Question

find the equation of two lines through the origin which intersect the lines (x-3)/2=(y-3)/1=z/1 at angle of 60^0) each

Answer 1

equation of the line is
(x-3)/2 = (y-3)/1 = z/1              (given)
let A is the point on the given line, then its co-ordinate is
A = (2r+3, r+3, r)  ----------------(1)
O = (0, 0, 0)
direction ratios of the line OA are = 2r+3-0 , r+3-0, r-0
                                                 = 2r+3, r+3, r
angle between line and OA is 60
cos60 = [2(2r+3) + 1(r+3) + 1(r)]/√(2²+1²+1²){(2r+3)²+(r+3)²+(r)²}
1/2 = (6r+9)/√{6(6r²+18r+18)}
r²+3r+3 = 4r²+12r+9
r²+3r+2 = 0
r = -1, -2
on putting the value of r in equation (1)we get two point
point A = (1, 2, -1)
point B = (-1, 1, -2)
equation of line OA and OB are
OA:
 (x-0)/1 = (y-0)/2 = (z-0)/-1
2x = y = -2z
OB:
(x-0)/-1  = (y-0)/1 = (z-0)/-2
-2x = 2y = -z