JEE Mains · Maths · STD 12 - 11. three dimension geometry
A symmetrical form of the line of intersection of the planes \(x = ay + b\) and \(z = cy + d\) is
- A \(\frac{{x - b}}{a} = \frac{{y - 1}}{1} = \frac{{z - d}}{c}\)
- B \(\frac{{x - b - a}}{a} = \frac{{y - 1}}{1} = \frac{{z - d - c}}{c}\)
- C \(\frac{{x - a}}{b} = \frac{{y - 0}}{1} = \frac{{z - c}}{d}\)
- D \(\frac{{x - b - a}}{b} = \frac{{y - 1}}{0} = \frac{{z - d - c}}{d}\)
Answer & Solution
Correct Answer
(B) \(\frac{{x - b - a}}{a} = \frac{{y - 1}}{1} = \frac{{z - d - c}}{c}\)
Step-by-step Solution
Detailed explanation
Given two planes: \(x-a y-b=0\) and \(c y-z+d=0\) Let, \(l, m, n\) be the direction ratio of the required line. since the required line is perpendicular to normal of both the plane, therefore \(l-a m=0\) and \(c m-n=0\) \(\Rightarrow l-a m+0 . n=0\) and \(0 . l+c m-n=0\)…
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