Find an equation for the plane that is perpendicular to the line l(t) = (2, 0, 9)t + (3, −1, 1) and passes through (9, −1, 0).
1 answer:
The plane will have its normal vector equal to the direction vector of the line (2, 0, 9). Its equation is
2x +9z = (2, 0, 9)•(9, -1, 0)
2x +9z = 18
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