TS EAMCET · Maths · Three Dimensional Geometry
The shortest distance between the lines \(\mathbf{r}=3 \mathbf{i}+5 \mathbf{j}+7 \mathbf{k}+\lambda(\mathbf{i}+2 \mathbf{j}+\mathbf{k}) \quad\) and \(\mathbf{r}=-\mathbf{i}-\mathbf{j}-\mathbf{k}+\mu(7 \mathbf{i}-6 \mathbf{j}+\mathbf{k})\) is
- A \(\frac{16}{5 \sqrt{5}}\)
- B \(\frac{26}{5 \sqrt{5}}\)
- C \(\frac{36}{5 \sqrt{5}}\)
- D \(\frac{46}{5 \sqrt{5}}\)
Answer & Solution
Correct Answer
(D) \(\frac{46}{5 \sqrt{5}}\)
Step-by-step Solution
Detailed explanation
The given lines are \(\mathbf{r}=\mathbf{a}_1+\lambda \mathbf{b}_1, \mathbf{r}=\mathbf{a}_2+\mu \mathbf{b}_2\) where,…
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