Question

Consider the line parametrized by r(t) = (3,-1,9) + t(8, 12, -6). Identify the parametrizations equivalent to r(t). r1(t) = (11, 11, 3) + t(8,12,-6) r2(t) = (4,0, 10) +1(8,12,-6) r3(t) = (6,-2, 18) + (8,12,-6) r4(t) = (3,-1,9) + f(-4, -6,3) r5(t) = (8,12,-6) + t(3,-1,9)

r6(t) = (-1, -7,12) + f(4,6,-3)

Answer 1

The line parametrized by r(t) = (3,-1,9) + t(8, 12, -6) can be equivalent to various other parametrizations.

Step-by-step explanation:

The line parametrized by r(t) = (3,-1,9) + t(8, 12, -6) can be identified as equivalent to r1(t) = (11, 11, 3) + t(8,12,-6), r2(t) = (4,0, 10) + t(8,12,-6), r3(t) = (6,-2, 18) + t(8,12,-6), r4(t) = (3,-1,9) + f(-4, -6,3), r5(t) = (8,12,-6) + t(3,-1,9), and r6(t) = (-1, -7,12) + f(4,6,-3).

Answer 2

The given line is parametrized by r(t) = (3,-1,9) + t(8, 12, -6). It passes through the point (3, -1, 9) and has a direction vector (8, 12, -6). The given parametrizations are all equivalent to the original parametrization.

Step-by-step explanation:

The given line parametrized by r(t) = (3,-1,9) + t(8, 12, -6) represents a line in three-dimensional space. This line passes through the point (3, -1, 9) and has a direction vector given by (8, 12, -6). The parameter t represents the distance traveled along the line from the initial point (3, -1, 9).

The parametrizations r1(t) = (11, 11, 3) + t(8,12,-6), r2(t) = (4,0, 10) + t(8,12,-6), r3(t) = (6,-2, 18) + t(8,12,-6), r4(t) = (3,-1,9) + f(-4, -6,3), r5(t) = (8,12,-6) + t(3,-1,9), and r6(t) = (-1, -7,12) + f(4,6,-3) are all equivalent parametrizations of the same line. The only difference among them is the choice of the initial point or the scale factor.