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
\(x – x_1\over a\) = \(y – y_1\over b\) = \(z – z_1\over c\) ………….(i)
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.
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 \(\vec{m}\) = \(a\hat{i} + b\hat{j} + c\hat{k}\).
So the vector equation of line (i) is
\(\vec{r}\) = \(\vec{a}\) + \(\lambda \vec{m}\)
or, \(\vec{r}\) = (\(x_1\hat{i} + y_1\hat{j} + z_1\hat{k}\)) + \(\lambda\) (\(a\hat{i} + b\hat{j} + c\hat{k}\))
where \(\lambda\) is a parameter.
Reduction of Vector Form of Line to Cartesian Form
Let the vector equation of line be
\(\vec{r}\) = \(\vec{a}\) + \(\lambda \vec{m}\) …………(ii)
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.
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.
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
\(x\hat{i} + y\hat{j} + z\hat{k}\) = (\(x_1\hat{i} + y_1\hat{j} + z_1\hat{k}\)) + \(\lambda\)(\(a\hat{i} + b\hat{j} + c\hat{k}\))
On equating the coefficients of \(\hat{i}\), \(\hat{j}\) and \(\hat{k}\), we get
x = \(x_1 + a \lambda\), y = \(y_1 + b \lambda\), z = \(z_1 + c \lambda\)
\(\implies\) \(x – x_1\over a\) = \(y – y_1\over b\) = \(z – z_1\over c\) = \(\lambda\)
These are the cartesian equations of line.
Example : 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.
Solution : We have,
\(\vec{r}\) = (\(2\hat{i} – \hat{j} + 4\hat{k}\)) + \(\lambda\) (\(\hat{i} + \hat{j} – 2\hat{k}\)) ………..(i)
Putting \(\vec{r}\) = \(x\hat{i} + y\hat{j} + z\hat{k}\) in equation (i), we obtain
\(x\hat{i} + y\hat{j} + z\hat{k}\) = (\(2\hat{i} – \hat{j} + 4\hat{k}\)) + \(\lambda\) (\(\hat{i} + \hat{j} – 2\hat{k}\))
On equating the coefficients of \(\hat{i}\), \(\hat{j}\) and \(\hat{k}\), we get
x = \(2 + \lambda\), y = \(-1 + \lambda\), z = \(4 – 2\lambda\)
\(\implies\) x – 2 = \(\lambda\), y + 1 = \(\lambda\), \(z – 4\over -2\) = \(\lambda\)
On eliminating \(\lambda\), we get
\(x – 1\over 1\) = \(y + 1\over 1\) = \(z – 4\over -2\)
Hence, the cartesian form of equation (i) is
\(x – 1\over 1\) = \(y + 1\over 1\) = \(z – 4\over -2\)
