Find the vector equation of the line passing through (2,1,-1) and parallel to the line \vec{r}= \left ( \hat{i}+\hat{j} \right )+\lambda \left ( 2\hat{i}-\hat{j}+\hat{k} \right )\cdot Also ,find the distance between these two lines.

 

 

 

 
 
 
 
 

Answers (1)

for parallel lines, the d.r's are proportional so, required line through \left ( 2,1,-1 \right ) is 
\vec{r}= \left ( 2\hat{i}+\hat{j}-\hat{k} \right )+\lambda \left ( 2\hat{i}-\hat{j}+\hat{k} \right )
Now \vec{a}_{1}= \left ( \hat{i}+\hat{j} \right ),\vec{a}_{2}= \left ( 2\hat{i}+\hat{j}- \hat{k}\right )
         \vec{b}= \left ( 2\hat{i}-\hat{j}+\hat{k} \right )
\Rightarrow \bar{a}_{2}-\bar{a}_{1}= \hat{i}-\hat{k}
\Rightarrow \left ( \bar{a}_{2}-\bar{a}_{1} \right )\times \vec{b}= \begin{vmatrix} i & j & k\\ 1& 0 &-1 \\ 2& -1 &1 \end{vmatrix}= -\hat{i} -3\hat{j} -\hat{k}
so,\: S\cdot D= \frac{\left | \left ( \bar{a}_{2}-\bar{a}_{1} \right )\times \vec{b} \right |}{\left | b \right |}= \frac{\sqrt{1+9+1}}{\sqrt{4+1+1}}= \sqrt{\frac{11}{6}}units

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