y+z+1=0=2x-y+z+3 is 1/sqrt(3) then value of alpha is

Answer 1

Answer:

32/19

Step-by-step explanation:

Given If the shortest distance between the lines x-1/alpha =y+1/-1=z/1 and x+y+z+1=0=2x-y+z+3 is 1/sqrt(3) then value of alpha is

Now let us consider x + y + z + 1 = 0 and

                                  2 x - y + z + 3 = 0

 Let us take as x = 0

                                y + z + 1 = 0

                              - y + z + 3 = 0

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

                                   2 z + 4 = 0

                                   z = - 2

               and  y - 2 + 1 = 0 and y = 1

 Since line is in direction, we get values as 1, 1, 1 and 2, -1, 1 which is a, b, c

Line will be perpendicular when

 a + b + c = 0 and 2a - b + c = 0 so a/2 = -b/-1 = c/-1

  So a/2 = b/1 = c/-3

So we get 2,1,-3

Now we can write this as

    x/2 = y - 1/1 = z+2/-3

    x - x1/a1 = y - y1/b1 = z - z1/c1

  x - x1/a2 = y - y1/b2 = z - z2/c2

  By applying the formula we get

   d =  x2 - x1         y2 - y1       z2 - z1

             a1                 b1              c1

              a2               b2              c2

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

         √(a1b2 - a2b1)^2 + (b1c2 - c1b2)^2 + (c1a2 - c2a1)^2

 d =   -1       2         - 2

          α      - 1          1

          2         1       - 3

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

√(α + 2)^2 + 4 + (2 + 3α)^2

4α - 2/√α^2 + 4 + 4α + 4 + 4 + 9α^2 + 12α = 1/√3

 3(4α - 2)^2 = 10α^2 + 16α + 12

   38 α^2 = 64 α

   α = 32/19