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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Editor">Editor Staff answered 11 months ago

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}}