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