Final answer:
The vector equation for the line perpendicular to the plane 4x - 3y + 3z = 4 and passing through the point (9, 6, 1) is r(t) = <9, 6, 1> + t<4, -3, 3>. The parametric equations derived from this vector equation are x(t) = 9 + 4t, y(t) = 6 - 3t, and z(t) = 1 + 3t.
Step-by-step explanation:
To find the vector equation and parametric equations for a line that is perpendicular to a plane, we first need to understand that a line is perpendicular to a plane if the direction vector of the line is parallel to the normal vector of the plane. The normal vector of the plane 4x - 3y + 3z = 4 is (4, -3, 3). This normal vector will serve as the direction vector of our line.
Given a point through which the line passes, which is (9, 6, 1), we can write the vector equation of the line as r(t) = r0 + tv, where r0 is the position vector of a point on the line, v is the direction vector of the line, and t is the parameter. In our case, r0 = <9, 6, 1> and v = <4, -3, 3>
The vector equation of the line is r(t) = <9, 6, 1> + t<4, -3, 3>. To derive the parametric equations, we equate the corresponding components:
- x(t) = 9 + 4t
- y(t) = 6 - 3t
- z(t) = 1 + 3t
These are the parametric equations of the line where t is the parameter.