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

Concept: Vector Equation of a Line — a line through a fixed point with position vector $\vec{a}$, parallel to a direction vector $\vec{b}$, is given by $\vec{r} = \vec{a} + \lambda \vec{b}$, where $\lambda \in \mathbb{R}$.

Steps:

1. The fixed point is $(1, 2, 3)$, so $\vec{a} = \hat{i} + 2\hat{j} + 3\hat{k}$.
2. The direction vector is $\vec{b} = 3\hat{i} + 2\hat{j} - 2\hat{k}$. …

The core idea is beautifully simple: a line is just a straight path. If you know one point on it and the direction you're heading, you can describe every other point by starting at that known point and taking some number of steps in that direction.

Think of it like this. You're standing at a fixed location — that's your given point $(1, 2, 3)$. You have a compass that always points in the same direction — that's the vector $3\hat{i} + 2\hat{j} - 2\hat{k}$. If you take one step forward in that direction, you reach a new point. If you take two steps, you go further. If you step backward (a negative number of steps), you go the other way along the same line. Every possible point on the line corresponds to some number of steps, which we call a parameter.

This is exactly what the vector equation of a line captures.

Now let's apply this to the given problem.

1. Identify the position vector of the given point. The point is $(1, 2, 3)$. Its position vector (the vector from the origin to the point) is:

2. Identify the direction vector. The line is parallel to $3\hat{i} + 2\hat{j} - 2\hat{k}$, so this is our direction vector:

3. Write the vector equation. Substitute $\vec{a}$ and $\vec{b}$ into the formula: $\vec{r} = (\hat{i} + 2\hat{j} + 3\hat{k}) + \lambda (3\hat{i} + 2\hat{j} - 2\hat{k})$ …

Method: Write a line's equation from a point and a direction

A line is fixed by just two ingredients: one point on it (position vector $\vec a$) and any vector along it (direction $\vec b$). The vector equation packages them directly.

Steps

Step 1: Identify $\vec a$ — the position vector of the given point. A point $(x_0,y_0,z_0)$ becomes $\vec a=x_0\hat i+y_0\hat j+z_0\hat k$.

Step 2: Identify $\vec b$ — any vector parallel to the line. If the line is "parallel to" a given vector, that vector is $\vec b$; if "parallel to another line," copy that line's direction.

Step 3: Assemble with a free parameter:

$\vec r=\vec a+\lambda\,\vec b,\qquad \lambda\in\mathbb R.$ …

Common Mistakes

Mistake 1: Confusing the point with the direction.

Why it's wrong: the point $(1,2,3)$ supplies $\vec a$ (the constant term); the given vector $3\hat i+2\hat j-2\hat k$ supplies $\vec b$ (the $\lambda$-term). Swapping them describes a different line. Correct approach: keep each in its role — $\vec r=(\hat i+2\hat j+3\hat k)+\lambda(3\hat i+2\hat j-2\hat k)$. …