The parametric equations for a line L1 are as follows:
x = −1−2t
y = 4+4t
z = 3−2t
Let L2 be the line parallel to L1 and passing through the point (2, 5, −3). Find the point P on L2 whose x-coordinate is −3.
P = (−3, 0, 0)
The second equation L2 is given by
(x,y,z) = (2,5,-3) + t(-2,4,-2)
we want -3 = 2 -2t
2t = 5
t = 5/2
then (x,y,z) = (2,5,-3) + (5/2)(-2,4,-2)
= (2,5,-3) + (-5, 10, -5) = (-3, 15, -8)
the point is (-3, 15, -8)
To find the point P on line L2 whose x-coordinate is -3, we need to substitute -3 for the x-coordinate in the parametric equations of line L2.
Given that L2 is parallel to L1 and passes through the point (2, 5, -3), the direction vector of L2 will be the same as the direction vector of L1, which is (-2, 4, -2).
Using the point-slope form of a line, where the parametric equations for L2 can be written as:
x = 2 - 2t
y = 5 + 4t
z = -3 -2t
Now, let's find the value of t that corresponds to x = -3:
-3 = 2 - 2t
Subtracting 2 from both sides:
-5 = -2t
Dividing both sides by -2:
t = 5/2
Now we can substitute this value of t into the parametric equations to find the coordinates of point P:
x = 2 - 2(5/2) = -3
y = 5 + 4(5/2) = 15
z = -3 - 2(5/2) = -8
Therefore, the coordinates of point P on line L2 with x-coordinate -3 are P = (-3, 15, -8).
To find the point P on line L2 whose x-coordinate is -3, we need to find the values of t that satisfy the x-component of the parametric equations for line L2 being -3.
The x-component of the parametric equation for line L2 is given by:
x = -1 - 2t
To find the value of t that makes x equal to -3, we can substitute -3 for x in the equation and solve for t:
-3 = -1 - 2t
Adding 1 to both sides:
-2 = -2t
Dividing by -2:
t = 1
Now that we know the value of t, we can substitute it back into the parametric equations for line L1 to find the corresponding values for y and z:
For y:
y = 4 + 4t
y = 4 + 4(1)
y = 8
For z:
z = 3 - 2t
z = 3 - 2(1)
z = 1
Therefore, the point P on L2 with an x-coordinate of -3 is P = (-3, 8, 1).
Well, well, well, it seems like we've got a parallel line problem here. You're looking for the point P on line L2 with an x-coordinate of -3, and L2 is parallel to line L1. Alrighty then, let's dive in!
Given that L2 is parallel to L1, we can use the same parameter t for both lines. We need to find the corresponding values of t that will give us an x-coordinate of -3 for L2.
For line L1, we have:
x = -1 - 2t
To find the corresponding value of t for L2 when x = -3, we can set the x-coordinates of L1 and L2 equal to each other:
-1 - 2t = -3
Solving this equation, we find that t = 1.
Now we can plug t = 1 into the parametric equations of L2 to find the point P:
x = -1 - 2(1) = -3
y = 4 + 4(1) = 8
z = 3 - 2(1) = 1
So, the point P on L2 with an x-coordinate of -3 is P = (-3, 8, 1). And hey, you got it right! P = (-3, 0, 0) is not on L2, so good job on catching that.
Hope that helps! Keep those clown questions coming!