{"id":6122,"date":"2021-10-07T18:13:59","date_gmt":"2021-10-07T12:43:59","guid":{"rendered":"https:\/\/mathemerize.com\/?p=6122"},"modified":"2021-10-09T00:43:50","modified_gmt":"2021-10-08T19:13:50","slug":"distance-between-two-parallel-lines-in-3d","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/distance-between-two-parallel-lines-in-3d\/","title":{"rendered":"Distance Between Two Parallel Lines in 3d"},"content":{"rendered":"
Here you will learn formula to find the distance between two parallel lines in 3d with example.<\/p>\n
Let’s begin –<\/p>\n
Let \\(l_1\\) and \\(l_2\\) be two parallel lines having vector equations<\/p>\n
\\(l_1\\) : \\(\\vec{r}\\) = \\(\\vec{a_1}\\) + \\(\\lambda\\)\\(\\vec{b}\\)<\/p>\n
and \\(l_2\\) : \\(\\vec{r}\\) = \\(\\vec{a_2}\\) + \\(\\mu\\)\\(\\vec{b}\\).<\/p>\n
The shortest distance (S.D.) between two the parallel lines \\(\\vec{r}\\) = \\(\\vec{a_1}\\) + \\(\\lambda\\)\\(\\vec{b}\\) and \\(\\vec{r}\\) = \\(\\vec{a_2}\\) + \\(\\mu\\)\\(\\vec{b}\\) is given by<\/p>\n
\nd = |\\((\\vec{a_2} -\\vec{ a_1}) \\times \\vec{b}\\over |\\vec{b}|\\)|<\/p>\n<\/blockquote>\n
Example<\/strong><\/span> : Find the shortest distance between the lines whose vector equations are<\/p>\n
\\(\\vec{r}\\) = (\\(\\hat{i} + 2\\hat{j} + 3\\hat{k}\\)) + \\(\\lambda\\)(\\(2\\hat{i} + 3\\hat{j} + 4\\hat{k}\\))<\/p>\n
and \\(\\vec{r}\\) = (\\(2\\hat{i} + 4\\hat{j} + 5\\hat{k}\\)) + \\(\\mu\\)(\\(4\\hat{i} + 6\\hat{j} + 8\\hat{k}\\))<\/p>\n
Solution<\/strong><\/span> : The vector equations of given lines are<\/p>\n
\\(\\vec{r}\\) = (\\(\\hat{i} + 2\\hat{j} + 3\\hat{k}\\)) + \\(\\lambda\\)(\\(2\\hat{i} + 3\\hat{j} + 4\\hat{k}\\)) …………(i)<\/p>\n
and \\(\\vec{r}\\) = (\\(2\\hat{i} + 4\\hat{j} + 5\\hat{k}\\)) + 2\\(\\mu\\)(\\(2\\hat{i} + 3\\hat{j} + 4\\hat{k}\\)) ………..(ii)<\/p>\n
Equation (ii) can be re-written as<\/p>\n
\\(\\vec{r}\\) = (\\(2\\hat{i} + 4\\hat{j} + 5\\hat{k}\\)) + \\({\\mu}’\\)(\\(2\\hat{i} + 3\\hat{j} + 4\\hat{k}\\)) …………(iii)<\/p>\n
where \\({\\mu}’\\) = 2\\(\\mu\\)<\/p>\n
These two lines passes through the points having position vectors \\(\\vec{a_1}\\) = \\(\\hat{i} + 2\\hat{j} + 3\\hat{k}\\) and \\(a_2\\) = \\(2\\hat{i} + 4\\hat{j} + 5\\hat{k}\\) respectively and both are parallel to the vector \\(\\vec{b}\\) = (\\(2\\hat{i} + 3\\hat{j} + 4\\hat{k}\\)).<\/p>\n
So, the shortest distance between them is given by<\/p>\n
S.D. = |\\((\\vec{a_2} -\\vec{ a_1}) \\times \\vec{b}\\over |\\vec{b}|\\)| ………(iv)<\/p>\n
Now, \\((\\vec{a_2} -\\vec{ a_1}) \\times \\vec{b}\\) = \\(\\hat{i} + 2\\hat{j} + 2\\hat{k}\\) \\(\\times\\) (\\(2\\hat{i} + 3\\hat{j} + 4\\hat{k}\\))<\/p>\n
= \\(\\begin{vmatrix} \\hat{i} & \\hat{j} & \\hat{k} \\\\ 1 & 2 & 2 \\\\ 2 & 3 & 4 \\end{vmatrix}\\) = \\(2\\hat{i} – 0\\hat{j} – k\\hat{k}\\)<\/p>\n
\\(\\therefore\\) |\\((\\vec{a_2} -\\vec{ a_1}) \\times \\vec{b}\\)| = \\(\\sqrt{4 + 0 + 1}\\) = \\(\\sqrt{5}\\)<\/p>\n
and |\\(\\vec{b}\\)| = \\(\\sqrt{4 + 9 + 16}\\) = \\(\\sqrt{29}\\)<\/p>\n
Substituting the values of |\\((\\vec{a_2} -\\vec{ a_1}) \\times \\vec{b}\\) | and |\\(\\vec{b}\\)| in (iv), we obtain<\/p>\n
S.D. = \\(\\sqrt{5}\\over \\sqrt{29}\\)<\/p>\n\n\n