Question

The line \( \frac{x-3}{4} = \frac{y-4}{5} = \frac{z-5}{6} \) is parallel to the plane.

• A. \( 3x + 4y + 5z = 7 \)
• B. \( x + y + z = 2 \)
• C. \( x - 2y + z = 0 \)
• D. \( 2x + 3y + 4z = 0 \)

Hint:

When the line is parallel to the plane, the direction ratios of the line are the same as the normal vector of the plane. Use this to find the equation of the plane.

Answer 1

The Correct Option is C

We are given the equation of a line \( \frac{x-3}{4} = \frac{y-4}{5} = \frac{z-5}{6} \). The direction ratios of the line are \( 4, 5, 6 \), which represent the components of the direction vector of the line. These direction ratios also represent the normal vector to the plane because the line is parallel to the plane. Thus, the normal vector of the plane is \( \mathbf{n} = (4, 5, 6) \). The general equation of the plane is: \[ 4x + 5y + 6z = D \] Now, to find \( D \), we substitute any point on the line into the equation of the plane. Let's use the point \( (3, 4, 5) \) which lies on the line. Substituting \( x = 3, y = 4, z = 5 \) into the equation: \[ 4(3) + 5(4) + 6(5) = D \] \[ 12 + 20 + 30 = D \] \[ D = 62 \] Thus, the equation of the plane is: \[ 4x + 5y + 6z = 62 \] Now, comparing this with the options, we find that option (C) is the correct one.