Find the distance between the point (10, 100, 1000) and the plane [x, y, z] = [1, 1, 1] + s[5, −5, 3] + t[−1, 1, 3]. [5 marks]

To find the distance between a point and a plane in 3D space, we can use the following steps:

### Step 1: Understand the Problem
- **Point:** \( P(10, 100, 1000) \)
- **Plane Equation:** Given in parametric form as \( \mathbf{r}(s, t) = \mathbf{r}_0 + s\mathbf{v}_1 + t\mathbf{v}_2 \), where:
  - \( \mathbf{r}_0 = [1, 1, 1] \) (a point on the plane)
  - \( \mathbf{v}_1 = [5, -5, 3] \) (direction vector 1)
  - \( \mathbf{v}_2 = [-1, 1, 3] \) (direction vector 2)

### Step 2: Find the Normal Vector to the Plane
The normal vector \( \mathbf{n} \) to the plane can be found by taking the cross product of the direction vectors \( \mathbf{v}_1 \) and \( \mathbf{v}_2 \):

\[
\mathbf{n} = \mathbf{v}_1 \times \mathbf{v}_2
\]

Calculating the cross product:

\[
\mathbf{n} = 
\begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
5 & -5 & 3 \\
-1 & 1 & 3
\end{vmatrix}
= \mathbf{i} \left((-5)(3) - (3)(1)\right) - \mathbf{j} \left((5)(3) - (3)(-1)\right) + \mathbf{k} \left((5)(1) - (-5)(-1)\right)
\]

\[
\mathbf{n} = \mathbf{i}(-15 - 3) - \mathbf{j}(15 + 3) + \mathbf{k}(5 - 5)
\]

\[
\mathbf{n} = \mathbf{i}(-18) - \mathbf{j}(18) + \mathbf{k}(0)
\]

\[
\mathbf{n} = [-18, -18, 0]
\]

### Step 3: Calculate the Distance
The formula for the distance \( d \) from a point \( P(x_1, y_1, z_1) \) to a plane given by the equation \( ax + by + cz + d = 0 \) is:

\[
d = \frac{|ax_1 + by_1 + cz_1 + d|}{\sqrt{a^2 + b^2 + c^2}}
\]

First, we need to convert the plane to its standard form \( ax + by + cz + d = 0 \). The equation of the plane can be derived from:

\[
\mathbf{n} \cdot (\mathbf{r} - \mathbf{r}_0) = 0
\]

Substituting the normal vector \( \mathbf{n} = [-18, -18, 0] \) and \( \mathbf{r}_0 = [1, 1, 1] \):

\[
-18(x - 1) - 18(y - 1) + 0(z - 1) = 0
\]

\[
-18x + 18 - 18y + 18 = 0
\]

\[
-18x - 18y + 36 = 0 \quad \text{(or equivalently \( 18x + 18y - 36 = 0 \))}
\]

Simplifying:

\[
x + y - 2 = 0
\]

So the equation of the plane is:

\[
x + y - 2 = 0
\]

Now, substitute the coordinates of the point \( P(10, 100, 1000) \) into the distance formula:

\[
d = \frac{|10 + 100 - 2|}{\sqrt{1^2 + 1^2 + 0^2}}
\]

\[
d = \frac{|108|}{\sqrt{2}} = \frac{108}{\sqrt{2}} = 54\sqrt{2}
\]

### Final Answer:
The distance between the point \( (10, 100, 1000) \) and the plane is \( 54\sqrt{2} \) units.

---

Would you like any further details or have any questions?

Here are some related questions:
1. How would the distance change if the point was on the plane?
2. What are the coordinates of the point on the plane closest to \( (10, 100, 1000) \)?
3. How would you find the angle between the plane and a given line?
4. What is the geometric interpretation of the cross product in this context?
5. How can you find the projection of a vector onto the plane?

**Tip:** The cross product of two vectors not only gives a normal vector to the plane they form but also represents an area when visualized geometrically.

Learn how to calculate the distance between the point (10, 100, 1000) and the plane x + y - 2 = 0 using vector mathematics. This problem involves finding the normal vector to the plane, converting the plane equation to standard form, and applying the distance formula. Suitable for advanced high school students or anyone studying vectors and plane geometry.

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