Question

The Cartesian equation of the line passing through (7, 5, 3) and perpendicular to the plane 3x + 2y + z = 6 is

• A. \(\frac{x-7}{3}=\frac{y-5}{2}=\frac{z-3}{1}\)
• B. \(\frac{x-3}{7}=\frac{y-2}{5}=\frac{z-1}{3}\)
• C. \(\frac{x-3}{7}=\frac{y-2}{5}=\frac{z}{3}\)
• D. \(\frac{x-7}{3}=\frac{y-5}{1}=\frac{z-3}{2}\)
• E. \(\frac{x-4}{4}=\frac{y-3}{3}=\frac{z-2}{2}\)

Answer 1

The Correct Option is A

Given: The plane equation is: 

\[ 3x + 2y + z = 6 \]

Step 1: Find the direction ratios

The line perpendicular to the plane has its direction ratios given by the coefficients of the plane equation:

\[ (3, 2, 1) \]

Step 2: Use the symmetric form equation

The equation of the required line passing through (7, 5, 3) and having direction ratios (3, 2, 1) is:

\[ \frac{x - 7}{3} = \frac{y - 5}{2} = \frac{z - 3}{1} \]

Final Answer:

\[ \frac{x-7}{3} = \frac{y-5}{2} = \frac{z-3}{1} \]