JEE Mains · Maths · STD 12 - 11. three dimension geometry
The equation of line passing through \((-4, 1, 3)\), parallel to the plane \(x + 2y - z - 5 = 0\) and intersecting the line \(\frac{{x + 1}}{{ - 3}} = \frac{{y - 3}}{2} = \frac{{z - 2}}{{ - 1}}\) is
- A \(\frac{{x + 4}}{2} = \frac{{y + 3}}{1} = \frac{{z + 1}}{4}\)
- B \(\frac{{x + 4}}{1} = \frac{{y - 3}}{1} = \frac{{z - 1}}{3}\)
- C \(\frac{{x + 4}}{3} = \frac{{y - 3}}{{ - 1}} = \frac{{z - 1}}{1}\)
- D \(\frac{{x + 4}}{{ - 1}} = \frac{{y - 3}}{1} = \frac{{z - 1}}{1}\)
Answer & Solution
Correct Answer
(C) \(\frac{{x + 4}}{3} = \frac{{y - 3}}{{ - 1}} = \frac{{z - 1}}{1}\)
Step-by-step Solution
Detailed explanation
\(\text { Let the line L be } \frac{x+4}{a}=\frac{y-3}{b}=\frac{z-1}{c}\) \(\mathrm{L} \| \mathrm{x}+2 \mathrm{y}-\mathrm{z}-5=0\) \(\text { L intersects } \frac{x+1}{-3}=\frac{y-3}{2}=\frac{z-2}{-1}\)…
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