{"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
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
\nx = \\(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