AP EAMCET · Maths · Three Dimensional Geometry
The shortest distance between the skew lines \(\vec{r}=(2 \hat{i}-\hat{j})+t(\hat{i}+2 \hat{k})\) and \(\vec{r}=(-2 \hat{i}+\hat{k})+s(\hat{i}-\hat{j}-\hat{k})\) is
- A \(\frac{3 \sqrt{2}}{\sqrt{7}}\)
- B \(\frac{3}{\sqrt{7}}\)
- C \(\frac{3}{\sqrt{14}}\)
- D \(\frac{4}{\sqrt{14}}\)
Answer & Solution
Correct Answer
(A) \(\frac{3 \sqrt{2}}{\sqrt{7}}\)
Step-by-step Solution
Detailed explanation
\begin{aligned} & \text { } \vec{r}=(2 \hat{i}-\hat{j})+t(\hat{i}+2 \hat{k}) \Rightarrow \vec{a}_1=2 \hat{i}-\hat{j}, \vec{b}_1=\hat{i}+2 \hat{k} \\ & \vec{r}=(-2 \hat{i}+\hat{k})+s(\hat{i}-\hat{j}-\hat{k}) \\ & \Rightarrow \vec{a}_2=-2 \hat{i}+\hat{k},…
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