Target Exam

CUET

Subject

Section B1

Chapter

Three-dimensional Geometry

Question:

Find the vector equation of a line passing through a point with position vector $2\hat{i} - \hat{j} + \hat{k}$ and parallel to the line joining the points $-\hat{i} + 4\hat{j} + \hat{k}$ and $\hat{i} + 2\hat{j} + 2\hat{k}$.

Options:

$\vec{r} = (2\hat{i} - \hat{j} + \hat{k}) + \lambda(2\hat{i} - 2\hat{j} + \hat{k})$

$\vec{r} = (-\hat{i} + 4\hat{j} + \hat{k}) + \lambda(2\hat{i} - \hat{j} + \hat{k})$

$\vec{r} = (2\hat{i} - 2\hat{j} + \hat{k}) + \lambda(2\hat{i} - \hat{j} + \hat{k})$

$\vec{r} = (2\hat{i} - \hat{j} + \hat{k}) + \lambda(\hat{i} + 2\hat{j} + 2\hat{k})$

Correct Answer:

$\vec{r} = (2\hat{i} - \hat{j} + \hat{k}) + \lambda(2\hat{i} - 2\hat{j} + \hat{k})$

Explanation:

The correct answer is Option (1) → $\vec{r} = (2\hat{i} - \hat{j} + \hat{k}) + \lambda(2\hat{i} - 2\hat{j} + \hat{k})$ ##

Let $A, B$ and $C$ be the points with position vectors $2\hat{i} - \hat{j} + \hat{k}, -\hat{i} + 4\hat{j} + \hat{k}$ and $\hat{i} + 2\hat{j} + 2\hat{k}$, respectively. We have to find the equation of a line passing through the point $A$ and parallel to vector $BC$.

Now,

$\vec{BC} = \text{position vector of } C - \text{position vector of } B$

$= (\hat{i} + 2\hat{j} + 2\hat{k}) - (-\hat{i} + 4\hat{j} + \hat{k}) = 2\hat{i} - 2\hat{j} + \hat{k}$

We know that, the equation of a line passing through a position vector $\vec{a}$ and parallel to vector $\vec{b}$ is

$\vec{r} = \vec{a} + \lambda\vec{b}$

$∴ \vec{r} = (2\hat{i} - \hat{j} + \hat{k}) + \lambda(2\hat{i} - 2\hat{j} + \hat{k})$ is the required equation of line in vector form. $[ \text{Here, } \vec{BC} = \vec{b} ]$