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find the distance between the line x=y-1/2=z-2/3 and x+1=y+2/2=z-1/3 |
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Answer» \(\frac {x- 0} 1 = \frac {y - 1}2 = \frac {z-2}3\) .....(1) \(\frac{x -(-1)}1 = \frac {y - (-2)}2 = \frac {z -1}3\) .....(2) Since, both lines (1) & (2) are parallel to a vector whose direction cosines is same. \(\therefore\) Both lines are parallel. \(\therefore\) Distance between them = Distance between points (0, 1, 2) & (-1, -2, 1) \(= \sqrt{(-1 - 0)^2 + (-2 -1)^2 + (1 - 2)^2}\) \(= \sqrt{1 + 9 + 1}\) \(= \sqrt{11}\) units. |
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