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

The square of the distance of the point \(\left(\frac{15}{7}, \frac{32}{7}, 7\right)\) from the line \(\frac{x+1}{3}=\frac{y+3}{5}=\frac{z+5}{7}\) in the direction of the vector \(\hat{i}+4 \hat{j}+7 \hat{k}\) is :

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

Answer & Solution

Correct Answer

(D) \(66\)

Step-by-step Solution

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…