# Vectors – Equation of Straight Line

## Vectors – Equation of Straight Line

Equation of Straight Line passing through a given Point and Parallel to a given Vector

Vector Form: Line passing through point $$A(\overline{a})$$ and parallel to vector $$\overline{b}$$.

Let A be the given point and let AP be the given line through A.

Let  $$\overline{b}$$ be any vector parallel to the given line.

Position vector of point A is $$\overline{a}$$.

Let P be any point on line AP, and let its position vector be $$\overline{r}$$.

Then, we have

$$\overline{r}=\overline{OP}=\overline{OA}+\overline{AP}$$

$$\overrightarrow{r}=\overrightarrow{a}+\lambda \overrightarrow{b}$$ … (1)

Here, $$\overline{r}$$ is the position vector of any point P (x, y) on the line. So $$\overrightarrow{r}=x\hat{i}+y\hat{j}+z\hat{k}$$.

In, particular, the equation of straight line through origin and parallel to $$\overline{b}$$ is $$\overrightarrow{r}=\lambda \overrightarrow{b}$$.

Cartesian Form: Let the coordinates of the given point A be (x₁, y₁, z₁) and the direction of the line be a, b and c. Consider the coordinates of any point P be (x, y, z). then

$$\overrightarrow{r}=x\hat{i}+y\hat{j}+z\hat{k}$$,

$$\overrightarrow{a}={{x}_{1}}\hat{i}+{{y}_{1}}\hat{j}+{{z}_{1}}\hat{k}$$ and

$$\overrightarrow{b}=a\hat{i}+b\hat{j}+c\hat{k}$$.

Substituting these values in (1) and equating the coefficients of $$\hat{i},\ \hat{j}$$ and $$\hat{k}$$. We get

x = x₁ + λa

y = y₁ + λb

z = z₁ + λc

These are parametric equation of the line.

Eliminating the parameter λ, we get $$\frac{x-{{x}_{1}}}{a}=\frac{y-{{y}_{1}}}{b}=\frac{z-{{z}_{1}}}{c}$$.

Note:

• Here any point on the line is (x, y, z) ≡ (x₁ + λa, y₁ + λb, z₁ + λc)
• Since the x, y and z – axes pass through the origin and have direction cosine (1, 0, 0), (0, 1, 0) and (0, 0, 1), their equations are
Equation of x – axis: y = 0, z = 0,
Equation of y – axis: x = 0, z = 0,
Equation of z – axis: x = 0, y = 0.