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:

Direction ratios of a line are derived from the coefficients of the parameter \( t \) in the parametric equations of the line.

Answer 1

The Correct Option is D

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