JEE Mains · Maths · STD 12 - 11. three dimension geometry
If the line, \(\frac{{x\, - \,1}}{2}\, = \,\frac{{y\, + \,1}}{3}\, = \,\frac{{z\, - \,2}}{4}\) meets the plane, \(x + 2y + 3z = 15\) at a point \(P,\) then the distance of \(P\) from the origin is
- A \(\frac {\sqrt 5}{2}\)
- B \(2\sqrt 5\)
- C \(\frac {9}{2}\)
- D \(\frac {7}{2}\)
Answer & Solution
Correct Answer
(C) \(\frac {9}{2}\)
Step-by-step Solution
Detailed explanation
Any point on the given line can be \(\left(1+2 \lambda_{,}-1+3 \lambda, 2+4 \lambda\right) ; \lambda \in \mathrm{R}\) Put in plane \(1+2 \lambda+(-2+6 \lambda)+(6+12 \lambda)=15\) \(20 \lambda+5=15\) \(20 \lambda=10\) \(\lambda=\frac{1}{2}\) \(\therefore\) Point…
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