Final answer:
To find the equation for the plane that is perpendicular to a given line and passes through a point, the normal vector of the plane is the direction vector of the line. The plane equation can be derived by using this normal vector and solving for the constant by plugging in the point's coordinates.
Step-by-step explanation:
To find an equation for the plane that passes through the point (1, 6, −8) and is perpendicular to the line v = (0, −6, 5) + t(1, −2, 3), we need to determine the normal vector of the plane, which is given by the direction vector of the line it is perpendicular to. In this case, the direction vector of the perpendicular line is (1, −2, 3).
The general form of the equation of a plane is Ax + By + Cz = D, where (A, B, C) is the normal vector to the plane. Since our normal vector is (1, −2, 3), we can write the equation as x − 2y + 3z = D.
To find the value of D, we substitute the coordinates of the given point (1, 6, −8) into the equation: 1(1) − 2(6) + 3(−8) = D, which simplifies to −11 − 12 − 24 = D or D = −47.
Therefore, the equation of the plane is x − 2y + 3z = −47