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STD 12 - 11. three dimension geometry — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 12 - 11. three dimension geometry4 Marks
MCQ
Shortest dist ance between the lines

${L_1}:\bar r = \hat i + \hat j + \lambda \left( {\hat i + \hat j - \hat k} \right)$

${L_2}:\bar r = \hat j + \hat k + \mu \left( {\hat j + 2\hat k - \hat i} \right)$ equal to

  • ✓
    $\frac{1}{{\sqrt {14} }}$
  • B
    $\frac{2}{{\sqrt {14} }}$
  • C
    $\frac{3}{{\sqrt {14} }}$
  • D
    $\frac{4}{{\sqrt {14} }}$
✓

Answer

Correct option: A.
$\frac{1}{{\sqrt {14} }}$
a
Distance between the lines

$\mathrm{L}_{1}: \overline{\mathrm{r}}=\overline{\mathrm{a}}+\lambda \overline{\mathrm{b}}$

$\mathrm{L}_{2}: \overline{\mathrm{r}}=\overline{\mathrm{c}}+\mathrm{m} \overline{\mathrm{d}}$ is

$D = \frac{{\left| {(\bar a - \bar c) \cdot \left| {\bar b \times \bar d} \right|} \right|}}{{\left| {\bar b \times \bar d} \right|}} = \frac{1}{{\sqrt {14} }}$

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