Q.Find the vector equation of the line which is parallel to the vector $3\hat{i} - 2\hat{j} + 6\hat{k}$ and which passes through the point $(1, -2, 3)$.

Concept: Vector Equation Of Line — a line is written as $\vec{r} = \vec{a} + \lambda \vec{b}$, where $\vec{a}$ is the position vector of a fixed point and $\vec{b}$ is a direction vector parallel to the line.

Steps:

1. The given point $(1, -2, 3)$ gives $\vec{a} = \hat{i} - 2\hat{j} + 3\hat{k}$.
2. The direction vector is $\vec{b} = 3\hat{i} - 2\hat{j} + 6\hat{k}$.
3. Substitute into $\vec{r} = \vec{a} + \lambda \vec{b}$.

Why the vector equation works

A line in space is determined by two things: a point it passes through, and a direction it runs along. The vector equation captures this beautifully.

Think of $\vec{r}$ as the position vector of any point on the line. If you start at the origin, first go to the fixed point $A$ (position vector $\vec{a}$). Then, from $A$, move some distance along the direction $\vec{b}$ — but how much? That's where the scalar parameter $\lambda$ comes in. By letting $\lambda$ take all real values, you sweep out every point on the line.

This is the standard vector equation of a line. $\vec{a}$ is the position vector of a known point, and $\vec{b}$ is any vector parallel to the line.

Step-by-step solution

1. Identify the fixed point. The line passes through $(1, -2, 3)$. Its position vector is:

2. Identify the direction vector. The line is parallel to $3\hat{i} - 2\hat{j} + 6\hat{k}$. Since parallel lines share the same direction, we can take this vector directly as $\vec{b}$:

3. Write the vector equation. Substitute $\vec{a}$ and $\vec{b}$ into the form $\vec{r} = \vec{a} + \lambda \vec{b}$:

That's it — this is the required equation.

Method: Vector equation of a line from a point and a parallel vector

Use this to write a line's equation given one point on it and a vector it is parallel to.

Steps

Step 1: Identify the two ingredients.

A line needs a POSITION vector $\vec{a}$ of a known point and a DIRECTION vector $\vec{b}$ it runs along. Keep them separate — the point supplies $\vec{a}$, the "parallel to" vector supplies $\vec{b}$.

Step 2: Write the point as a position vector.

The point $(x_0,y_0,z_0)$ becomes $\vec{a}=x_0\hat{i}+y_0\hat{j}+z_0\hat{k}$.

Step 3: Assemble the equation.

Any non-zero multiple of $\vec{b}$ describes the same line, so the direction need not be simplified.

Common Mistakes

Mistake 1: Swapping the point and the direction vector.

Why it's wrong: the point $(1,-2,3)$ supplies $\vec{a}$, while $3\hat{i}-2\hat{j}+6\hat{k}$ is the direction $\vec{b}$; using one in place of the other describes a different line. Correct approach: $\vec{r}=(\hat{i}-2\hat{j}+3\hat{k})+\lambda(3\hat{i}-2\hat{j}+6\hat{k})$.

Mistake 2: Dropping the parameter $\lambda$ or its range.

Why it's wrong: without $\lambda\in\mathbb{R}$ the expression names a single point, not the whole line. Correct approach: always include $\lambda$ as a free real parameter.