Target Exam

CUET

Subject

Section B1

Chapter

Three-dimensional Geometry

Question:

The cartesian equation of a line passing through the point with position vector $\vec{a} = \hat{i} - \hat{j}$ and parallel to the line $\vec{r} = \hat{i} + \hat{k} + \mu(2\hat{i} - \hat{j})$, is

Options:

$\frac{x - 2}{1} = \frac{y + 1}{0} = \frac{z}{1}$

$\frac{x - 1}{2} = \frac{y + 1}{-1} = \frac{z}{0}$

$\frac{x + 1}{2} = \frac{y + 1}{-1} = \frac{z}{0}$

$\frac{x - 1}{2} = \frac{y}{-1} = \frac{z - 1}{0}$

Correct Answer:

$\frac{x - 1}{2} = \frac{y + 1}{-1} = \frac{z}{0}$

Explanation:

The correct answer is Option (2) → $\frac{x - 1}{2} = \frac{y + 1}{-1} = \frac{z}{0}$ ##

Here, $\vec{a} = \hat{i} - \hat{j}$

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

Let $\vec{m}$ be a line passing through $\vec{a}$ and parallel to $\vec{r}$

$\Rightarrow \vec{m}$ be a line passing through $\vec{a}$ and parallel to $(2\hat{i} - \hat{j}) = \vec{b}$ (say)

So we know that a line through a point with position vector $\vec{a}$ and parallel to $\vec{b}$ is given by the equation.

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

$\Rightarrow (x\hat{i} + y\hat{j} + z\hat{k}) = (\hat{i} - \hat{j}) + \lambda(2\hat{i} - \hat{j})$

$= \hat{i}(1 + 2\lambda) + \hat{j}(-1 - \lambda)$

So its Cartesian equation is

$\frac{x - 1}{2} = \frac{y + 1}{-1} = \frac{z - 0}{0} \quad (\text{After equating})$