# 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

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