JEE Mains · Maths · STD 12 - 11. three dimension geometry
The shortest distance between the line \(\frac{x-3}{4}=\frac{y+7}{-11}=\frac{z-1}{5} \text { and } \frac{x-5}{3}=\frac{y-9}{-6}=\frac{z+2}{1}\) is :
- A \(\frac{187}{\sqrt{563}}\)
- B \(\frac{178}{\sqrt{563}}\)
- C \(\frac{185}{\sqrt{563}}\)
- D \(\frac{179}{\sqrt{563}}\)
Answer & Solution
Correct Answer
(A) \(\frac{187}{\sqrt{563}}\)
Step-by-step Solution
Detailed explanation
\(\overrightarrow{\mathrm{n}}=\overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{q}}\) \(\vec{n}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 4 & -11 & 5 \\ 3 & -6 & 1\end{array}\right|=19 \hat{i}+11 \hat{j}+9 \hat{k}\) S.d. \(=\) projection of…
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