Final answer:
The parametric equations for the line through (-8, 2, 7) and parallel to the line 1/2x = 1/3y = z/1 are x(t) = -8 + 2t, y(t) = 2 + 3t, and z(t) = 7 + t.
Step-by-step explanation:
To find parametric equations for the line that passes through the point (-8, 2, 7) and is parallel to the given line, we first need to determine the direction vector of the given line. The line is given by the equations ½x = ⅓y = z/1. This indicates that the line's direction vector can be represented as (2, 3, 1) because the coefficients represent the proportional changes in x, y, and z, respectively, as the parameter changes.
The next step is to use the point (-8, 2, 7) as the base point and include the direction vector. The parametric equations of a line are given by x(t) = x0 + at, y(t) = y0 + bt, and z(t) = z0 + ct, where (x0, y0, z0) is a point on the line, (a, b, c) is the direction vector, and t is the parameter.
Applying this information, the parametric equations for the line we want are:
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- x(t) = -8 + 2t
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- y(t) = 2 + 3t
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- z(t) = 7 + t