TS EAMCET · Maths · Three Dimensional Geometry
The shortest distance between the line \(\mathbf{r}=2 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}+\lambda(\hat{\mathbf{i}}-\hat{\mathbf{j}}+4 \hat{\mathbf{k}})\) and the plane \(\mathbf{r} \cdot(\hat{\mathbf{i}}+5 \hat{\mathbf{j}}+\hat{\mathbf{k}})=5\) is
- A \(\frac{1}{3 \sqrt{3}}\)
- B \(\frac{5}{3 \sqrt{3}}\)
- C \(\frac{10}{3 \sqrt{3}}\)
- D \(\frac{11}{3 \sqrt{3}}\)
Answer & Solution
Correct Answer
(C) \(\frac{10}{3 \sqrt{3}}\)
Step-by-step Solution
Detailed explanation
We have, \(\begin{aligned} & L: \mathbf{r}=2 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}+\lambda(\hat{\mathbf{i}}-\hat{\mathbf{j}}+4 \hat{\mathbf{k}}) \\ & P: \mathbf{r} \cdot(\hat{\mathbf{i}}+5 \hat{\mathbf{j}}+\hat{\mathbf{k}})=5 \end{aligned}\) \(\quad\) Now,…
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