JEE Mains · Maths · STD 12 - 11. three dimension geometry
Let \(N\) be the foot of perpendicular from the point \(P\) \((1,-2,3)\) on the line passing through the points \((4,5,8)\) and \((1,-7,5)\). Then the distance of \(N\) from the plane \(2 x-2 y+z+5=0\) is \(.......\).
- A \(6\)
- B \(9\)
- C \(7\)
- D \(8\)
Answer & Solution
Correct Answer
(C) \(7\)
Step-by-step Solution
Detailed explanation
Equation of line \(\frac{x-4}{4-1}=\frac{y-5}{5-(-7)}=\frac{z-8}{8-5}\) \(\frac{x-4}{3}=\frac{y-5}{12}=\frac{z-8}{3}\) Let point \(N (3 \lambda+4,12 \lambda+5,3 \lambda+8)\) \(\overline{ PN }=(3 \lambda+4-1) \hat{i}+(12 \lambda+5-(-2)) \hat{ j }+(3 \lambda+8-3) \hat{ k }\)…
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