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Find the vector equation of a line passing through the point \left ( 2,3,2 \right )  and parallel to the line \vec{r}= \left ( -2\hat{i}+3\hat{j} \right )+\lambda \left ( 2\hat{i}-3\hat{j}+6\hat{k} \right ). Also ,find the distance between these two lines.

 

 

 

 
 
 
 
 

Answers (1)
R Ravindra Pindel

As the d.r's of parallel lines are proportional so,the equation of line passing through \left ( 2,3,2 \right )  and parallel to \vec{r}= \left ( -2\hat{i}+3\hat{j} \right )+\lambda \left ( 2\hat{i}-3\hat{j}+6\hat{k} \right ) is:
                      \vec{r}= \left ( 2\hat{i}+3\hat{j}+2\hat{k} \right )+\lambda \left ( 2\hat{i}-3\hat{j}+6\hat{k} \right )
       Now  \vec{a_{1}}= -2\hat{i}+3\hat{j},\vec{a_{2}}= 2\hat{i}+3\hat{j}+2\hat{k}
                \vec{b_{1}}= 2\hat{i}-3\hat{j}+6\hat{k}
     \Rightarrow \vec{a_{2}}-\vec{a_{1}}= \left ( 2\hat{i}+3\hat{j}+2\hat{k} \right )-\left ( -2\hat{i}+3\hat{j} \right )= 4\hat{i}+2\hat{k}
          and\Rightarrow \left (\vec{a_{2}}-\vec{a_{1}} \right )\times \vec{b}= \begin{vmatrix} \hat{i} & \hat{j} &\hat{k} \\ 4 & 0 & 2\\ 2& -3 & 6 \end{vmatrix}
                                                = 6\hat{i}-20\hat{j}-12\hat{k}   
           \therefore SD= \frac{\left | \left ( \vec{a_{2}}-\vec{a_{1}} \right )\times \vec{b} \right |}{\left | \vec{b} \right |}
\Rightarrow \frac{\left | 6\hat{i}-20\hat{j}-12\hat{k} \right |}{\left | 2\hat{i}-3\hat{j}+6\hat{k} \right |}\Rightarrow \frac{\sqrt{36+400+144}}{\sqrt{4+9+36}}
\Rightarrow \frac{\sqrt{580}}{7} units

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