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If lines `(x -1)/(2) = (y+1)/(3) = (z-1)/(4)` and `(x-3)/(1) =(y -lambda)/(2) = (z)/(1)` intersect each other then `lambda`= …..A. `(7)/(2)`B. `(3)/(2)`C. `(9)/(2)`D. `(5)/(2)` |
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Answer» Correct Answer - C Given lines are `(x - 1)/(2) = (y + 1)/(3) = (z - 1)/(4) = lambda_(1)` let ….(i) and `(x - 3)/(1) = (y - lambda)/(2) = (z)/(1) = lambda_(2)` (let) …(ii) Then, any point on line (i) is `(2 lambda_(1) + 1, 3 lambda_(1) - 1, 4 lambda_(1) + 1)` and any point on line (ii) is `(lambda_(2) + 3, 2 lambda_(2) + lambda, lambda_(2))` Clearly, then lines (i) and (ii) will intersect if `(2 lambda_(1) + 1, 3 lambda_(1) - 1, 4 lambda_(1)) = (lambda_(2) + 3, 2 lambda_(2) + lambda, lambda_(2))` for some particular value of `lambda_(1)` and `lambda_(2)` `implies 2 lambda_(1) + 1 = lambda_(2) + 3, 3 lambda_(1) - 1 = 2 lambda_(2) + lambda` and `4 lambda_(1) + 1 = lambda_(2)` `implies 2 lambda_(1) - lambda_(2) = 2, 3 lambda_(1) - 2 lambda_(2) + 1` and `4 lambda_(1) - lambd_(2) =- 1` On solving `2 lambda_(1) - lambda_(2) = 2` and `4 lambda_(1) - lambda_(2) = - 1` We get `lambda_(1) = - (3)/(2)` and `lambda_(2) = - 5` Now, putting the values of `lambda_(1)` and `lambda_(2)` in `3 lambda_(1) - 2 lambda_(2) = lambda + 1` `implies 3((-3))/(2) - 2 (-5) = lambda + 1 implies (-9)/(2) + 10 = lambda + 1` `implies lambda + 1 = (11)/(2) implies lambda = (9)/(2)` |
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