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प्रश्न
Find the value of λ, so that the lines `(1-"x")/(3) = (7"y" -14)/(λ) = (z -3)/(2) and (7 -7"x")/(3λ) = ("y" - 5)/(1) = (6 -z)/(5)` are at right angles. Also, find whether the lines are intersecting or not.
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उत्तर
Given lines are `(1-x)/(3) = (7y -14)/(λ) = (z -3)/(2) and (7 -7x)/(3λ) = (y - 5)/(1) = (6 -z)/(5)`
Converting them into standard form, we have `(x - 1)/(-3) = ("y" - 2)/((λ/7)) = (z - 3)/(2) and (x-1)/((-3λ/7)) = (y - 5)/(1) = (z -6)/(-5)`
Corresponding d.r.'s are `(-3, λ/7, 2) and ((-3λ)/7, 1, -5)`
Since the angle between the lines is right angle so, cos 90° = `|((-3) ((-3λ)/7) + (λ/7) (1) + (2) (-5))/(sqrt((-3)^2 + (λ/7)^2 + 2^2) sqrt(((-3λ)/7)^2+ 1^2 + (-5)^2)))|`
⇒ 0 = `|(9λ/7 + λ/7 - 10)/(sqrt(λ^2/49 + 13) sqrt((9λ^2)/49 + 26)) |`
Squaring and cross-multiplying
⇒ `(10λ/7 - 10)^2 = 0`
⇒ `(10λ)/(7) = 10`
⇒ λ = 7.
Substituting the value λ, of the lines are ` (x - 1)/(-3) = (y - 2)/(1) = (z - 3)/(2)` = a (let) and `(x-1)/(-3) = (y - 5)/(1) = (z -6)/(-5)` = b (let)
From first equation, `(x, y, z) = ( -3a + 1 ,a + 2, 2a+ 3) and "from second equation", (x, y, z) = (-3b + 1, b + 5, -5b + 6)`
Equating the corresponding values of coordinates, we have
- 3a + 1 = - 3b + 1, a + 2 = b + 5 and 2a + 3 = -5b + 6
Or, - 3a + 3b = 0, a - b = 3 and 2a + 5b = 3
Solving the second and third equations of the above, we get a `= (18)/(7)` and b` = (-3)/(7)`
Substituting these values of a and b in the first one
`-3 (18/7) + 3 (-3)/(7) = -9`
Thus, it is clear that the first equation is not satisfied so the lines are not intersecting.
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