Q.The vector equation of line $2x-1=3y+2=z-2$ is
(A) $\vec{r}=\left(\dfrac{1}{2}\hat{i}-\dfrac{2}{3}\hat{j}+2\hat{k}\right)+\lambda(3\hat{i}+2\hat{j}+6\hat{k})$
(B) $\vec{r}=\hat{i}-\hat{j}+(2\hat{i}+\hat{j}+\hat{k})$
(C) $\vec{r}=\left(\dfrac{1}{2}\hat{i}-\hat{j}\right)+\lambda(\hat{i}-2\hat{j}+6\hat{k})$
(D) $\vec{r}=(\hat{i}+\hat{j})+\lambda(\hat{i}-2\hat{j}+6\hat{k})$

Vector equation of a line through a point $A(\bar a)$ parallel to $\bar b$ is $\bar r=\bar a+\lambda\bar b$.

The line passes through the point with position vector $\bar a = 4\hat i-\hat j+2\hat k$ and is parallel to $\bar b=-2\hat i-\hat j+\hat k$.

The vector equation of a line through a fixed point $\bar a$ and parallel to $\bar b$ is

$\bar r=\bar a+\lambda \bar b$

Substituting,

$\bar r=(4\hat i-\hat j+2\hat k)+\lambda(-2\hat i-\hat j+\hat k)$

This matches option (d) exactly (options (a), (b), (c) alter a sign in either the point or the direction vector).

Vector equation: (point) + $\lambda$(direction).

Line through position vector $\bar a=4\hat i-\hat j+2\hat k$, parallel to $\bar b=-2\hat i-\hat j+\hat k$.

Vector equation: $\bar r=\bar a+\lambda\bar b=(4\hat i-\hat j+2\hat k)+\lambda(-2\hat i-\hat j+\hat k)$, $\lambda\in\mathbb R$

Vector equation of a line through a point $\bar a$ in the direction $\bar b$ is $\bar r = \bar a + \lambda \bar b$.

The line passes through the point with position vector $\bar a = 4\hat i - \hat j + 2\hat k$ and has direction vector $\bar b = -2\hat i + \hat j + \hat k$.

The vector equation of a line through a given point with position vector $\bar a$ along direction $\bar b$ is:

$\bar r = \bar a + \lambda \bar b, \qquad \lambda \in \mathbb{R}$

Substituting:

$\bar r = (4\hat i - \hat j + 2\hat k) + \lambda(-2\hat i + \hat j + \hat k)$