JEE Mains · Maths · STD 12 - 11. three dimension geometry
Consider the line \(\mathrm{L}\) passing through the points \((1,2,3)\) and \((2,3,5)\). The distance of the point \(\left(\frac{11}{3}, \frac{11}{3}, \frac{19}{3}\right)\) from the line \(\mathrm{L}\) along the line \(\frac{3 x-11}{2}=\frac{3 y-11}{1}=\frac{3 z-19}{2}\) is equal to :
- A \(3\)
- B \(5\)
- C \(4\)
- D \(6\)
Answer & Solution
Correct Answer
(A) \(3\)
Step-by-step Solution
Detailed explanation
\( \frac{x-1}{2-1}=\frac{y-2}{3-2}=\frac{z-3}{5-3} \) \( \Rightarrow \frac{x-1}{1}=\frac{y-2}{1}=\frac{z-3}{2}=\lambda\) \( \mathrm{B}(1+\lambda, 2+\lambda, 3+2 \lambda) \) \( \text { D.R. of } \mathrm{AB}=<\frac{3 \lambda-8}{3}, \frac{3 \lambda-5}{3}, \frac{6 \lambda-10}{3}> \)…
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