Final answer:
To find a point contained in both q(t) and r(t), we need to set the corresponding components of the two equations equal to each other and solve for t. By solving the system of equations, we find that t = 1 and the point (4, 4, 6) is common to both equations. To find the equation of the plane containing both lines and the point, we use the point-normal form of the equation of a plane. If a different point on line q was used in the calculations, the equation of the plane would be different.
Step-by-step explanation:
A. To find a point that is contained on both q(t) and r(t), we need to set the corresponding components of the two equations equal to each other and solve for t.
5t - 1 = t + 4
3t + 2 = 2t + 5
6t = 6
By solving the system of equations, we find that t = 1. Substituting this value back into q(t) or r(t), we find a point that is common to both equations: (4, 4, 6).
B. To find the equation of the plane that contains both lines and the point (4, 4, 6), we can use the point-normal form of the equation of a plane.
Taking r(t) as the direction vector of one of the lines, and the cross product of the direction vectors of both lines as the normal vector of the plane, we can write the equation of the plane as:
(ax - 4) + b(y - 4) + c(z - 6) = 0, where a, b, and c are determined from the cross product of the direction vectors of both lines.
C. If the calculations in part B were based on a different point on line q, the equation of the plane would not be the same. This is because the point-normal form of the equation of a plane relies on both the direction vector of one of the lines and a common point between the two lines to determine the equation. Using a different point on line q would lead to a different equation of the plane.
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