For the points X (-2,2,-3), Y(2,4,5), find the point Z(4,a,b) that makes X,Y and Z colinear

To find the values of \(a\) and \(b\) such that the points \(X(-2, 2, -3)\), \(Y(2, 4, 5)\), and \(Z(4, a, b)\) are collinear, we can use the property that the vectors formed by any two pairs of these points must be parallel.

Let's consider the vectors \(\overrightarrow{XY}\) and \(\overrightarrow{XZ}\):

1. **Vector \(\overrightarrow{XY}\)** is obtained by subtracting the coordinates of \(X\) from those of \(Y\):
   \[
   \overrightarrow{XY} = (2 - (-2), 4 - 2, 5 - (-3)) = (4, 2, 8)
   \]

2. **Vector \(\overrightarrow{XZ}\)** is obtained by subtracting the coordinates of \(X\) from those of \(Z\):
   \[
   \overrightarrow{XZ} = (4 - (-2), a - 2, b - (-3)) = (6, a - 2, b + 3)
   \]

For the points to be collinear, the vectors \(\overrightarrow{XY}\) and \(\overrightarrow{XZ}\) must be parallel, which means there must exist a scalar \(k\) such that:
\[
\overrightarrow{XZ} = k \cdot \overrightarrow{XY}
\]

This gives us the following system of equations:
\[
6 = 4k \quad \text{(1)}
\]
\[
a - 2 = 2k \quad \text{(2)}
\]
\[
b + 3 = 8k \quad \text{(3)}
\]

Now, let's solve these equations:

1. From equation (1): 
   \[
   k = \frac{6}{4} = \frac{3}{2}
   \]

2. Substitute \(k = \frac{3}{2}\) into equation (2):
   \[
   a - 2 = 2 \times \frac{3}{2} = 3 \implies a = 5
   \]

3. Substitute \(k = \frac{3}{2}\) into equation (3):
   \[
   b + 3 = 8 \times \frac{3}{2} = 12 \implies b = 9
   \]

Thus, the coordinates of point \(Z\) are \(Z(4, 5, 9)\).

### Final Answer:
The point \(Z(4, 5, 9)\) makes the points \(X(-2, 2, -3)\), \(Y(2, 4, 5)\), and \(Z\) collinear.

Would you like more details or have any questions? Here are some related questions you might find interesting:

1. How to verify if three points in space are collinear using the cross product?
2. How do you calculate the distance between two points in 3D space?
3. How to find the equation of the line passing through two points in 3D?
4. What is the general condition for vectors to be parallel in 3D space?
5. How does the concept of collinearity extend to higher dimensions?

**Tip:** When dealing with vectors, remember that parallel vectors are scalar multiples of each other. This property is key in determining collinearity in 3D space.

For the points X (-2,2,-3), Y(2,4,5), find the point Z(4,a,b) that makes X,Y and Z colinear.

Explore how to determine the point Z(4,a,b) that makes the points X(-2,2,-3), Y(2,4,5), and Z collinear in 3D space. This problem involves vector subtraction and the condition for parallel vectors, suitable for Grades 10-12.

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