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: 
 

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To convert from parametric to Cartesian form for a line, eliminate the parameter by solving for it in terms of one variable and substituting into the others.
  • \( \frac{x - 2}{1} = \frac{y + 1}{0} = \frac{z}{1} \)
  • \( \frac{x - 1}{2} = \frac{y + 1}{-1} = \frac{z}{0} \)
  • \( \frac{x + 1}{2} = \frac{y + 1}{-1} = \frac{z}{0} \)
  • \( \frac{x - 1}{2} = \frac{y}{-1} = \frac{z - 1}{0} \)
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The Correct Option is B

Solution and Explanation

The parametric form of the line passing through the point with position vector 
\( \vec{a} = \hat{i} - \hat{j} \) 
and parallel to the given line 
\( \vec{r} = \hat{i} + \hat{k} + \mu(2\hat{i} - \hat{j}) \) is: \[ \vec{r} = \vec{a} + \lambda \vec{d} \] where \( \vec{a} = \hat{i} - \hat{j} \) is the point on the line and \( \vec{d} = 2\hat{i} - \hat{j} \) is the direction vector of the line. 
Thus, the parametric equations of the line are: \[ x = 1 + 2\lambda, \quad y = -1 - \lambda, \quad z = 0 \] 
To convert this to the Cartesian form, eliminate \( \lambda \) from the equations: 
From the equation for 
\( x \): \[ \lambda = \frac{x - 1}{2} \] 
Substitute this into the equation for \( y \): \[ y = -1 - \frac{x - 1}{2} \] 
Simplifying: \[ y = -1 - \frac{x - 1}{2} = \frac{-2 - (x - 1)}{2} = \frac{-x - 1}{2} \] 
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 matches option (B). 
 

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