Question

Find the equations of the two lines through the origin which intersect the line x-3/2 = y-3/1 = z/1 at angles of pi/3 each

Answer 1

Given equation of line : (x - 3//2 = (y - 3)/1 = z = r ( Let )
x = 2r + 3
y = r + 3
z = r
so, { (2r + 3), (r + 3) , r } is the direction ratio of two lines which intersects at π/3 with the given line and passes through origin.
so, angle between line and unknown lines are π/3
direction ratio of given line is {2, 1 , 1}
|a| = √{2² + 1² + 1²} = √6
|b| = √{(2r + 3)² + (r + 3)² + r²} = √{6r² + 18r + 18}
now, if any line makes π/3 .
so, cosπ/3 = (a.b)/|a|.|b| or, cos/3 = (a.b)/|a|.|b|
1/2 = {4r + 6 + r + 3 + r }/√6.√{6r² + 18r + 18}
1/2 = (6r + 9)/6√{r² + 3r + 3}
3√{r² + 3r + 3 } = (6r + 9)
r² + 3r + 3 = 4r² + 12r + 9
3r² + 9r + 6 = 0
r² + 3r + 2 = 0
(r + 1)(r + 2) = 0 ⇒ r = -1 and -2
so, direction ratio of lines : ( 2r + 3, r + 3, r ) : ( -1, 1 , -2) and (1, 2, -1)
Lines passes through origin so,
Equations of lines :
(x -0)/-1 = (y - 0)/1 = (z - 0)/2
(x - 0)/1 =(y - 0)/2 = (z - 0)/-1