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:
The parametric form of the 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 given by \( \vec{r} = \vec{a} + \lambda \vec{d} \). Here, \( \vec{a} = \hat{i} - \hat{j} \) represents a point on the line, and \( \vec{d} = 2\hat{i} - \hat{j} \) is the direction vector. The parametric equations for this line are: \[ x = 1 + 2\lambda, \quad y = -1 - \lambda, \quad z = 0 \]. To obtain the Cartesian form, we eliminate \( \lambda \). From the equation for \( x \), we get \( \lambda = \frac{x - 1}{2} \). Substituting this into the equation for \( y \): \[ y = -1 - \frac{x - 1}{2} \]. Simplifying this expression yields: \[ y = -1 - \frac{x - 1}{2} = \frac{-2 - (x - 1)}{2} = \frac{-x - 1}{2} \]. Thus, the Cartesian form of the line is: \[ \frac{x - 1}{2} = \frac{y + 1}{-1} = \frac{z}{0} \]. Step 2: {Verify the options} This result corresponds to option (B).