Vector Algebra Question 123
Question: The shortest distance between the lines $ {{\mathbf{r}}_1}=4\mathbf{i}-3\mathbf{j}-\mathbf{k}+\lambda (\mathbf{i}-4\mathbf{j}+7\mathbf{k}) $ and $ {{\mathbf{r}}_2}=\mathbf{i}-\mathbf{j}-10\mathbf{k}+\lambda (2\mathbf{i}-3\mathbf{j}+8\mathbf{k}) $ is
[J & K 2005]
Options:
A) 3
B) 1
C) 2
D) 0
Show Answer
Answer:
Correct Answer: D
Solution:
- The Given lines are $ {{\mathbf{r}}_1}={{\mathbf{a}}_1}+\lambda ,{{\mathbf{b}}_1},{{\mathbf{r}}_2}={{\mathbf{a}}_2}+\mu {{\mathbf{b}}_2} $ Where $ {{\mathbf{a}}_1}=4\mathbf{i}-3\mathbf{j}-\mathbf{k};{{\mathbf{b}}_1}=\mathbf{i}-4\mathbf{j}+7\mathbf{k} $ $ {{\mathbf{a}}_2}=\mathbf{i}-\mathbf{j}-10\mathbf{k};{{\mathbf{b}}_2}=2\mathbf{i}-3\mathbf{j}+8\mathbf{k} $ $ |{{\mathbf{b}}_1}\times {{\mathbf{b}}_2}|= \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & -4 & 7 \\ 2 & -3 & 8 \\ \end{vmatrix} =-11\mathbf{i}+6\mathbf{j}+5\mathbf{k} $ 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}) $ $ =(-3\mathbf{i}+2\mathbf{j}-9\mathbf{k})(-11\mathbf{i}+6\mathbf{j}+5\mathbf{k})=0 $ Therefore, shortest distance $ =\frac{[({{\mathbf{a}}_2}-{{\mathbf{a}}_1}){{\mathbf{b}}_1}{{\mathbf{b}}_2}]}{|{{\mathbf{b}}_1}\times {{\mathbf{b}}_2}|}=0 $ .
BETA
