Question

Direction ratios of a vector parallel to the line \( \frac{x - 1}{2} = -y = \frac{2z + 1}{6} \) are:

• A. \( 2, -1, 6 \)
• B. \( 2, 1, 6 \)
• C. \( 2, 1, 3 \)
• D. \( 2, -1, 3 \)

Hint:

To find the direction ratios of a line, identify the coefficients of the parameter \( t \) in the parametric equations of the line.

Answer 1

The Correct Option is D

Step 1: Parametrize the given line.
The given equation of the line is: \[ \frac{x - 1}{2} = -y = \frac{2z + 1}{6}. \] Let \( t \) be the parameter. From each equation, we solve for \( x \), \( y \), and \( z \): \[ \frac{x - 1}{2} = t \quad \Rightarrow \quad x = 2t + 1, \] \[ -y = t \quad \Rightarrow \quad y = -t, \] \[ \frac{2z + 1}{6} = t \quad \Rightarrow \quad z = 3t - \frac{1}{2}. \] Step 2: Identify the direction ratios.
The direction ratios are the coefficients of \( t \) in the parametric equations: \[ \text{Direction ratios} = 2, -1, 3. \] Step 3: Conclusion.
The direction ratios of the line are: \[ \boxed{2, -1, 3}. \]