Find the equation between the two parallel lines l1 and l2 whose equations is given below. \vec{r}=3\hat{i}+2\hat{j}-\hat{k}+λ(3\hat{i}-2\hat{j}+\hat{k}) \vec{r}=2\hat{i}-\hat{j}+\hat{k}+μ(3\hat{i}-2\hat{j}+\hat{k})

Category: QuestionsFind the equation between the two parallel lines l1 and l2 whose equations is given below. \vec{r}=3\hat{i}+2\hat{j}-\hat{k}+λ(3\hat{i}-2\hat{j}+\hat{k}) \vec{r}=2\hat{i}-\hat{j}+\hat{k}+μ(3\hat{i}-2\hat{j}+\hat{k})
Editor">Editor Staff asked 11 months ago

Find the equation between the two parallel lines l1 and l2 whose equations is given below.
 
\vec{r}=3\hat{i}+2\hat{j}-\hat{k}+λ(3\hat{i}-2\hat{j}+\hat{k})
 
\vec{r}=2\hat{i}-\hat{j}+\hat{k}+μ(3\hat{i}-2\hat{j}+\hat{k})
 
(a) \sqrt{\frac{172}{14}}
 
(b) \sqrt{\frac{145}{14}}
 
(c) \sqrt{\frac{171}{14}}
 
(d) \sqrt{\frac{171}{134}}
 
The question was asked in an internship interview.
 
Enquiry is from Three Dimensional Geometry topic in chapter Three Dimensional Geometry of Mathematics – Class 12
NCERT Solutions for Subject Clas 12 Math Select the correct answer from above options 
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1 Answers
Editor">Editor Staff answered 11 months ago

Right choice is (c) \sqrt{\frac{171}{14}}
 
Easy explanation: The distance between two parallel lines is given by
 
d=\left |\frac{\vec{b}×(\vec{a_2}-\vec{a_1})}{|\vec{b}|}\right |
 
=\left |\frac{((3\hat{i}-2\hat{j}+\hat{k})×((3\hat{i}+2\hat{j}-\hat{k})-(2\hat{i}-\hat{j}+\hat{k})))}{|\sqrt{3^2+(-2)^2+1^2}|}\right |
 
=\left |\frac{(3\hat{i}-2\hat{j}+\hat{k})×(\hat{i}+3\hat{j}-2\hat{k}))}{\sqrt{14}}\right |
 
(3\hat{i}-2\hat{j}+\hat{k})×(\hat{i}+3\hat{j}-2\hat{k})=\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\3&-2&1\\1&3&-2\end{vmatrix}
 
=\hat{i}(4-3)-\hat{j}(-6-1)+\hat{k}(9+2)
 
=\hat{i}+7\hat{j}+11\hat{k}
 
∴d=\frac{|\hat{i}+7\hat{j}+11\hat{k}|}{\sqrt{14}}=\frac{\sqrt{1+49+121}}{\sqrt{14}}=\sqrt{\frac{171}{14}}.