Final answer:
To find the plane equation that contains point (2, 3, −7) and is parallel to the plane x y − 5z = 1, we use the same normal vector (1, 1, −5), and apply it to the point-normal form of a plane equation to derive x + y − 5z + 30 = 0 as the equation of the desired plane.
Step-by-step explanation:
To find an equation of the plane that contains the point (2, 3, −7) and is parallel to the plane given by the equation x y − 5z = 1, we start by understanding that parallel planes will have the same normal vector. The normal vector can be derived from the coefficients of x, y, and z in the equation. Since no coefficient is multiplied with x and y while −5 is multiplied with z, the normal vector is (1, 1, −5).
We know a point on our plane, (2, 3, −7), and we have our normal vector (1, 1, −5). Now, we can write the plane equation in the general form:
A(x − x1) + B(y − y1) + C(z − z1) = 0,
where (x1, y1, z1) is a point on the plane and (A, B, C) is the normal vector to the plane. By plugging in our specific values we get:
(1)(x − 2) + (1)(y − 3) + (−5)(z + 7) = 0,
which simplifies to:
x + y − 5z − 2 − 3 + 35 = 0,
or, the final equation of the plane:
x + y − 5z + 30 = 0.