Reduction of Cartesian Form of Line to Vector Form and Vice-Versa

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

\(x – 1\over 1\) = \(y + 1\over 1\) = \(z – 4\over -2\)

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