{"id":6112,"date":"2021-10-07T17:52:25","date_gmt":"2021-10-07T12:22:25","guid":{"rendered":"https:\/\/mathemerize.com\/?p=6112"},"modified":"2021-10-09T00:41:17","modified_gmt":"2021-10-08T19:11:17","slug":"reduction-of-cartesian-form-of-line-to-vector-form-and-vice-versa","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/reduction-of-cartesian-form-of-line-to-vector-form-and-vice-versa\/","title":{"rendered":"Reduction of Cartesian Form of Line to Vector Form and Vice-Versa"},"content":{"rendered":"

Here you will learn reduction of cartesian form of line to vector form and vice-versa with examples.<\/p>\n

Let’s begin –<\/p>\n

Reduction of Cartesian Form of Line to Vector Form<\/h2>\n

Let the cartesian equations of a line be<\/p>\n

\\(x – x_1\\over a\\) = \\(y – y_1\\over b\\) = \\(z – z_1\\over c\\)                       ………….(i)<\/p>\n

These are the equation of a line passing through the point A(\\(x_1, y_1, z_1\\)) and its direction ratios are proportional to a, b, c. <\/p>\n

In vector form this means the line passes through point having position vector \\(\\vec{a}\\) = \\(x_1\\hat{i} + y_1\\hat{j} + z_1\\hat{k}\\) and is parallel to the vector \\(\\vec{m}\\) = \\(a\\hat{i} + b\\hat{j} + c\\hat{k}\\).<\/p>\n

So the vector equation of line (i) is<\/p>\n

\n

\\(\\vec{r}\\) = \\(\\vec{a}\\) + \\(\\lambda \\vec{m}\\)<\/p>\n

or, \\(\\vec{r}\\) = (\\(x_1\\hat{i} + y_1\\hat{j} + z_1\\hat{k}\\)) + \\(\\lambda\\) (\\(a\\hat{i} + b\\hat{j} + c\\hat{k}\\))<\/p>\n

where \\(\\lambda\\) is a parameter.<\/p>\n<\/blockquote>\n

Reduction of Vector Form of Line to Cartesian Form<\/strong><\/h4>\n

Let the vector equation of line be<\/p>\n

\\(\\vec{r}\\) = \\(\\vec{a}\\) + \\(\\lambda \\vec{m}\\)                            …………(ii)<\/p>\n

where \\(\\vec{a}\\) = \\(x_1\\hat{i} + y_1\\hat{j} + z_1\\hat{k}\\) and \\(\\vec{m}\\) = \\(a\\hat{i} + b\\hat{j} + c\\hat{k}\\) and \\(\\lambda\\) is a parameter.<\/p>\n

In order to reduce equation (ii) to cartesian form we put \\(\\vec{r}\\) = \\(x\\hat{i} + y\\hat{j} + z\\hat{k}\\) and equate the coefficient of \\(\\hat{i}\\), \\(\\hat{j}\\) and \\(\\hat{k}\\) as discussed below.<\/p>\n

Putting \\(\\vec{r}\\) = \\(x\\hat{i} + y\\hat{j} + z\\hat{k}\\), \\(\\vec{a}\\) = \\(x_1\\hat{i} + y_1\\hat{j} + z_1\\hat{k}\\) and \\(\\vec{m}\\) = \\(a\\hat{i} + b\\hat{j} + c\\hat{k}\\) in (ii), we get<\/p>\n

\\(x\\hat{i} + y\\hat{j} + z\\hat{k}\\) = (\\(x_1\\hat{i} + y_1\\hat{j} + z_1\\hat{k}\\)) + \\(\\lambda\\)(\\(a\\hat{i} + b\\hat{j} + c\\hat{k}\\))<\/p>\n

On equating the coefficients of \\(\\hat{i}\\), \\(\\hat{j}\\) and \\(\\hat{k}\\), we get<\/p>\n

\n

x = \\(x_1 + a \\lambda\\), y = \\(y_1 + b \\lambda\\), z = \\(z_1 + c \\lambda\\)<\/p>\n

\\(\\implies\\) \\(x – x_1\\over a\\) = \\(y – y_1\\over b\\) = \\(z – z_1\\over c\\) = \\(\\lambda\\)<\/p>\n<\/blockquote>\n

These are the cartesian equations of line.<\/p>\n

Example<\/strong><\/span> : Reduce the following vector equation of line \\(\\vec{r}\\) = (\\(2\\hat{i} – \\hat{j} + 4\\hat{k}\\)) + \\(\\lambda\\) (\\(\\hat{i} + \\hat{j} – 2\\hat{k}\\)) in cartesian form.<\/p>\n

Solution<\/span><\/strong> : We have,<\/p>\n

\\(\\vec{r}\\) = (\\(2\\hat{i} – \\hat{j} + 4\\hat{k}\\)) + \\(\\lambda\\) (\\(\\hat{i} + \\hat{j} – 2\\hat{k}\\))            ………..(i)<\/p>\n

Putting \\(\\vec{r}\\) = \\(x\\hat{i} + y\\hat{j} + z\\hat{k}\\) in equation (i), we obtain<\/p>\n

\\(x\\hat{i} + y\\hat{j} + z\\hat{k}\\) = (\\(2\\hat{i} – \\hat{j} + 4\\hat{k}\\)) + \\(\\lambda\\) (\\(\\hat{i} + \\hat{j} – 2\\hat{k}\\))<\/p>\n

On equating the coefficients of \\(\\hat{i}\\), \\(\\hat{j}\\) and \\(\\hat{k}\\), we get<\/p>\n

x = \\(2 + \\lambda\\), y = \\(-1 +  \\lambda\\), z = \\(4 –  2\\lambda\\)<\/p>\n

\\(\\implies\\) x – 2 = \\(\\lambda\\), y + 1 = \\(\\lambda\\), \\(z – 4\\over -2\\) = \\(\\lambda\\)<\/p>\n

On eliminating \\(\\lambda\\), we get<\/p>\n

\\(x – 1\\over 1\\) = \\(y + 1\\over 1\\) = \\(z – 4\\over -2\\)<\/p>\n

Hence, the cartesian form of equation (i) is<\/p>\n

\\(x – 1\\over 1\\) = \\(y + 1\\over 1\\) = \\(z – 4\\over -2\\)<\/p>\n\n\n

\n
Next – Angle Between Two Lines in 3d<\/a><\/div>\n<\/div>\n\n\n\n
\n
Previous – Cartesian Equation of a Line<\/a><\/div>\n<\/div>\n","protected":false},"excerpt":{"rendered":"

Here you will learn reduction of cartesian form of line to vector form and vice-versa with examples. Let’s begin – Reduction of Cartesian Form of Line to Vector Form Let the cartesian equations of a line be \\(x – x_1\\over a\\) = \\(y – y_1\\over b\\) = \\(z – z_1\\over c\\)          …<\/p>\n

Reduction of Cartesian Form of Line to Vector Form and Vice-Versa<\/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":"\nReduction of Cartesian Form of Line to Vector Form and Vice-Versa<\/title>\n<meta name=\"description\" content=\"In this post you will learn reduction of cartesian form of line to vector form and vice-versa 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