JEE Mains · Maths · STD 12 - 11. three dimension geometry
If the square of the shortest distance between the lines \(\frac{x-2}{1}=\frac{y-1}{2}=\frac{z+3}{-3}\) and \(\frac{x+1}{2}=\frac{y+3}{4}=\frac{z+5}{-5}\) is \(\frac{\mathrm{m}}{\mathrm{n}}\), where \(\mathrm{m}, \mathrm{n}\) are coprime numbers, then \(\mathrm{m}+\mathrm{n}\) is equal to :
- A 21
- B 9
- C 14
- D 6
Answer & Solution
Correct Answer
(B) 9
Step-by-step Solution
Detailed explanation
\begin{aligned} & \overrightarrow{\mathrm{a}}=(2,1,-3) \\ & \overrightarrow{\mathrm{b}}=(-1,-3,-5) \\ & \overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{q}}=\left|\begin{array}{ccc}\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 1 & 2 & -3 \\ 2 & 4 &…
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