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JEE Mains · Maths · STD 12 - 11. three dimension geometry

Let the line \(L\) pass through the point \((0,1,2)\), intersect the line \(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}\) and be parallel to the plane \(2 x+y-3 z=4\). Then the distance of the point \(P(1,-9,2)\) from the line \(L\) is

  1. A \(9\)
  2. B \(\sqrt{54}\)
  3. C \(\sqrt{69}\)
  4. D \(\sqrt{74}\)
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Answer & Solution

Correct Answer

(D) \(\sqrt{74}\)

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Detailed explanation

\(\overline{ AB } \cdot \overrightarrow{ n }\) \(\Rightarrow[(1+2 \lambda) \hat{ i }+(1+3 \lambda) \hat{ j }+(1+4 \lambda) \hat{ k }] \cdot(2 \hat{ i }+\hat{ j }-3 \hat{ k })\) \(2+4 \lambda+1+3 \lambda-3-12 \lambda=0\) \(5 \lambda=0 \Rightarrow \lambda=0\)…
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