Question

The vector equation of the line \[ \frac{x+3}{2}=\frac{2y-3}{5}, \quad z=-1 \] is

• A. \(\vec r=(3\hat i-\tfrac{3}{2}\hat j-\hat k)+\lambda(4\hat i+5\hat j)\)
• B. \(\vec r=(-3\hat i+\tfrac{3}{2}\hat j-\hat k)+\lambda(4\hat i+5\hat j)\)
• C. \(\vec r=(-3\hat i+\tfrac{3}{2}\hat j+\hat k)+\lambda(4\hat i+5\hat j)\)
• D. \(\vec r=(3\hat i+\tfrac{3}{2}\hat j-\hat k)+\lambda(4\hat i+\tfrac{5}{2}\hat j)\)

Hint:

Convert symmetric form to parametric form to read the point and direction vector easily.

Answer 1

The Correct Option is B

Step 1: Write parametric form.
From \[ \frac{x+3}{2}=\frac{2y-3}{5}=t \] we get \[ x=2t-3,\quad y=\frac{5t+3}{2},\quad z=-1 \]
Step 2: Identify position vector and direction vector.
At \(t=0\): point on the line is \[ (-3,\tfrac{3}{2},-1) \] Direction ratios are proportional to \((2,\tfrac{5}{2},0)\), i.e. \((4,5,0)\).
Step 3: Write the vector equation.
\[ \vec r=(-3\hat i+\tfrac{3}{2}\hat j-\hat k)+\lambda(4\hat i+5\hat j) \]