{"id":6120,"date":"2021-10-07T18:11:46","date_gmt":"2021-10-07T12:41:46","guid":{"rendered":"https:\/\/mathemerize.com\/?p=6120"},"modified":"2021-10-09T00:43:37","modified_gmt":"2021-10-08T19:13:37","slug":"distance-between-two-lines-in-3d","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/distance-between-two-lines-in-3d\/","title":{"rendered":"Distance Between Two Lines in 3d"},"content":{"rendered":"
Here you will learn formula to find the distance between two lines in 3d in both vector form and cartesian form with example.<\/p>\n
Let’s begin –<\/p>\n
Let \\(l_1\\) and \\(l_2\\) be two lines having vector equations<\/p>\n
\\(l_1\\) : \\(\\vec{r}\\) = \\(\\vec{a_1}\\) + \\(\\lambda\\)\\(\\vec{b_1}\\)<\/p>\n
and \\(l_2\\) : \\(\\vec{r}\\) = \\(\\vec{a_2}\\) + \\(\\mu\\)\\(\\vec{b_2}\\)<\/p>\n
The shortest distance (S.D.) between two the two non-parallel lines \\(\\vec{r}\\) = \\(\\vec{a_1}\\) + \\(\\lambda\\)\\(\\vec{b_1}\\) and \\(\\vec{r}\\) = \\(\\vec{a_2}\\) + \\(\\mu\\)\\(\\vec{b_2}\\) is given by<\/p>\n
\nS.D. = |\\((\\vec{b_1} \\times \\vec{b_2}).(\\vec{a_2} -\\vec{ a_1})\\over |\\vec{b_1} \\times \\vec{b_2}|\\)|<\/p>\n<\/blockquote>\n
Condition for two given lines to intersect : <\/strong>If given lines intersect, then the shortest distance between them is zero.<\/p>\n
\n\\((\\vec{b_1} \\times \\vec{b_2}).(\\vec{a_2} -\\vec{ a_1})\\) = 0<\/p>\n<\/blockquote>\n
(b) Cartesian Form<\/strong><\/h4>\n
Let the cartesian equation of two skew lines be<\/p>\n
\\(x – x_1\\over l_1\\) = \\(y – y_1\\over m_1\\) = \\(z – z_1\\over n_1\\)<\/p>\n
and \\(x – x_2\\over l_2\\) = \\(y – y_2\\over m_2\\) = \\(z – z_2\\over n_2\\)<\/p>\n
\nShortest Distance (S.D.) = \\(\\begin{vmatrix} x_2 – x_1 & y_2 – y_1 & z_2 – z_1 \\\\ l_1 & m_1 & n_1 \\\\ l_2 & m_2 & n_2 \\end{vmatrix}\\over \\sqrt{(m_1n_2 – m_2n_1)^2 + (n_1l_1 – l_1n_2)^2 + (l_1m_2 – l_2m_1)^2}\\)<\/p>\n<\/blockquote>\n
Condition for two given lines to intersect : <\/strong>If given lines intersect, then the shortest distance between them is zero.<\/p>\n
\n\\(\\begin{vmatrix} x_2 – x_1 & y_2 – y_1 & z_2 – z_1 \\\\ l_1 & m_1 & n_1 \\\\ l_2 & m_2 & n_2 \\end{vmatrix}\\) = 0<\/p>\n<\/blockquote>\n
Example<\/span><\/strong> : Find the shortest distance between the lines<\/p>\n
\\(\\vec{r}\\) = (\\(4\\hat{i} – \\hat{j}\\)) + \\(\\lambda\\)(\\(\\hat{i} + 2\\hat{j} – 3\\hat{k}\\))<\/p>\n
and \\(\\vec{r}\\) = (\\(\\hat{i} – \\hat{j} + 2\\hat{k}\\)) + \\(\\mu\\)(\\(2\\hat{i} + 4\\hat{j} – 5\\hat{k}\\))<\/p>\n
Solution<\/span><\/strong> : We know that the shortest distance between two lines \\(\\vec{r}\\) = \\(\\vec{a_1}\\) + \\(\\lambda\\)\\(\\vec{b_1}\\) and \\(\\vec{r}\\) = \\(\\vec{a_2}\\) + \\(\\mu\\)\\(\\vec{b_2}\\) is given by<\/p>\n
d = |\\((\\vec{b_1} \\times \\vec{b_2}).(\\vec{a_2} -\\vec{ a_1})\\over |\\vec{b_1} \\times \\vec{b_2}|\\)|<\/p>\n
Comparing the given equation with the equations \\(\\vec{r}\\) = \\(\\vec{a_1}\\) + \\(\\lambda\\)\\(\\vec{b_1}\\) and \\(\\vec{r}\\) = \\(\\vec{a_2}\\) + \\(\\mu\\)\\(\\vec{b_2}\\) respectively.<\/p>\n
\\(\\vec{a_1}\\) = \\(4\\hat{i} – \\hat{j}\\), \\(\\vec{a_2}\\) = \\(\\hat{i} – \\hat{j} + 2\\hat{k}\\)<\/p>\n
and \\(\\vec{b_1}\\) = \\(\\hat{i} + 2\\hat{j} – 3\\hat{k}\\), \\(\\vec{b_2}\\) = \\(2\\hat{i} + 4\\hat{j} – 5\\hat{k}\\)<\/p>\n
\\(a_2 – a_1\\) = \\(-3\\hat{i} + 0\\hat{j} + 2\\hat{k}\\)<\/p>\n
and \\(b_1\\times b_2\\) = \\(\\begin{vmatrix} \\hat{i} & \\hat{j} & \\hat{k} \\\\ 1 & 2 & -3 \\\\ 2 & 4 & -5 \\end{vmatrix}\\) = \\(2\\hat{i} – \\hat{j} + 0\\hat{k}\\)<\/p>\n
(\\(a_2 – a_1\\)).(\\(b_1\\times b_2\\)) = \\(-3\\hat{i} + 0\\hat{j} + 2\\hat{k}\\).\\(2\\hat{i} – \\hat{j} + 0\\hat{k}\\) = -6 + 0 + 0 = 0<\/p>\n
and | \\(b_1\\times b_2\\) | = \\(\\sqrt{4 + 1 + 0}\\) = \\(\\sqrt{5}\\)<\/p>\n
\\(\\therefore\\) Shortest Distance = |\\((\\vec{b_1} \\times \\vec{b_2}).(\\vec{a_2} -\\vec{ a_1})\\over |\\vec{b_1} \\times \\vec{b_2}|\\)| <\/p>\n
= |\\(-6\\over \\sqrt{5}\\)| = \\(6\\over \\sqrt{5}\\)<\/p>\n\n\n