Find the shortest distance between the following set of parallel lines. \vec{r}=6\hat{i}+2\hat{j}-\hat{k}+λ(\hat{i}+2\hat{j}-4\hat{k}) \vec{r}=\hat{i}+\hat{j}+\hat{k}+μ(\hat{i}+2\hat{j}-4\hat{k})

Category: QuestionsFind the shortest distance between the following set of parallel lines. \vec{r}=6\hat{i}+2\hat{j}-\hat{k}+λ(\hat{i}+2\hat{j}-4\hat{k}) \vec{r}=\hat{i}+\hat{j}+\hat{k}+μ(\hat{i}+2\hat{j}-4\hat{k})
Editor">Editor Staff asked 11 months ago

Find the shortest distance between the following set of parallel lines.
 
\vec{r}=6\hat{i}+2\hat{j}-\hat{k}+λ(\hat{i}+2\hat{j}-4\hat{k})
 
\vec{r}=\hat{i}+\hat{j}+\hat{k}+μ(\hat{i}+2\hat{j}-4\hat{k})
 
(a) d=\sqrt{\frac{324}{45}}
 
(b) d=\sqrt{\frac{405}{21}}
 
(c) d=\sqrt{\frac{24}{21}}
 
(d) d=\sqrt{\frac{21}{567}}
 
The question was posed to me in a national level competition.
 
This interesting question is from Three Dimensional Geometry in portion Three Dimensional Geometry of Mathematics – Class 12
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1 Answers
Editor">Editor Staff answered 11 months ago

Right answer is (b) d=\sqrt{\frac{405}{21}}
 
Explanation: The shortest distance between two parallel lines is given by:
 
d=\left |\frac{\vec{b}×(\vec{a_2}-\vec{a_1})}{|\vec{b}|}\right |
 
∴d=\left|\frac{(\hat{i}+2\hat{j}-4\hat{k})×(6\hat{i}+2\hat{j}-\hat{k})-(\hat{i}+\hat{j}+\hat{k})}{\sqrt{1^2+2^2+(-4)^2}}\right |
 
=\left |\frac{(\hat{i}+2\hat{j}-4\hat{k})×(5\hat{i}+\hat{j}-2\hat{k})}{\sqrt{21}} \right |
 
(\hat{i}+2\hat{j}-4\hat{k})×(5\hat{i}+\hat{j}-2\hat{k})=\begin{vmatrix}\hat{i}&\hat{j}& \hat{k}\\1&2&-4\\5&1&-2\end{vmatrix}
 
=\hat{i}(-4+4)-\hat{j}(-2+20)+\hat{k}(1-10)
 
=-18\hat{j}-9\hat{k}
 
⇒d=\left|\frac{\sqrt{(-18)^2+(-9)^2}}{√21}\right|
 
d=\sqrt{\frac{405}{21}}