Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • NCERT
  • Find the shortest distance between the lines whose vector equations are r vector =i caret+ 2 j caret + 3 k caret+ lambda i caret- 3 j caret+ 2 k caret and r vector= 4 i caret+ 5 j caret + 6 k caret+ mu 2 i caret

16   Find the shortest distance between the lines whose vector equations are \overrightarrow{r}=(\widehat{i}+2\widehat{j}+3\widehat{k})+ \lambda (\widehat{i}-3\widehat{j}+2\widehat{k}) and 

             \overrightarrow{r}=(4\widehat{i}+5\widehat{j}+6\widehat{k})+ \mu (2\widehat{i}+3\widehat{j}+\widehat{k}).                               

Answers (1)

best_answer

Given two equations of line 

\overrightarrow{r}=(\widehat{i}+2\widehat{j}+3\widehat{k})+ \lambda (\widehat{i}-3\widehat{j}+2\widehat{k}) \overrightarrow{r}=(4\widehat{i}+5\widehat{j}+6\widehat{k})+ \mu (2\widehat{i}+3\widehat{j}+\widehat{k}) in the vector form.

So, we will apply the distance formula  for knowing the distance between two lines \vec{r} =\vec{a_{1}}+\lambda{b_{1}}   and   \vec{r} =\vec{a_{2}}+\lambda{b_{2}}

d= \left | \frac{\left ( \vec{b_{1}}\times\vec{b_{2}} \right ).\left ( \vec{a_{2}}-\vec{a_{1}} \right )}{\left | \vec{b_{1}}\times \vec{b_{2}} \right |} \right |

After comparing the given equations, we obtain

\vec{a_{1}} = \widehat{i}+2\widehat{j}+3\widehat{k}                \vec{b_{1}} = \widehat{i}-3\widehat{j}+2\widehat{k}

\vec{a_{2}} = 4\widehat{i}+5\widehat{j}+6\widehat{k}             \vec{b_{2}} = 2\widehat{i}+3\widehat{j}+\widehat{k}

\vec{a_{2}}-\vec{a_{1}} = (4\widehat{i}+5\widehat{j}+6\widehat{k}) - (\widehat{i}+2\widehat{j}+3\widehat{k})

                 = 3\widehat{i}+3\widehat{j}+3\widehat{k} 

Then calculating the determinant value numerator.

\vec{b_{1}}\times\vec{b_{2}} = \begin{vmatrix} \widehat{i} &\widehat{j} &\widehat{k} \\ 1& -3 &2 \\ 2& 3& 1 \end{vmatrix}

= (-3-6)\widehat{i}-(1-4)\widehat{j}+(3+6)\widehat{k} = -9\widehat{i}+3\widehat{j}+9\widehat{k}

That implies, \left | \vec{b_{1}}\times\vec{b_{2}} \right | = \sqrt{(-9)^2+(3)^2+(9)^2}

= \sqrt{81+9+81} = \sqrt{171} =3\sqrt{19}

\left ( \vec{b_{1}}\times\vec{b_{2}} \right ).\left ( \vec{a_{2}}-\vec{a_{1}} \right )=(-9\widehat{i}+3\widehat{j}+9\widehat{k})(3\widehat{i}+3\widehat{j}+3\widehat{k})

= (-9\times3)+(3\times3)+(9\times3) = 9

Now, after substituting the value in the above formula we get,

d= \left | \frac{9}{3\sqrt{19}} \right | = \frac{3}{\sqrt{19}}

Therefore,\frac{3}{\sqrt{19}} is the shortest distance between the two given lines. 

 

 

 

Posted by

Divya Prakash Singh

View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support
cuet_ads

Similar Questions

Trending Articles/News

anam-khan

PM Modi's Pariksha Pe Charcha 2025 to be held next month, registrations open till January 14
anam-khan

CBSE Board Exams 2025: Online portal for availing exemptions by CWSN students opens
anam-khan

Board Exams: States agree to equivalence; no question paper ‘jumbling’ from next year, says PARAKH CEO

Latest Question