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

बिंदु \(\left(\frac{15}{7}, \frac{32}{7}, 7\right)\) की रेखा \(\frac{x+1}{3}=\frac{y+3}{5}=\frac{z+5}{7}\) से सदिश \(\hat{i}+4 \hat{j}+7 \hat{k}\) की दिशा में दूरी का वर्ग ___ है।

  1. A \(54\)
  2. B \(44\)
  3. C \(41\)
  4. D \(66\)
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Answer & Solution

Correct Answer

(D) \(66\)

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Detailed explanation

\(\begin{aligned} & L=\frac{x+1}{3}=\frac{y+3}{5}=\frac{z+5}{7} \\ & P Q=\frac{x-\frac{15}{7}}{1}=\frac{y-\frac{32}{7}}{4}=\frac{z-7}{7}=\lambda\end{aligned}\) \(\Rightarrow \mathrm{Q}\left(\lambda+\frac{15}{7}, 4 \lambda+\frac{32}{7}, 7 \lambda+7\right)\) Since Q lies on line L…
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