JEE Mains · Maths · STD 12 - 11. three dimension geometry
Let \(O\) be the origin, and \(\mathrm{M}\) and \(\mathrm{N}\) be the points on the lines \(\frac{x-5}{4}=\frac{y-4}{1}=\frac{z-5}{3}\) and \(\frac{\mathrm{x}+8}{12}=\frac{\mathrm{y}+2}{5}=\frac{\mathrm{z}+11}{9}\) respectively such that \(\mathrm{MN}\) is the shortest distance between the given lines. Then \(\overrightarrow{\mathrm{OM}} \cdot \overrightarrow{\mathrm{ON}}\) is equal to ...........
- A \(10\)
- B \(9\)
- C \(4\)
- D \(6\)
Answer & Solution
Correct Answer
(B) \(9\)
Step-by-step Solution
Detailed explanation
\(L_1: \frac{\mathrm{x}-5}{4}=\frac{\mathrm{y}-4}{1}=\frac{\mathrm{z}-5}{3}=\lambda \quad \text { drs }(4,1,3)=\mathrm{b}_1 \) \(M(4 \lambda+5, \lambda+4,3 \lambda+5) \) \(L_2: \frac{\mathrm{x}+8}{12}=\frac{\mathrm{y}+2}{5}=\frac{\mathrm{z}+11}{9}=\mu \)…
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