CUET · MATHS · PYQ PAPER 2025
The shortest distance between the following lines :
\(\vec{r}=(\hat{i}+\hat{j}-\hat{k})+s(2 \hat{i}+\hat{j}+\hat{k})\)
\(\vec{r}=(\hat{i}+\hat{j}+2 \hat{k})+t(4 \hat{i}+2 \hat{j}+2 \hat{k})\), where \(s\) and \(t\) are scalars, is :
- A \(\frac{3 \sqrt{6}}{\sqrt{5}}\)
- B \(\sqrt{\frac{15}{2}}\)
- C \(\frac{\sqrt{5}}{3 \sqrt{6}}\)
- D \(\sqrt{\frac{2}{15}}\)
Answer & Solution
Correct Answer
(B) \(\sqrt{\frac{15}{2}}\)
Step-by-step Solution
Detailed explanation
\( \vec{a}_1 = \hat{i}+\hat{j}-\hat{k} \), \( \vec{b}_1 = 2 \hat{i}+\hat{j}+\hat{k} \) \( \vec{a}_2 = \hat{i}+\hat{j}+2 \hat{k} \), \( \vec{b}_2 = 4 \hat{i}+2 \hat{j}+2 \hat{k} \) \( \vec{b}_2 = 2 \vec{b}_1 \), so the lines are parallel. Shortest distance…
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