{"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

Distance Between Two Lines in 3d<\/h2>\n

(a) Vector Form<\/strong><\/h4>\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

\n

S.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

\n

Shortest 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

\n
Next – Distance Between Two Parallel Lines in 3d<\/a><\/div>\n<\/div>\n\n\n\n
\n
Previous – Perpendicular Distance of a Point From a Line in 3d<\/a><\/div>\n<\/div>\n","protected":false},"excerpt":{"rendered":"

Here you will learn formula to find the distance between two lines in 3d in both vector form and cartesian form with example. Let’s begin – Distance Between Two Lines in 3d (a) Vector Form Let \\(l_1\\) and \\(l_2\\) be two lines having vector equations \\(l_1\\) : \\(\\vec{r}\\) = \\(\\vec{a_1}\\) + \\(\\lambda\\)\\(\\vec{b_1}\\) and \\(l_2\\) : \\(\\vec{r}\\) …<\/p>\n

Distance Between Two Lines in 3d<\/span> Read More »<\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"default","ast-global-header-display":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","theme-transparent-header-meta":"default","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":""},"categories":[33],"tags":[],"yoast_head":"\nDistance Between Two Lines in 3d - Mathemerize<\/title>\n<meta name=\"description\" content=\"In this post you will learn formula to find the distance between two lines in 3d in both vector form and cartesian form with example.\" \/>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/mathemerize.com\/distance-between-two-lines-in-3d\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Distance Between Two Lines in 3d - 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