Question

the line l in r3 is parameterized by x(t) y(t) 3t 6. find x(t) and y(t) if the line passes through the points (0,1,3) and (8,7,9)

Answer 1

To determine the parametric equations for line l in R3, we utilize the two given points it passes through. By solving for t using the z-coordinates, we establish the relationships x(t) = 4t + 4 and y(t) = 3t + 4.

Step-by-step explanation:

To find the parametric equations x(t) and y(t) for the line l in R3, we need to use the two given points the line passes through, which are (0,1,3) and (8,7,9). Since we are told that z is parameterized by 3t + 6, we can find t values corresponding to the z-coordinates of the points by solving 3t + 6 = z. For the point (0,1,3), t is -1, and for the point (8,7,9), t is 1. With these t values, we can find the equations for x(t) and y(t).

The line passing through (0,1,3) when t is -1 and (8,7,9) when t is 1 indicates that the change in x is 8 units over the change in t of 2 units, which gives us the slope for x(t) as 4. Since x is 0 when t = -1, x(t) = 4t + 4. Similarly, the change in y is 6 units over the change in t of 2 units, yielding a slope for y(t) of 3. Since y is 1 when t = -1, y(t) = 3t + 4.

Answer 2

To find the parameterization of the line, we can equate the x-coordinate and the y-coordinate of the points to x(t) and y(t) respectively, and solve for t to find the values of t that correspond to these points. Then, substitute those t-values into 3t + 6 to find z(t).

Step-by-step explanation:

To find the parameterization of the line in R3 passing through the points (0,1,3) and (8,7,9), we need to determine the equations for x(t), y(t), and z(t). Since the line is given by x(t), y(t), 3t + 6, we can equate the x-coordinate and the y-coordinate of the points to x(t) and y(t) respectively, and solve for t to find the values of t that correspond to these points. Then, substitute those t-values into 3t + 6 to find z(t).

For the point (0,1,3):

1. Equating the x-coordinate: x(t) = 0 ➔ t = 0
2. Equating the y-coordinate: y(t) = 1 ➔ y(0) = 1
3. Substituting t = 0 into z(t): z(t) = 3t + 6 ➔ z(0) = 6

For the point (8,7,9):

1. Equating the x-coordinate: x(t) = 8 ➔ 8 = 3t + 6 ➔ t = 2
2. Equating the y-coordinate: y(t) = 7 ➔ y(2) = 7
3. Substituting t = 2 into z(t): z(t) = 3t + 6 ➔ z(2) = 12

Therefore, the parameterization of the line passing through the points (0,1,3) and (8,7,9) is:
x(t) = t, y(t) = t + 1, z(t) = 3t + 6