Vector Algebra Question 49
Question: The shortest distance between the lines $ \mathbf{r}=(3\mathbf{i}-2\mathbf{j}-2\mathbf{k})+\mathbf{i}t $ and $ \mathbf{r}=\mathbf{i}-\mathbf{j}+2\mathbf{k}+\mathbf{j}s $ (t and s being parameters) is
[AMU 1999]
Options:
A) $ \sqrt{21} $
B) $ \sqrt{102} $
C) 4
D) 3
Show Answer
Answer:
Correct Answer: C
Solution:
- The given lines are $ \mathbf{r}={{\mathbf{a}}_1}+\lambda {{\mathbf{b}}_1},\mathbf{r}={{\mathbf{a}}_2}+\mu {{\mathbf{b}}_2} $ , Where $ {{\mathbf{a}}_1}=3\mathbf{i}-2\mathbf{j}-2\mathbf{k},,{{\mathbf{b}}_1}=\mathbf{i} $ $ {{\mathbf{a}}_2}=\mathbf{i}-\mathbf{j}+2\mathbf{k},,{{\mathbf{b}}_2}=\mathbf{j} $ $ |{{\mathbf{b}}_1}\times {{\mathbf{b}}_2}|,=,|\mathbf{i}\times \mathbf{j}|,=,|\mathbf{k}|=1 $ Now, $ [({{\mathbf{a}}_2}-{{\mathbf{a}}_1})\ {{\mathbf{b}}_1}\ {{\mathbf{b}}_2}]=({{\mathbf{a}}_2}-{{\mathbf{a}}_1}).({{\mathbf{b}}_1}\times {{\mathbf{b}}_2}) $ $ =(-2\mathbf{i}+\mathbf{j}+4\mathbf{k})(\mathbf{k})=4 $ \ Shortest distance $ =\frac{[({{\mathbf{a}}_2}-{{\mathbf{a}}_1})({{\mathbf{b}}_1}-{{\mathbf{b}}_2})]}{|{{\mathbf{b}}_1}\times {{\mathbf{b}}_2}|}=\frac{4}{1}=4 $ .
BETA
