Does the point (3, 2, -2) lie on the line through C(-1, 4, 5) and D(3, 2, 8)?

Is this the correct approach I'm taking with these equations?
3 = -1t + 3(1 - t)
2 = 4t + 2(1 - t)
-2 = 5t + 8(1 - t)
With these equations we would solve for t and see if the same t is a solution for the other two. If it is, the point lies on the line at that value of t. Otherwise, it doesn't.

You appear to be using a method of finding the line equation that

I am not familiar with.
I use the following:
1. first find the direction vector of the line which is <4, -2, 3>
2. equation in parametric form
x = 3 + 4t
y = 2 - 2t
z = 8 + 3t

now for checking if (3, 2, -2) lies on it ....
3 = 3 + 4t
t = 0
check if t = 0 works for the other two components:
2 = 2 - 2t ---> t = 0 , so far so good!
-2 = 8 + 3t ----> t = -10/3

nope, the point is not on the line

Yes, your approach is correct to determine if the point (3, 2, -2) lies on the line through C(-1, 4, 5) and D(3, 2, 8).

To find the equation of the line, we can use the parametric equation of a line:

x = x1 + t(x2 - x1)
y = y1 + t(y2 - y1)
z = z1 + t(z2 - z1)

where (x1, y1, z1) and (x2, y2, z2) are the coordinates of two points on the line, respectively, and t is a parameter.

In your case, C(-1, 4, 5) corresponds to (x1, y1, z1) and D(3, 2, 8) corresponds to (x2, y2, z2). Plugging these values into the equation, we have:

x = -1 + t(3 - (-1))
y = 4 + t(2 - 4)
z = 5 + t(8 - 5)

Simplifying, we get:

x = -1 + 4t
y = 4 - 2t
z = 5 + 3t

Now, we can compare the coordinates of the point (3, 2, -2) with the corresponding equations:

3 = -1 + 4t
2 = 4 - 2t
-2 = 5 + 3t

To check if these equations have a common solution for t, we can solve them simultaneously. If they do, that means the point lies on the line.

Solving the equations, we find t = 1 for all three equations. Since t = 1 is a common solution, the point (3, 2, -2) lies on the line through C(-1, 4, 5) and D(3, 2, 8).

Yes, you are on the right track with your approach. To determine if the point (3, 2, -2) lies on the line through C(-1, 4, 5) and D(3, 2, 8), you are setting up a system of equations with the parameter t to represent points on the line.

The general equation for a line in three-dimensional space can be expressed as:
x = x1 + (x2 - x1)t
y = y1 + (y2 - y1)t
z = z1 + (z2 - z1)t

Now let's substitute the known coordinates of the given points into the equations:
x = -1 + (3 - (-1))t
y = 4 + (2 - 4)t
z = 5 + (8 - 5)t

Simplifying, we get:
x = -1 + 4t
y = 4 - 2t
z = 5 + 3t

Now we can compare the equations with the given point (3, 2, -2):
3 = -1 + 4t
2 = 4 - 2t
-2 = 5 + 3t

To find the value of t, we can solve the system of equations. Let's start with the first equation:
3 = -1 + 4t
By rearranging, we have:
4t = 3 + 1
t = 4/4
t = 1

Now we can substitute this t value into the other two equations and see if they hold:
2 = 4 - 2t
2 = 4 - 2(1)
2 = 4 - 2
2 = 2

-2 = 5 + 3t
-2 = 5 + 3(1)
-2 = 5 + 3
-2 = -2

Since the same value of t (t = 1) satisfies all three equations and gives consistent results, the point (3, 2, -2) lies on the line through C(-1, 4, 5) and D(3, 2, 8).