MHT CET · Maths · Three Dimensional Geometry
If the shortest distance between the lines
\(\overline{\mathrm{r}}_1=\alpha \hat{i}+2 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}+\lambda(\hat{i}-2 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}), \lambda \in \mathbb{R}, \alpha>0 \text { and }\)
\(\overline{\mathrm{r}}_2=-4 \hat{i}-\hat{\mathrm{k}}+\mu(3 \hat{i}-2 \hat{\mathrm{j}}-2 \hat{\mathrm{k}}), \mu \in \mathrm{R}\), is 9, then the value of \(\alpha\) is
- A \(4\)
- B \(6\)
- C \(8\)
- D \(3\)
Answer & Solution
Correct Answer
(B) \(6\)
Step-by-step Solution
Detailed explanation
\(\mathbf{a}_2 - \mathbf{a}_1 = (-4-\alpha)\hat{i} - 2\hat{j} - 3\hat{k}\) \(\mathbf{b}_1 \times \mathbf{b}_2 = (1\hat{i}-2\hat{j}+2\hat{k}) \times (3\hat{i}-2\hat{j}-2\hat{k}) = 8\hat{i}+8\hat{j}+4\hat{k}\)
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