JEE Mains · Maths · STD 12 - 11. three dimension geometry
Equation of the line of the shortest distance between the lines \(\frac{x}{1} = \frac{y}{{ - 1}} = \frac{z}{1}\) and \(\frac{{x - 1}}{0} = \frac{{y + 1}}{{ - 2}} = \frac{z}{1}\) is
- A \(\frac{x}{1} = \frac{y}{{ - 1}} = \frac{z}{{ - 2}}\)
- B \(\frac{{x - 1}}{1} = \frac{{y + 1}}{{ - 1}} = \frac{z}{{ - 2}}\)
- C \(\frac{{x - 1}}{1} = \frac{{y + 1}}{{ - 1}} = \frac{z}{1}\)
- D \(\frac{x}{{ - 2}} = \frac{y}{1} = \frac{z}{2}\)
Answer & Solution
Correct Answer
(B) \(\frac{{x - 1}}{1} = \frac{{y + 1}}{{ - 1}} = \frac{z}{{ - 2}}\)
Step-by-step Solution
Detailed explanation
Let equation of the required line be \(\frac{{x - {x_1}}}{a} = \frac{{y - {y_1}}}{b} = \frac{{z - {z_1}}}{c}\) .....\((i)\) Given two lines \(\frac{x}{1}=\frac{y}{-1}=\frac{z}{1}\) .........\((ii)\) and \(\frac{x-1}{0}=\frac{y+1}{0}=\frac{z}{1}\) .....\((iii)\) since the line…
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