To find the intersection point of the line $\mathbf{L}$ with the plane $\mathbf{P}$, we can follow these steps:

1. Determine the equation of the plane $\mathbf{P}$.
2. Find the parameter $s$ for which the line intersects the plane.
3. Substitute the parameter $s$ back into the line equation to find the intersection point.

Let's go through each step in detail.

Step 1: Determine the Equation of the Plane

The plane $\mathbf{P}$ passes through the points:

• $A = [2, 0, 0]$
• $B = [-3, 3, -2]$
• $C = [2, -1, 1]$

To find the equation of the plane, we need a normal vector $\mathbf{n}$ to the plane. We can obtain this by calculating the cross product of two vectors lying on the plane.

Calculate vectors $\mathbf{AB}$ and $\mathbf{AC}$:

[ \mathbf{AB} = B - A = [-3, 3, -2] - [2, 0, 0] = [-5, 3, -2] ]

[ \mathbf{AC} = C - A = [2, -1, 1] - [2, 0, 0] = [0, -1, 1] ]

Compute the cross product $\mathbf{n} = \mathbf{AB} \times \mathbf{AC}$:

\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -5 & 3 & -2 \\ 0 & -1 & 1 \end{vmatrix} = \mathbf{i}(3 \cdot 1 - (-2) \cdot (-1)) - \mathbf{j}(-5 \cdot 1 - (-2) \cdot 0) + \mathbf{k}(-5 \cdot (-1) - 3 \cdot 0)$$ $$= \mathbf{i}(3 - 2) - \mathbf{j}(-5) + \mathbf{k}(5)$$ $$= \mathbf{i}(1) + \mathbf{j}(5) + \mathbf{k}(5)$$ $$= [1, 5, 5]$$ The normal vector $$\mathbf{n}$$ is $$[1, 5, 5]$$. #### Equation of the plane: Using the point $$A = [2, 0, 0]$$ and the normal vector $$\mathbf{n} = [1, 5, 5]$$, the equation of the plane is: $$1(x - 2) + 5(y - 0) + 5(z - 0) = 0$$ Simplifying, we get: $$x + 5y + 5z = 2$$ ### Step 2: Find the Parameter $$s$$ for Line Intersection The line $$\mathbf{L}$$ is given by: \[ \mathbf{r}(s) = [2, 3, 3] + s[2, 3, -1] \] This implies the parametric equations: $$x = 2 + 2s, \quad y = 3 + 3s, \quad z = 3 - s$$ Substitute these into the plane equation $$x + 5y + 5z = 2$$: $$(2 + 2s) + 5(3 + 3s) + 5(3 - s) = 2$$ Simplify: $$2 + 2s + 15 + 15s + 15 - 5s = 2$$ $$2s + 15s - 5s + 2 + 15 + 15 = 2$$ $$12s + 32 = 2$$ $$12s = 2 - 32$$ $$12s = -30$$ $$s = -\frac{30}{12} = -\frac{5}{2}$$ ### Step 3: Find the Intersection Point Substitute $$s = -\frac{5}{2}$$ into the parametric equations of the line: $$x = 2 + 2\left(-\frac{5}{2}\right) = 2 - 5 = -3$$ $$y = 3 + 3\left(-\frac{5}{2}\right) = 3 - \frac{15}{2} = 3 - 7.5 = -4.5$$ $$z = 3 - \left(-\frac{5}{2}\right) = 3 + \frac{5}{2} = 3 + 2.5 = 5.5$$ Thus, the intersection point is: $$\boxed{[-3, -4.5, 5.5]}$$ ### Verification: To ensure accuracy, we verify if this point satisfies the plane equation: $$x + 5y + 5z = 2 \Rightarrow (-3) + 5(-4.5) + 5(5.5) = 2$$ $$-3 - 22.5 + 27.5 = 2$$ $$2 = 2$$ The calculations are verified as correct, confirming that the intersection point is indeed $$\boxed{[-3, -4.5, 5.5]}$$. --- Feel free to ask if you have any questions or need further clarification! Here are some questions you might find useful: 1. **Can you explain the concept of a normal vector again?** 2. **Why do we use the cross product to find the normal vector?** 3. **How would the process change if we were given different points?** 4. **What if the line was parallel to the plane?** 5. **Is there a quicker method for solving plane intersection problems?** ### Tip: When checking your work in problems like this, substitute back into both the plane and line equations to ensure consistency and correctness.