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 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).
List-I | List-II |
(A) Absolute maximum value | (I) 3 |
(B) Absolute minimum value | (II) 0 |
(C) Point of maxima | (III) -5 |
(D) Point of minima | (IV) 4 |