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 :
- A \(54\)
- B \(44\)
- C \(41\)
- D \(66\)
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…
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