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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.
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Solution
Given, equation of line:
`(x - 3)/2 = (y - 3)/1 = z/1` = x
∴ x = 2r + 3
y = r + 3
z = r
So, (2r + 3, r + 3, r) is the direction ratio of two lines that intersect at `pi/3` with given line and passes through (0,0).
∴ Angle between the line and unknown lines is `pi/3`.
Direction ratio of line is (2,1,1)
`|a| = sqrt(2^2 + 1^2 + 1^2) = sqrt(6)`
`|b| = sqrt((2r + 3)^2 + (r + 3)^2 + r^2)`
= `sqrt(6r^2 + 18 + 18)`
`cos pi/3 = (a*b)/(|a||b|)`
= `1/2 = (4r + 6 + r + 3 + r)/(sqrt(6)sqrt(6r^2 + 18r + 18)`
= `1/2 (6r + 9)/(6sqrt(r^2 + 3r + 3)`
∴ `sqrt(r^2 + 3r + 3) = 3r + 3`
= `r^2 + 3r + 3`
= `4r^2 + 9 + 12r` = 0
= `3r^2 + 6 + 9r` = 0
= `r^2 + 3r + 2`
∴ `(r + 1)(r + 2)` = 0
So, direction ratios are (−1, 1, −2) and (1, 2, −1)
Lines are `(x - 0)/(-1) = (y - 0)/1 = (z - n)/(-2)` and `(x - 0)/1 = (y - 0)/2 = (z - 0)/(-1)`.
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