Find the distance between the lines l1 and l2 with the following vector equations. \(\vec{r}=2\hat{i}+2\hat{j}-2\hat{k}+λ(3\hat{i}+2\hat{j}+5\hat{k})\) \(\vec{r}=4 \hat{i}-\hat{j}+5\hat{k}+μ(3\hat{i}-2\hat{j}+4\hat{k})\)

Category: QuestionsFind the distance between the lines l1 and l2 with the following vector equations. \(\vec{r}=2\hat{i}+2\hat{j}-2\hat{k}+λ(3\hat{i}+2\hat{j}+5\hat{k})\) \(\vec{r}=4 \hat{i}-\hat{j}+5\hat{k}+μ(3\hat{i}-2\hat{j}+4\hat{k})\)
Editor">Editor Staff asked 11 months ago

Find the distance between the lines l1 and l2 with the following vector equations.
 
\vec{r}=2\hat{i}+2\hat{j}-2\hat{k}+λ(3\hat{i}+2\hat{j}+5\hat{k})
 
\vec{r}=4 \hat{i}-\hat{j}+5\hat{k}+μ(3\hat{i}-2\hat{j}+4\hat{k})
 
(a) \frac{57}{\sqrt{47}}
 
(b) \frac{57}{\sqrt{77}}
 
(c) \frac{7}{\sqrt{477}}
 
(d) \frac{57}{\sqrt{477}}
 
I got this question in semester exam.
 
The query is from Three Dimensional Geometry in section 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 option is (d) \frac{57}{\sqrt{477}}
 
To explain I would say: We know that, the shortest distance between two skew lines is given by
 
d=\left |\frac{(\vec{b_1}×\vec{b_2}).(a_2-a_1)}{|\vec{b_1}×\vec{b_2}|}\right |
 
The vector equations of the two lines is
 
\vec{r}=2\hat{i}+2\hat{j}-2\hat{k}+λ(3\hat{i}+2\hat{j}+5\hat{k})
 
\vec{r}=4 \hat{i}-\hat{j}+5\hat{k}+μ(3\hat{i}-2\hat{j}+4\hat{k})
 
∴d=\left|\frac{((3\hat{i}+2\hat{j}+5\hat{k})×(3\hat{i}-2\hat{j}+4\hat{k}).(4\hat{i}-\hat{j}+5\hat{k})-(2\hat{i}+2\hat{j}-2\hat{k}))}{|(3\hat{i}+2\hat{j}+5\hat{k})×(3\hat{i}-2\hat{j}+4\hat{k})|}\right |
 
(3\hat{i}+2\hat{j}+5\hat{k})×(3\hat{i}-2\hat{j}+4\hat{k})=\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\3&2&5\\3&-2&4\end{vmatrix}
 
=\hat{i}(8+10)-\hat{j}(12-15)+\hat{k}(-6-6)
 
=18\hat{i}+3\hat{j}-12\hat{k}
 
d=\left|\frac{(18\hat{i}+3\hat{j}-12\hat{k}).(2\hat{i}-3\hat{j}+7\hat{k})}{\sqrt{18^2+3^2+(-12)^2}}\right |
 
d=\left|\frac{36-9-84}{\sqrt{477}}\right |=\frac{57}{\sqrt{477}}.