JEE Mains · Maths · STD 12 - 11. three dimension geometry
Let the values of p , for which the shortest distance between the lines \(\frac{x+1}{3}=\frac{y}{4}=\frac{z}{5}\) and \(\overrightarrow{\mathrm{r}}=(\mathrm{p} \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}})+\lambda(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+4 \hat{\mathrm{k}})\) is \(\frac{1}{\sqrt{6}}\), be \(\mathrm{a}, \mathrm{b}\), \((a \lt b)\). Then the length of the latus rectum of the ellipse \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\) is :-
- A \(9\)
- B \(\frac{3}{2}\)
- C \(\frac{2}{3}\)
- D \(18\)
Answer & Solution
Correct Answer
(C) \(\frac{2}{3}\)
Step-by-step Solution
Detailed explanation
\(\text { shortest distance }=\frac{|(\overline{\mathrm{a}}-\overline{\mathrm{b}})| \cdot(\overline{\mathrm{p}} \times \overline{\mathrm{q}})}{|\overline{\mathrm{p}} \times \overline{\mathrm{q}}|}\) where…
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