Q.Find the equation of the line in vector and in cartesian form that passes through the point with position vector $2\hat{i} - \hat{j} + 4\hat{k}$ and is in the direction $\hat{i} + 2\hat{j} - \hat{k}$.

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

Steps:

1. Here, $\vec{a} = 2\hat{i} - \hat{j} + 4\hat{k}$ and $\vec{b} = \hat{i} + 2\hat{j} - \hat{k}$.
2. Vector form: $\vec{r} = (2\hat{i} - \hat{j} + 4\hat{k}) + \lambda(\hat{i} + 2\hat{j} - \hat{k})$.
3. For cartesian form, write $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$ and equate components: $x = 2 + \lambda$, $y = -1 + 2\lambda$, $z = 4 - \lambda$. …

The core idea: a line is just a point moving in a fixed direction. If you know where it starts (a given point) and which way it goes (a direction vector), you can describe every point on the line by starting at that point and adding some multiple of the direction vector. That multiple, usually called $\lambda$ (or $t$), is a parameter — each value of $\lambda$ gives a different point on the line.

Why this works: Think of walking along a straight road. You begin at a landmark (the given point). Every step you take is in the same direction (the direction vector). If you take $\lambda$ steps, your position is: starting point + $\lambda$ × (step direction). That’s the vector equation in a nutshell.

Now let’s build it step by step.

1. Identify the given point and direction vector

   The point has position vector $\vec{a} = 2\hat{i} - \hat{j} + 4\hat{k}$.

   The direction vector is $\vec{b} = \hat{i} + 2\hat{j} - \hat{k}$.

2. Write the vector equation

   The general vector equation of a line through point $\vec{a}$ in direction $\vec{b}$ is:

Substituting:

That’s the vector form. Done.

3. Convert to cartesian form Let $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$. Then the vector equation becomes:

Equate components:

4. Eliminate the parameter $\lambda$ From $x = 2 + \lambda$, we get $\lambda = x - 2$. From $y = -1 + 2\lambda$, we get $\lambda = \frac{y + 1}{2}$. …

Method: From point-and-direction to both vector and Cartesian forms

Given a point and a direction, a line can be written in either standard form; the two are the same statement — one carrying a parameter, one with the parameter eliminated.

Steps

Step 1: Read off $\vec a$ and $\vec b$. The position vector of the point is $\vec a$; the given direction is $\vec b=a\hat i+b\hat j+c\hat k$.

Step 2: Vector form — write directly:

Step 3: Cartesian form — set $\vec r=x\hat i+y\hat j+z\hat k$, match components ($x=x_0+a\lambda$, and so on) and solve each for $\lambda$:

$\frac{x-x_0}{a}=\frac{y-y_0}{b}=\frac{z-z_0}{c}.$ …

Common Mistakes

Mistake 1: Wrong sign in the $z$-denominator of the Cartesian form.

Why it's wrong: the direction's $\hat k$ component is $-1$, so the denominator under $z$ is $-1$: $\frac{z-4}{-1}$, not $\frac{z-4}{1}$. Correct approach: copy each denominator from the direction vector, sign and all.

Mistake 2: A sign slip in a point coordinate.

Why it's wrong: the point is $(2,-1,4)$, so the $y$-term is $y-(-1)=y+1$. Correct approach: write $\frac{x-2}{1}=\frac{y+1}{2}=\frac{z-4}{-1}$. …