TS EAMCET · Maths · Three Dimensional Geometry
The shortest distance between the lines \(\mathbf{r}=(-2 \hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}})+\mathbf{r}(2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}-\hat{\mathbf{k}})\) and \(\mathbf{r}=(\hat{\mathbf{i}}-\hat{\mathbf{j}}+2 \hat{\mathbf{k}})+\mathbf{k}(-\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+4 \hat{\mathbf{k}})\) is
- A 0
- B \(\frac{10}{\sqrt{6}}\)
- C \(\frac{11}{\sqrt{6}}\)
- D \(\frac{13}{\sqrt{6}}\)
Answer & Solution
Correct Answer
(C) \(\frac{11}{\sqrt{6}}\)
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