The parametric equations for a line L1 are as follows:
x = 2+2t
y = 2+2t
z = −3+2t
Let L2 be the line parallel to L1 and passing through the point (1, −4, −3). Find the point P on L2 whose x-coordinate is −4.
P = P(−4, 0, 0)
Thanks
Seems to me that L2 is just
x = 1+2t
y = -4+2t
z = -3+2t
If x=-4, t = -5/2
So, P=(-4,-9,-8)
To find the point P on line L2 whose x-coordinate is -4, we need to determine the corresponding y-coordinate and z-coordinate.
Since L2 is parallel to L1, its parametric equations will have the same direction vector as L1, which is given by the coefficients of t in the equations for x, y, and z. In this case, the direction vector is (2, 2, 2).
To find the corresponding y-coordinate and z-coordinate for the point P(-4, 0, 0) on L2, we can use the following formula:
(x - x1) / a = (y - y1) / b = (z - z1) / c
where (x1, y1, z1) is a point on L2 (in this case, (1, -4, -3)) and (a, b, c) is the direction vector.
Substituting the values, we have:
(-4 - 1) / 2 = (y - (-4)) / 2 = (z - (-3)) / 2
Simplifying, we get:
-5 / 2 = (y + 4) / 2 = (z + 3) / 2
From the first equation, we can solve for y:
y + 4 = -5
y = -9
From the second equation, we can solve for z:
z + 3 = -5
z = -8
Therefore, the point P on L2 whose x-coordinate is -4 is P(-4, -9, -8).