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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
The shortest distance between the lines ${r_1} = 4i - 3j - k + \lambda (i - 4j + 7k)$ and ${r_2} = i - j - 10k + \lambda (2i - 3j + 8k)$ is
  • A
    $3$
  • B
    $1$
  • C
    $2$
  • ✓
    $0$
✓

Answer

Correct option: D.
$0$
d
(d) The Given lines are ${r_1} = {a_1} + \lambda \,{b_1},\,\,\,\,{r_2} = {a_2} + \mu {b_2}$

Where ${a_1} = 4i - 3j - k;\,\,\,\,{b_1} = i - 4j + 7k$

${a_2} = i - j - 10k;\,\,\,\,{b_2} = 2i - 3j + 8k$

$|{b_1} \times {b_2}| = \left| {\begin{array}{*{20}{c}}i&j&k\\1&{ - 4}&7\\2&{ - 3}&8\end{array}} \right| = - 11i + 6j + 5k$

Now $[({a_2} - {a_1})\,\,{b_1}\,\,{b_2}] = ({a_2} - {a_1}).({b_1} \times {b_2})$

$ = ( - 3i + 2j - 9k)( - 11i + 6j + 5k) = 0$

Therefore, shortest distance $ = \frac{{[({a_2} - {a_1})\,\,{b_1}\,\,{b_2}]}}{{|{b_1} \times {b_2}|}} = 0$.

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